Analytic — Definition, Formula & Examples

Analytic is a property of a function meaning it can be represented by a convergent power series in a neighborhood of each point in its domain. In simpler terms, a function is analytic at a point if you can perfectly approximate it near that point using an infinite polynomial.

A function $f$ is analytic (or real-analytic) at a point $a$ if there exists a positive radius $r$ and a sequence of coefficients $\{c_n\}$ such that $f(x) = \sum_{n=0}^{\infty} c_n(x - a)^n$ for all $x$ with $|x - a| < r$ . In complex analysis, a function $f: \mathbb{C} \to \mathbb{C}$ is analytic (holomorphic) at $z_0$ if it is complex-differentiable in an open neighborhood of $z_0$ .

Key Formula

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

Where:

$f^{(n)}(a)$ = The nth derivative of f evaluated at the point a
$a$ = The center point around which the power series is expanded
$n!$ = n factorial, the product of all positive integers up to n

How It Works

To determine whether a function is analytic at a point

a

, you check whether its Taylor series around

a

actually converges to the function in some interval (or disk, in the complex case) around

a

. Every analytic function is infinitely differentiable, and its power series coefficients are

c_n = \frac{f^{(n)}(a)}{n!}

. In complex analysis, something remarkable happens: a single complex derivative at every point in a region automatically guarantees the function is analytic there. This equivalence does not hold in real analysis — a real function can be infinitely differentiable yet fail to be analytic.

Worked Example

Problem: Show that f(x) = e^x is analytic at a = 0 by writing its power series.

Step 1: Compute the derivatives of f(x) = e^x at a = 0. Since every derivative of e^x is e^x, we get f^{(n)}(0) = 1 for all n.

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

Step 2: Substitute into the Taylor series formula centered at a = 0.

e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

Step 3: The ratio test confirms this series converges for all real x (radius of convergence is infinite), and it equals e^x everywhere.

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

Answer: Since the Taylor series converges to e^x for all x ∈ ℝ, the function f(x) = e^x is analytic at 0 (and in fact is analytic everywhere on ℝ).

Why It Matters

Analyticity is central to complex analysis, where it underpins results like Cauchy's integral theorem and the residue theorem used in physics and engineering. In differential equations, knowing a solution is analytic lets you use power series methods to find it explicitly. Understanding analyticity also clarifies why some infinitely smooth real functions, like the classic example

e^{-1/x^2}

, resist Taylor series approximation.

Common Mistakes

Mistake: Assuming every infinitely differentiable (smooth) function is analytic.

Correction: In real analysis, smoothness does not imply analyticity. The function f(x) = e^{-1/x^2} (with f(0) = 0) is infinitely differentiable at 0, but its Taylor series at 0 is identically zero and does not converge to f(x) for x ≠ 0. Always verify the Taylor series converges to the function.