Real Analytic Function¶

Definition¶

Let $f\colon E \to \mathbb{R}$ be a function where $E \subseteq \mathbb{R}\text{.}$ We say:

• $f$ is real analytic at $x=a$ if $a$ is an interior point of $E$ and there exists a power series $\displaystyle \sum_{n=0}^\infty c_n (x-a)^n$ centred at $a$ and an open interval $(a-r, a+r) \subseteq E$ (for some $r > 0$$0$) where this series converges to $f\text{.}$
• $f$ is real analytic on $E$ if $E$ is an open set and $f$ is real analytic at every point $a \in E\text{.}$

The results on Convergence of Power Series justify differentiating the local representation term by term inside its radius. Analyticity requires the resulting series to equal $f$, not merely to exist formally.

Interpretation¶

A function is real analytic at a point $a$ if near $a$ it equals the sum of a power series centred at $a$: not only approximated by it, but equal to it on a whole open interval $(a-r,a+r)$$r)$.

The point $a$ must be an interior point of $E$ so that such an interval fits inside the domain, and "real analytic on $E$" is defined for an open set $E$ so that every point has room for its own interval. Once a series represents $f$ near $a$, its coefficients are forced: by convergence of power series the series can be differentiated term by term, and evaluating the $n$-th derivative at $a$ gives $c_n=f^{(n)}(a)/n!$$f^{(n)}(a)/n!$. So $f$ is real analytic at $a$ if and only if $f$ is infinitely differentiable near $a$ and the series $\displaystyle \sum_n f^{(n)}(a)(x-a)^n/n!$ converges to $f$ on some interval around $a$.

One series need not serve the whole domain, which is why the definition asks for a series at every point. The function $1/(1+x^2)$ is real analytic on $\mathbb{R}$, yet its power series at $0$, the geometric series $\displaystyle \sum_n(-1)^nx^{2n}$, has radius $1$ and represents it only on $(-1,1)$; around other centres, other series take over.

A common slip is to think that an infinitely differentiable function is automatically real analytic. The function in the Examples is infinitely differentiable, but all its derivatives at $0$ vanish, so the series built from them is the zero series: it converges, but to $0$ rather than to $f$. The series built from the derivatives returns later as the Taylor series.

Key Results¶

• The power series converges to $f$ pointwise on $(a-r,a+r)$ and uniformly on every closed subinterval $[a-s,a+s]$ with $0<s<r$$r$.
• Sums, products, and compositions of real analytic functions are also real analytic.
• A real analytic function is always infinitely differentiable, and all of its higher-order derivatives are real analytic.

Examples¶

• Polynomials, $\mathrm{e}^x$, $\sin x$ and $\cos x$ are real analytic on $\mathbb{R}$, and $\ln x$ is real analytic on $(0,+\infty)$.
• An infinitely differentiable function need not be real analytic. The function $f$ with $f(x)=0$ for $x\le0$ and $f(x)=\mathrm{e}^{-1/x}$ for $x>0$ is infinitely differentiable on $\mathbb{R}$ but not real analytic at $0$.

Every derivative of $f$ at $0$ is $0$, so its Taylor series at $0$ is the zero series, yet $f(x)>0$ for $x>0$$0$, so that series does not represent $f$ on any interval around $0$.

Prerequisites¶

• Open and Closed Sets in a Metric Space
• Convergence of Power Series

Related Concepts¶

• Differentiability Class
• Power Series

Sources¶

• Math III: Analysis · Lecture 2

Aliases: analytic function Symbols: element of, infinity, real numbers, subset, sum

Last updated: 20 October 2026 03:02