Maclaurin Series

⚡ Key Takeaways

The Maclaurin series is the Taylor series of a function expanded about the point $x = 0$ : it represents a function $f(x)$ as an infinite power series $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ . Useful for approximating functions near the origin, evaluating limits, and deriving identities. Named after Colin Maclaurin (18th century), though the technique predates him.

💡 The Core Insight

Near a point, a smooth function "looks like" its tangent line; looking closer, it looks like a parabola; closer still, a cubic — and so on. The Maclaurin series captures all these progressively better polynomial approximations at once by matching every derivative of $f$ at $x=0$ . The $n$ -th coefficient $a_n = f^{(n)}(0)/n!$ is precisely chosen so that the series and $f$ agree in value and in all derivatives at the origin.

DefinitionMaclaurin Series

Let $f$ be a function that is infinitely differentiable in a neighborhood of $x = 0$ . The Maclaurin series of $f$ is the power series

1

\quad f(x) \;\sim\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Derivation hint: Specialize the Taylor series $\sum \frac{f^{(n)}(a)}{n!}(x-a)^n$ to the center $a = 0$ . Here:

$f^{(n)}(0)$ is the $n$ -th derivative of $f$ evaluated at $0$ (with $f^{(0)} = f$ ),
$n!$ is the factorial of $n$ (with $0! = 1$ ),
$x^n$ is the $n$ -th power of the variable,
the symbol $\sim$ signals that we have associated a series to $f$ — equality holds only where the series converges to $f$ .

TheoremConvergence and Taylor's Theorem with Remainder

Let $f$ be $(N+1)$ -times differentiable on an interval containing $0$ and $x$ . Then

2

\quad f(x) = \sum_{n=0}^{N} \frac{f^{(n)}(0)}{n!}x^n + R_N(x),

Derivation hint: This is Taylor's theorem; the partial sum is the degree- $N$ Maclaurin polynomial. where the Lagrange form of the remainder is

3

\quad R_N(x) = \frac{f^{(N+1)}(\xi)}{(N+1)!}\,x^{N+1}, \qquad \xi \text{ strictly between } 0 \text{ and } x.

Derivation hint: Apply the Cauchy mean value theorem / generalized Rolle's theorem to the remainder function; $\xi$ exists by the intermediate-value behavior of derivatives. The Maclaurin series converges to $f(x)$ at a point $x$ if and only if $\displaystyle\lim_{N\to\infty} R_N(x) = 0$ .

ProofWhy the coefficients must be

f^{(n)}(0)/n!

Suppose $f$ can be represented near $0$ by some convergent power series:

4

\quad f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots

Derivation hint: Assume term-by-term differentiation is valid inside the radius of convergence — a theorem for power series.

Step 1. Evaluate at $x = 0$ . Every term with $x^n$ ( $n \ge 1$ ) vanishes, leaving $f(0) = a_0$ .

Step 2. Differentiate [4] term by term:

5

\quad f'(x) = a_1 + 2a_2 x + 3a_3 x^2 + \cdots = \sum_{n=1}^{\infty} n\,a_n x^{n-1}.

Derivation hint: Power series may be differentiated term-by-term within their open interval of convergence. Setting $x = 0$ gives $f'(0) = a_1$ .

Step 3. Differentiate again:

6

\quad f''(x) = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}.

Derivation hint: Repeat term-by-term differentiation; the constant and linear terms drop. At $x = 0$ , $f''(0) = 2a_2$ , so $a_2 = f''(0)/2!$ .

Step 4. By induction, differentiating $n$ times and setting $x = 0$ kills all but the constant term, which equals $n!\,a_n$ :

7

\quad f^{(n)}(0) = n!\,a_n \quad\Longrightarrow\quad a_n = \frac{f^{(n)}(0)}{n!}.

Derivation hint: The $n$ -th derivative of $a_k x^k$ at $0$ is $0$ unless $k = n$ , in which case it is $n!\,a_n$ . Thus the coefficients are forced — if a power-series representation exists, it must be the Maclaurin series. $\blacksquare$

∎

mathematician · Colin Maclaurin | 1698–1746

A Scottish mathematician and a child prodigy who entered the University of Glasgow at age eleven and held a professorship at Aberdeen by nineteen. He published the series bearing his name in his 1742 Treatise of Fluxions, a rigorous defense of Newton's calculus against George Berkeley's philosophical attacks. Maclaurin freely credited the more general result to Brook Taylor, presenting the $a=0$ case as a useful special tool. He also made deep contributions to the theory of the gravitational attraction of ellipsoids, geometry, and actuarial science.

🧠 Intuition

Think of the Maclaurin polynomial of degree $N$ as the "best possible" polynomial of that degree for matching $f$ at the single point $x=0$ . The constant term pins the height, the linear term pins the slope, the quadratic term pins the curvature, and each successive term corrects one more order of local shape. Adding more terms is like progressively sharpening a lens focused on the origin — the approximation hugs the true curve over an ever-wider window. The factorial in the denominator reflects how much "information" each derivative carries: high derivatives grow fast, but dividing by $n!$ tames them so the series can converge.

Figure

Successive Maclaurin polynomials of

\sin x

converging to the curve near the origin

ExampleMaclaurin series of

e^x

and approximating

e^{0.2}

Given: $f(x) = e^x$ . We want its Maclaurin series and a numerical estimate of $e^{0.2}$ .

Step 1 — Derivatives at 0. Since $\frac{d}{dx}e^x = e^x$ , every derivative is $e^x$ , so $f^{(n)}(0) = e^0 = 1$ for all $n$ .

Step 2 — Build the series. Coefficients are $a_n = 1/n!$ :

8

\quad e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, \quad \text{valid for all } x\in\mathbb{R}.

Derivation hint: Use $f^{(n)}(0)=1$ in equation [1]; the ratio test gives infinite radius of convergence.

Step 3 — Evaluate at $x = 0.2$ through 4 terms:

1 + 0.2 + \frac{0.04}{2} + \frac{0.008}{6} = 1 + 0.2 + 0.02 + 0.001333 = 1.221333\ldots

Step 4 — Error bound. With $N=3$ , $R_3 = \frac{e^{\xi}}{4!}(0.2)^4 \le \frac{e^{0.2}}{24}(0.0016) \approx 8.1\times10^{-5}$ .

\boxed{e^{0.2} \approx 1.22133 \quad (\text{true value } 1.221403,\ \text{error} < 10^{-4})}

ExampleMaclaurin series of

\cos x

and a limit

Given: $f(x) = \cos x$ . Find the series and use it to evaluate $\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}$ .

Step 1 — Derivatives at 0. Cyclic: $\cos, -\sin, -\cos, \sin, \ldots$ at $0$ give $1, 0, -1, 0, 1, \ldots$ Odd derivatives vanish.

Step 2 — Series:

9

\quad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

Derivation hint: Only even powers survive; signs alternate from the derivative cycle, giving the $(-1)^n/(2n)!$ pattern.

Step 3 — Substitute into the limit.

1 - \cos x = \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots \quad\Rightarrow\quad \frac{1-\cos x}{x^2} = \frac{1}{2} - \frac{x^2}{24} + \cdots

Step 4 — Take $x \to 0$ : all terms with $x$ vanish.

\boxed{\lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}}

⚠️ Common mistake

A function is not always equal to its Maclaurin series. Three independent things can go wrong: 1. The series may have a finite radius of convergence $R$ , so it equals $f$ only on $(-R, R)$ . Example: $\frac{1}{1-x} = \sum x^n$ converges only for $|x|<1$ , even though $f$ is smooth at $x=2$ . 2. The series may converge but to the wrong value (see the counterexample below). 3. Existence of all derivatives at $0$ is required — $\sqrt{|x|}$ and $|x|$ are not differentiable at $0$ , so they have no Maclaurin series. Also do not confuse the Maclaurin polynomial (a finite truncation, always a polynomial) with the Maclaurin series (the infinite object). Convergence questions only arise for the series.

CounterexampleA smooth function whose Maclaurin series is identically zero

Define

10

\quad f(x) = \begin{cases} e^{-1/x^2}, & x \neq 0,\\ 0, & x = 0. \end{cases}

Derivation hint: Repeated differentiation produces $P(1/x)\,e^{-1/x^2}$ for polynomials $P$ ; the exponential decay dominates, forcing each derivative to $0$ at the origin. This function is infinitely differentiable everywhere, and every derivative at $0$ equals $0$ : $f^{(n)}(0) = 0$ for all $n$ . Hence its Maclaurin series is $0 + 0\cdot x + 0\cdot x^2 + \cdots = 0$ , which converges everywhere — but to the zero function, not to $f$ , since $f(x) > 0$ for $x \neq 0$ . A function equal to its Maclaurin series on a neighborhood is called real-analytic; this $f$ is smooth but not analytic at $0$ .

connection

The Maclaurin series is the $a=0$ instance of the Taylor Series, inheriting all its convergence machinery. The coefficient extraction relies on Term-by-Term Differentiation of Power Series and the underlying theory of the Radius of Convergence (via the ratio and root tests). In Complex Analysis, a function holomorphic on a disk centered at $0$ automatically equals its Maclaurin series there — the analytic pathology of the real counterexample cannot occur, revealing a deep rigidity. The series also connects to Generating Functions in combinatorics, where coefficients encode counting sequences, and to L'Hôpital's Rule as an alternative method for evaluating indeterminate limits.

🔬 Where this appears

⏳ History

- c. 1665–1671: Isaac Newton derives power-series expansions for

\sin

,

\cos

,

\arcsin

, and the binomial series, computing with them freely though without modern convergence theory.

- 1668: Nicolaus Mercator publishes the series for

\ln(1+x)

— an early named special expansion.

- 1671: James Gregory independently obtains Taylor-type expansions, including the series for

\arctan x

.

- 1715: Brook Taylor publishes the general Taylor theorem in Methodus Incrementorum Directa et Inversa.

- 1730s: Continental mathematicians, notably Euler, exploit such expansions for myriad results (e.g. the Basel problem solution and Euler's formula).

- 1742: Colin Maclaurin presents the

a = 0

special case prominently in his Treatise of Fluxions, after which it bears his name.

- 1755–1770s: Euler systematizes power-series manipulation in his Introductio and Institutiones, treating series almost as polynomials.

- 1797–1813: Lagrange formalizes the remainder term, attempting to base all of calculus on Taylor series (Théorie des fonctions analytiques).

- 1821–1823: Cauchy provides rigorous convergence theory, defines analyticity, and gives the famous

e^{-1/x^2}

counterexample showing smoothness does not imply analyticity.

open-problem

The boundary between smooth ( $C^\infty$ ) and real-analytic functions remains conceptually rich. While the $e^{-1/x^2}$ example shows the gap, classifying exactly which smooth functions are analytic, and characterizing the growth of derivatives (the Denjoy–Carleman theorem delineates "quasi-analytic" classes between the two), occupies a deep corner of analysis. In several variables and on manifolds, questions about analytic continuation, Borel's theorem (every sequence of values is realizable as the derivatives of some smooth function), and the resurgence/Borel-summation of divergent asymptotic Maclaurin series (common in physics, e.g. perturbative QED) are active research areas. Determining the exact radius of convergence or asymptotic behavior of series coefficients for implicitly or recursively defined functions can be genuinely hard.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also