Power Series and Analytic Functions

⚡ Key Takeaways

A power series is an infinite series of the form $\sum_{n=0}^\infty a_n (x-c)^n$ . It converges on an interval centered at $c$ , with a radius of convergence $R$ determined by its coefficients. A function equal to a convergent power series near every point of its domain is called analytic — such functions are infinitely differentiable, term-by-term differentiable/integrable, and uniquely determined by their values on any small interval. Applies throughout calculus, complex analysis, differential equations, and numerical computation. Developed by Newton, Taylor, Maclaurin, Cauchy, Abel, and Weierstrass (17th–19th centuries).

💡 The Core Insight

A convergent power series behaves exactly like a polynomial on the interior of its disc of convergence — you may differentiate and integrate it term by term, and the result has the same radius of convergence. This converts hard analytic operations on functions into mechanical operations on coefficient sequences. Analyticity packages this: an analytic function is locally nothing more than a power series, and its entire local behavior is encoded in the numbers $a_n = f^{(n)}(c)/n!$ .

DefinitionPower Series

A power series centered at $c$ is a series

1

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

Derivation hint: This is the definition; $a_n \in \mathbb{R}$ (or $\mathbb{C}$ ) are coefficients, $c$ the center, $x$ the variable. Here $a_n$ is the $n$ -th coefficient, $c$ the center, and $(x-c)^n$ the $n$ -th basis monomial. The series is a function of $x$ defined precisely at those $x$ for which the limit of partial sums $s_N = \sum_{n=0}^N a_n(x-c)^n$ exists.

DefinitionRadius and Interval of Convergence

The radius of convergence $R$ is the unique value in $[0,\infty]$ such that the series converges absolutely for $|x-c| < R$ and diverges for $|x-c| > R$ . It is given by the Cauchy–Hadamard formula:

2

\quad \frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}

Derivation hint: Apply the root test to $\sum a_n(x-c)^n$ ; convergence requires $\limsup |a_n(x-c)^n|^{1/n} = |x-c|\limsup|a_n|^{1/n} < 1$ . When the limit exists, an often-easier ratio formula applies:

3

\quad R = \lim_{n\to\infty} \left| \frac{a_n}{a_{n+1}} \right| \quad \text{(when the limit exists)}

Derivation hint: Apply the ratio test; $\lim |a_{n+1}(x-c)^{n+1}/a_n(x-c)^n| = |x-c|\lim|a_{n+1}/a_n| < 1$ . The interval of convergence is $(c-R, c+R)$ together with whichever endpoints (if any) yield convergence; these must be tested separately.

TheoremTerm-by-Term Differentiation and Integration

Let $f(x) = \sum_{n=0}^\infty a_n (x-c)^n$ have radius of convergence $R > 0$ . Then $f$ is infinitely differentiable on $(c-R, c+R)$ , and

4

\quad f'(x) = \sum_{n=1}^{\infty} n\, a_n (x-c)^{n-1}, \qquad \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}

Derivation hint: Uniform convergence on compact subsets of $(c-R,c+R)$ justifies interchanging $\frac{d}{dx}$ and $\sum$ ; the root test gives both differentiated/integrated series the same $R$ since $n^{1/n}\to 1$ . Both series have the same radius of convergence $R$ .

TheoremTaylor Coefficients are Forced

If $f(x) = \sum_{n=0}^\infty a_n(x-c)^n$ on $(c-R,c+R)$ , then the coefficients are uniquely determined:

5

\quad a_n = \frac{f^{(n)}(c)}{n!}

Derivation hint: Differentiate [1] $n$ times using [4], then set $x=c$ ; all terms but the $n!\,a_n$ term vanish. Hence a power series representation, if it exists, is unique — this is the identity theorem for power series.

DefinitionAnalytic Function

A function $f$ is analytic at $c$ if there exists $R>0$ such that

6

\quad f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n \quad\text{for all } |x-c| < R.

Derivation hint: This is the definition — $f$ equals its own Taylor series in a neighborhood of $c$ . The right side is the Taylor series of $f$ . $f$ is analytic on an open set $U$ if it is analytic at every point of $U$ .

ProofCauchy–Hadamard and Term-by-Term Operations

Step 1 — Convergence region. Apply the root test to $\sum a_n (x-c)^n$ . Set

L(x) = \limsup_{n\to\infty} |a_n (x-c)^n|^{1/n} = |x-c| \cdot \limsup_{n\to\infty} |a_n|^{1/n} = \frac{|x-c|}{R}.

The root test gives absolute convergence when $L(x) < 1$ , i.e. $|x-c| < R$ , and divergence when $L(x) > 1$ , i.e. $|x-c| > R$ . This proves [2].

Step 2 — Abel's lemma (key inequality). If $\sum a_n r^n$ converges for some $r>0$ , then $a_n r^n \to 0$ , so $|a_n r^n|$ is bounded by some $M$ . For $|x-c| = \rho < r$ ,

|a_n (x-c)^n| \le M \left(\frac{\rho}{r}\right)^n,

a convergent geometric series. By the Weierstrass $M$ -test the original series converges uniformly on $|x-c| \le \rho$ .

Step 3 — Smoothness. The differentiated series $\sum n a_n (x-c)^{n-1}$ has radius $1/\limsup (n|a_n|)^{1/(n-1)}$ . Since $n^{1/n} \to 1$ , this equals $R$ . Uniform convergence on compact subsets lets us interchange limit and derivative (standard theorem on uniform limits of derivatives), proving [4]. Iterating gives $f \in C^\infty$ .

Step 4 — Forced coefficients. From [4] applied $k$ times,

f^{(k)}(x) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1)\, a_n (x-c)^{n-k}.

Setting $x = c$ kills every term except $n=k$ , leaving $f^{(k)}(c) = k!\,a_k$ , which is [5]. $\blacksquare$

∎

⚠️ Smooth ≠ Analytic

The Taylor series of a $C^\infty$ function need not converge to the function, even when it converges. Analyticity is a strictly stronger property than being infinitely differentiable. The remainder term in Taylor's theorem must be shown to vanish — possessing all derivatives is not enough.

mathematician · Augustin-Louis Cauchy | 1789–1857

Cauchy placed power series on a rigorous footing, introducing precise definitions of convergence and limits and proving the convergence criteria that bear his name (Cauchy–Hadamard, after Hadamard sharpened it with the limsup in 1888). In complex analysis he established that a complex-differentiable function is automatically analytic — the Cauchy integral formula yields the power series expansion directly — a result with no real-variable analogue. His 1821 Cours d'analyse is the foundational text where modern rigor for series began. Cauchy also identified the subtle distinction between pointwise and uniform convergence (an error in his original work later corrected by Abel and Weierstrass) that is essential for justifying term-by-term operations.

🧠 Intuition

Picture a power series as a recipe that builds a function out of ever-finer polynomial corrections. The constant $a_0$ sets the value at $c$ ; $a_1$ sets the slope; $a_2$ sets the curvature; and so on — each coefficient tunes one more derivative. The radius of convergence is the distance from $c$ to the nearest obstruction. For real functions like $\frac{1}{1+x^2}$ , expanded at $0$ , the obstruction is invisible on the real line but lives at $x = \pm i$ in the complex plane — which is exactly why $R = 1$ even though the function is perfectly smooth everywhere on $\mathbb{R}$ . This is the deepest single intuition: the radius of convergence is the distance to the nearest singularity in the complex plane.

Figure

Convergence regions of a power series centered at c, with radius R

ExampleGeometric Series and Its Radius

Given: $f(x) = \sum_{n=0}^\infty x^n$ , the geometric power series centered at $c=0$ with $a_n = 1$ .

Step 1 — Radius. Use [3]: $R = \lim |a_n/a_{n+1}| = \lim |1/1| = 1$ .

Step 2 — Endpoints. At $x=1$ : $\sum 1$ diverges. At $x=-1$ : $\sum (-1)^n$ diverges (terms don't $\to 0$ ). So the interval is $(-1,1)$ , open at both ends.

Step 3 — Closed form. Partial sum $s_N = \frac{1-x^{N+1}}{1-x}$ ; for $|x|<1$ , $x^{N+1}\to 0$ , so

f(x) = \frac{1}{1-x}.

Step 4 — Differentiate term by term (using [4]): $\sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2}$ , valid on the same interval $(-1,1)$ .

\boxed{\;\sum_{n=0}^\infty x^n = \frac{1}{1-x},\quad R = 1,\quad \text{interval } (-1,1)\;}

ExampleSeries Representation of arctan and a Numerical Estimate

Given: $f(x) = \arctan x$ , find its power series at $c=0$ and estimate $\arctan(0.5)$ .

Step 1 — Start from a known series. Since $\frac{1}{1+t^2} = \sum_{n=0}^\infty (-1)^n t^{2n}$ for $|t|<1$ (geometric series with ratio $-t^2$ ).

Step 2 — Integrate term by term (using [4]) from $0$ to $x$ :

\arctan x = \int_0^x \frac{dt}{1+t^2} = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots

Radius $R = 1$ (same as the integrand's series).

Step 3 — Plug in $x=0.5$ :

$n=0$ : $0.5$
$n=1$ : $-\frac{0.125}{3} = -0.0416667$
$n=2$ : $\frac{0.03125}{5} = 0.00625$
$n=3$ : $-\frac{0.0078125}{7} = -0.0011161$

Partial sum: $0.5 - 0.0416667 + 0.00625 - 0.0011161 = 0.4634672$ .

Step 4 — Error bound. This is an alternating series, so the error is less than the first omitted term, $|a_4| = \frac{0.5^9}{9} \approx 0.000217$ radians.

\boxed{\;\arctan(0.5) \approx 0.46347 \text{ rad},\quad \text{error} < 2.2\times10^{-4}\text{ rad}\;}

(True value: $0.463648$ rad.)

⚠️ Endpoints Are Not Decided by R

The radius of convergence tells you nothing about the endpoints $c \pm R$ . The four series $\sum x^n$ , $\sum x^n/n$ , $\sum (-1)^n x^n/n$ , $\sum x^n/n^2$ all have $R=1$ but different endpoint behavior. Always test $x = c\pm R$ separately using a non-power-series convergence test (alternating series test, $p$ -series, etc.).

CounterexampleA Smooth Non-Analytic Function

Define

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

Every derivative $g^{(n)}(0) = 0$ (the exponential decay beats every polynomial). So the Taylor series of $g$ at $0$ is identically $0$ — yet $g(x) \neq 0$ for $x \neq 0$ . The Taylor series converges (to $0$ ) everywhere but equals $g$ only at the single point $x=0$ . Therefore $g$ is $C^\infty$ but not analytic at $0$ . This shows that analyticity in [6] genuinely requires the remainder term to vanish, not merely the existence of derivatives.

⚠️ Reindexing and the n=0 Trap

When differentiating, the constant term vanishes, so the differentiated sum starts at $n=1$ , not $n=0$ — see [4]. Forgetting to shift the index (or to drop the $n=0$ term) is the single most common algebra error in solving ODEs by power series. After differentiating, always reindex with $m = n-1$ to align powers before combining series.

connection

Power series are the engine of Taylor Series and Taylor's Theorem — Taylor's theorem provides the remainder estimate that proves a smooth function actually is its power series. In Complex Analysis, analyticity becomes equivalent to complex differentiability (holomorphy), and the radius of convergence equals the distance to the nearest pole, illuminating real-line mysteries via the complex plane. The term-by-term operations rely on Uniform Convergence and the Weierstrass M-Test, without which interchanging sum and derivative is invalid. Power series solve Ordinary Differential Equations near ordinary points (the Frobenius method extends this to regular singular points), and generating functions in Combinatorics are power series whose coefficients encode counting sequences.

🔬 Where this appears

⏳ History

- 1665–1671 — Isaac Newton develops infinite series for

\ln(1+x)

,

\sin

,

\cos

, and the binomial series, using them as his primary computational tool in De Analysi.

- 1715 — Brook Taylor publishes the general expansion

\sum f^{(n)}(a)(x-a)^n/n!

in Methodus Incrementorum, formalizing the Taylor series.

- 1742 — Colin Maclaurin popularizes the special case centered at

0

in his Treatise of Fluxions; the "Maclaurin series" bears his name.

- 1797 — Lagrange attempts to base all of calculus on power series in Théorie des fonctions analytiques, coining "analytic function," though his program was logically incomplete.

- 1821 — Cauchy's Cours d'analyse introduces rigorous convergence, the ratio and root tests, and the radius of convergence concept.

- 1826 — Niels Henrik Abel proves the continuity of power series up to convergent boundary points (Abel's theorem) and corrects Cauchy's flawed claim about sums of continuous functions.

- 1851–1860s — Bernhard Riemann and Karl Weierstrass build complex analysis; Weierstrass defines analytic functions rigorously via convergent power series and develops analytic continuation.

- 1872 — Weierstrass exhibits a continuous nowhere-differentiable function, sharpening the gap between continuity, smoothness, and analyticity.

- 1888 — Jacques Hadamard establishes the definitive

\limsup

formula [2], completing the theory of the radius of convergence.

open-problem

The boundary behavior of power series remains subtle and not fully tamed. Abel's theorem gives convergence-implies-continuity at a boundary point, but its converse fails without Tauberian side conditions (Tauber's and Littlewood's theorems give partial converses). The structure of the set of boundary points where a series converges can be remarkably complicated, and natural boundaries — circles beyond which analytic continuation is utterly impossible, e.g. the lacunary series $\sum z^{2^n}$ on $|z|=1$ — show that the circle of convergence can be an impenetrable wall. The Fabry gap theorem and the Hadamard gap theorem characterize when this occurs, but a complete classification of boundary singularity sets, and the analytic continuation of generic series, involves deep open questions in analytic number theory (where, e.g., the analytic continuation and zero distribution of $L$ -series remain among the hardest problems in mathematics).

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also