Taylor Series of Analytic Functions

A central miracle of complex analysis is that holomorphy — mere complex differentiability on an open set — is enormously stronger than its real cousin. A real function can be differentiable once and never again; it can be infinitely differentiable and still not equal its Taylor series. A complex function that is holomorphic on an open disc suffers neither fate. It is automatically infinitely differentiable there, and it is equal to a convergent power series centred at the disc's centre — its Taylor series. Differentiability once forces differentiability infinitely often, and more: exact agreement with an explicit analytic expression. The engine that delivers this rigidity is Cauchy's integral formula. Writing $f$ as a contour integral and expanding the Cauchy kernel $1/(\zeta-z)$ as a geometric series in the ratio $(z-z_0)/(\zeta-z_0)$ produces, after one legitimate interchange of sum and integral, a power series whose coefficients are exactly $a_n=f^{(n)}(z_0)/n!$ . The interchange is licensed by uniform convergence on the compact contour, and the resulting series converges on the largest open disc about $z_0$ that stays inside the region of holomorphy. That radius has a strikingly geometric meaning: it is the distance from the centre to the nearest point where $f$ stops being holomorphic. This single fact demystifies the otherwise baffling behaviour of real Taylor series, whose convergence is secretly governed by complex singularities that never touch the real axis. It is the reason the innocuous function $1/(1+x^2)$ , smooth and bounded across all of $\mathbb{R}$ , nonetheless has a Taylor series that dies at $|x|=1$ — the culprits are the poles at $z=\pm i$ . Everything downstream — the identity theorem, the order of zeros, Laurent series, and residues — is built on this foundation.

Key takeaways

A function holomorphic on an open disc $|z-z_0|<R$ equals its Taylor series $\sum a_n(z-z_0)^n$ there, with coefficients $a_n=f^{(n)}(z_0)/n!$ — no gap between smoothness and analyticity.
The coefficients are Cauchy's contour integrals, $a_n=\frac{1}{2\pi i}\oint f(\zeta)/(\zeta-z_0)^{n+1}\,\mathrm{d}\zeta$ , so holomorphy on a disc forces power-series representability.
The radius of convergence equals the distance from $z_0$ to the nearest singularity of $f$ in the complex plane — an obstruction invisible from the real line alone.
This is why $1/(1+x^2)=\sum(-1)^n x^{2n}$ diverges for $|x|>1$ despite being perfectly tame on $\mathbb{R}$ : the poles at $z=\pm i$ sit at distance $1$ .
Uniqueness of the coefficients, term-by-term differentiation and integration inside the disc, Cauchy's coefficient estimates, and zeros of finite order all follow at once.
The Taylor series is the singularity-free special case of the Laurent series — the same construction with no negative powers.
Key idea: Holomorphy on a disc is equivalent to representability by a convergent power series there, and the radius of convergence is exactly the distance from the centre to the nearest singularity.

Where this lives today

Taylor expansion of analytic functions is a live computational and conceptual tool. In physics the sharpest modern use is the analysis of perturbation series: quantum field theory and statistical mechanics routinely produce power series in a coupling constant whose radius of convergence is zero, and locating the complex singularities (in the coupling, or via Borel summation in a dual variable) is how one extracts finite predictions from a formally divergent expansion — the same "singularity sets the radius" logic that governs $1/(1+x^2)$ , applied now to the anomalous magnetic moment or the anharmonic-oscillator ground-state energy. In numerical computation the Cauchy coefficient integral (2) underlies spectrally accurate methods: to obtain the $n$ -th derivative of an analytic function one evaluates it on a circle and takes an FFT, sidestepping the catastrophic cancellation of finite differences. And the whole apparatus is the direct on-ramp to the Laurent series, residues, and the residue theorem that evaluate real integrals, sum series, and count zeros and poles.

The physical picture

A holomorphic function is so tightly constrained that its value and all its derivatives at a single point $z_0$ pin it down throughout the largest disc that avoids every singularity. The Taylor series is nothing but the ledger of that rigidity: derivative $f^{(n)}(z_0)$ buys coefficient $a_n$ , and the whole function is reassembled from those local data. Picture the disc of convergence as a balloon inflating outward from $z_0$ . It swells freely through holomorphic territory and stops the instant its boundary kisses a singularity — it cannot swallow one, because a power series converges on a disc, and a disc containing a singularity would make the sum holomorphic where the function is not. The radius is therefore a literal distance in the complex plane. The first figure makes this concrete: for $1/(1+z^2)$ the balloon about the origin stops at $R=1$ , but re-centring it enlarges the disc precisely by how much farther the nearest pole recedes.

Figure

The disc of convergence grows until it touches the nearest singularity. For 1/(1+z²) the poles sit at z = ±i, so expanding about z₀ = 0 gives R = 1, while recentring at z₀ = 1 pushes the nearest pole to distance √2 ≈ 1.414 and enlarges the disc.

The same principle explains a genuine paradox of the real line. On $\mathbb{R}$ the function $1/(1+x^2)$ is a smooth, bounded, entirely well-behaved bell — yet its Taylor series about the origin converges only on $(-1,1)$ and blows up beyond. Nothing on the real axis warns you of the failure; the obstruction lives off the axis, at the complex poles $z=\pm i$ . This is the germ of the Runge phenomenon and of the way perturbation series in physics can diverge even where the physical quantity they compute is finite. The second figure shows the collision directly: the true function stays gentle while the truncated Taylor polynomial peels away and plunges the moment $|x|$ crosses the radius.

Figure

The real function 1/(1+x²) is smooth and bounded on the whole real line, yet its Taylor partial sums (amber) hug the curve only for x < 1 in magnitude and diverge beyond, because the complex poles at z = ±i fix the radius at R = 1.

The core equations

Taylor's theorem is the statement that a holomorphic function is its power series, together with the formula for the coefficients. Two sentences say it all: the value at any $z$ inside the disc is recovered from the derivatives at the centre, and each derivative supplies one coefficient after division by $n!$ .

f(z)=\sum_{n=0}^{\infty} a_n\,(z-z_0)^n,\qquad a_n=\frac{f^{(n)}(z_0)}{n!} \qquad (1)

where: $f$ — function holomorphic on the disc $|z-z_0|<R$ ; $z_0$ — centre of expansion; $a_n$ — the $n$ -th Taylor coefficient; $f^{(n)}$ — the $n$ -th complex derivative of $f$ . Reading it: knowing every derivative at one point reconstructs the whole function on the disc, so a holomorphic function carries no independent local information — it is rigid. The coefficients need not be computed by differentiating at all. Cauchy's integral formula for derivatives packages them as a single contour integral over any circle inside the disc, which is what makes them robust to compute and estimate.

a_n=\frac{1}{2\pi i}\oint_{C}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,\mathrm{d}\zeta \qquad (2)

where: $C$ — the positively oriented circle $|\zeta-z_0|=r<R$ ; $\zeta$ — integration variable running once around $C$ . Reading it: the coefficient depends only on the values of $f$ on a loop around $z_0$ , never on what $f$ does elsewhere — a boundary integral fixing an interior quantity. Everything rests on one universal identity, the geometric series. Substituting the right ratio into it turns a single line of algebra into the whole Taylor theorem.

\frac{1}{1-w}=\sum_{n=0}^{\infty} w^n,\qquad |w|<1 \qquad (3)

where: $w$ — any complex number with $|w|<1$ . Reading it: this is the seed. Setting $w=(z-z_0)/(\zeta-z_0)$ inside the Cauchy kernel and integrating term by term generates equation (1) directly, as the first derivation shows. The radius of convergence is not an abstract limit to be feared; it is a distance you can measure by locating singularities. Equivalently it is read off the coefficients themselves through the Cauchy–Hadamard formula, and the two answers always agree.

R=\operatorname{dist}\!\big(z_0,\ \text{nearest singularity}\big),\qquad \frac{1}{R}=\limsup_{n\to\infty}|a_n|^{1/n} \qquad (4)

where: $R$ — radius of convergence; $\limsup$ — the largest subsequential limit of $|a_n|^{1/n}$ . Reading it: the geometry (distance to a singularity) and the arithmetic (growth of the coefficients) are two windows on the same number. Bounding the coefficient integral by its length times its maximum gives Cauchy's estimates — the modest-looking inequalities that power Liouville's theorem and the fundamental theorem of algebra.

|a_n|\le\frac{M(r)}{r^{n}},\qquad M(r)=\max_{|\zeta-z_0|=r}|f(\zeta)| \qquad (5)

where: $M(r)$ — the maximum modulus of $f$ on the circle of radius $r$ ; $r$ — any radius with $0<r<R$ . Reading it: a function that stays small on large circles cannot have large high-order coefficients — growth of $f$ controls growth of the $a_n$ . Finally, the leading nonzero coefficient classifies a zero by its order, and the series may be differentiated and integrated term by term with the radius unchanged — the manipulations that let you build new expansions from known ones.

f(z)=(z-z_0)^m g(z),\ \ g(z_0)\neq0;\qquad f'(z)=\sum_{n=1}^{\infty} n\,a_n(z-z_0)^{n-1} \qquad (6)

where: $m$ — the order (multiplicity) of the zero at $z_0$ , so $a_0=\dots=a_{m-1}=0$ and $a_m\neq0$ ; $g$ — holomorphic with $g(z_0)=a_m\neq0$ ; the differentiated series has the same radius $R$ . Reading it: near a zero of order $m$ , $f(z)\approx a_m(z-z_0)^m$ , and inside the disc calculus commutes with summation.

Derivation ATaylor's theorem from Cauchy's integral formula

The goal is to prove that a function holomorphic on a disc is exactly equal to a convergent power series there, and to identify its coefficients, using nothing beyond Cauchy's integral formula and the geometric series. The strategy is to write $f(z)$ as a contour integral, expand the Cauchy kernel as a geometric series in a ratio that is genuinely small on the contour, and interchange the sum with the integral — an interchange that uniform convergence makes legitimate. Fix $z$ with $|z-z_0|=\rho$ , choose a circle $C:|\zeta-z_0|=r$ with $\rho<r<R$ , so that $z$ lies strictly inside $C$ .

Step 1. Start from Cauchy's integral formula, $f(z)=\frac{1}{2\pi i}\oint_C \frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta$ , which is valid because $z$ lies inside $C$ and $f$ is holomorphic on and within $C$ . Why: this is the only input that knows $f$ is holomorphic; every later step is algebra and convergence bookkeeping applied to this one representation.

Step 2. Rewrite the kernel by splitting off the centre, $\frac{1}{\zeta-z}=\frac{1}{(\zeta-z_0)-(z-z_0)}=\frac{1}{\zeta-z_0}\cdot\frac{1}{1-\tfrac{z-z_0}{\zeta-z_0}}$ . Why: this exposes a geometric ratio $w=(z-z_0)/(\zeta-z_0)$ whose modulus is $\rho/r<1$ for every $\zeta$ on $C$ , so the geometric series will converge.

Step 3. Expand geometrically, $\frac{1}{\zeta-z}=\sum_{n=0}^{\infty}\frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}}$ , a series that converges uniformly in $\zeta$ on $C$ because $|w|=\rho/r$ is a constant strictly below $1$ , independent of $\zeta$ . Why: uniformity — not merely pointwise convergence — is exactly the property that will license swapping the infinite sum past the integral in the next step.

Step 4. Interchange sum and integral, $f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\oint_C\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,\mathrm{d}\zeta\right)(z-z_0)^n$ , which is permitted because a uniformly convergent series on the compact contour $C$ may be integrated term by term. Why: this is the load-bearing move — it converts one integral of an infinite sum into an infinite sum of ordinary Cauchy integrals, each a constant coefficient.

Step 5. Identify the bracket as the coefficient. By Cauchy's integral formula for derivatives it equals $f^{(n)}(z_0)/n!$ , so $a_n=\frac{1}{2\pi i}\oint_C\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,\mathrm{d}\zeta=\frac{f^{(n)}(z_0)}{n!}$ . Since $r<R$ was arbitrary, the representation holds for every $z$ with $|z-z_0|<R$ . Why: this closes the loop — the abstract integral coefficients are precisely the familiar derivative coefficients, and the disc of validity is the full disc of holomorphy.

Go deeper · Derivation B · 1 minWhy the radius equals the distance to the nearest singularity

Derivation BWhy the radius equals the distance to the nearest singularity

The goal is to show that the radius of convergence $R$ is neither larger nor smaller than the distance $d$ from the centre $z_0$ to the closest point where $f$ fails to be holomorphic, and to confirm that the coefficient-based Cauchy–Hadamard formula returns the same value. The argument is a two-sided squeeze: convergence cannot fall short of $d$ because $f$ is holomorphic out to that distance, and it cannot exceed $d$ because a convergent series would illegitimately continue $f$ across its own singularity. Let $d=\inf\{|s-z_0|:s\text{ a singularity of }f\}$ .

Step 1. Show $R\ge d$ . On the open disc $|z-z_0|<d$ the function $f$ is holomorphic by definition of $d$ , so Taylor's theorem (the previous derivation) applies and the series converges throughout that disc; hence the radius is at least $d$ . Why: this direction is pure existence — wherever $f$ is holomorphic, the series is guaranteed to represent it, so convergence reaches at least as far as the singularity.

Step 2. Show $R\le d$ . Suppose instead $R>d$ . Then the power series would define a holomorphic function on the larger disc $|z-z_0|<R$ agreeing with $f$ near $z_0$ , hence continuing $f$ holomorphically across the singularity at distance $d$ — contradicting that $f$ is genuinely singular there. Why: a convergent power series is holomorphic on its whole open disc, so it can never contain a singularity of the function it represents; the singularity caps the radius.

Step 3. Conclude $R=d$ , and cross-check with Cauchy–Hadamard, $1/R=\limsup_n|a_n|^{1/n}$ , which extracts the same radius from the coefficients alone without reference to any singularity. Why: two independent computations — a geometric distance and an arithmetic growth rate — must agree, which is a powerful consistency check in practice.

Step 4. Test on $1/(1+z^2)=\sum_{n=0}^\infty(-1)^n z^{2n}$ . The singularities are simple poles at $z=\pm i$ , each at distance $1$ from the origin, so $d=1$ ; and $\limsup|a_n|^{1/n}=1$ since the nonzero coefficients have modulus $1$ , giving $R=1$ by both routes. Why: a concrete instance where the geometric and arithmetic answers visibly coincide, anchoring the abstract theorem to a computable example.

Go deeper · Derivation C · 1 minCauchy's coefficient estimates and Liouville's theorem

Derivation CCauchy's coefficient estimates and Liouville's theorem

The goal is to derive the inequality that bounds each Taylor coefficient by the size of $f$ on a circle, and then to run it once with $r\to\infty$ to prove Liouville's theorem — that a bounded entire function is constant — which in turn yields the fundamental theorem of algebra. The method is the estimation lemma: an integral is bounded by the length of the contour times the maximum of the integrand. Take a circle $C:|\zeta-z_0|=r$ with $0<r<R$ and let $M(r)=\max_{|\zeta-z_0|=r}|f(\zeta)|$ .

Step 1. Apply the estimation lemma to the coefficient integral (2): the contour has length $2\pi r$ and the integrand is bounded by $M(r)/r^{n+1}$ , so $|a_n|\le\frac{1}{2\pi}\cdot 2\pi r\cdot\frac{M(r)}{r^{n+1}}=\frac{M(r)}{r^n}$ . Why: this converts an exact integral identity into a clean one-line growth bound — the modulus of $f$ on a circle controls the modulus of every coefficient.

Step 2. Specialise to an entire function bounded by a constant $M$ , so $M(r)\le M$ for all $r$ . Then $|a_n|\le M/r^n$ for every $r>0$ ; letting $r\to\infty$ forces $a_n=0$ for all $n\ge1$ . Why: for $n\ge1$ the bound decays to zero as the circle grows, and a fixed number bounded by arbitrarily small quantities must vanish — only the constant term survives.

Step 3. Conclude Liouville's theorem: $f(z)=a_0$ is constant. Applied to $p(z)=1/P(z)$ for a non-constant polynomial $P$ with no root, this contradiction proves every non-constant polynomial has a complex root — the fundamental theorem of algebra. Why: the same estimate that looks like a technical inequality is the entire engine behind two of the deepest theorems in the subject.

Limiting cases

The radius of convergence is a barometer of the singularity structure: it reads $\infty$ when there is nothing to hit, and shrinks to the distance of whatever obstruction lies nearest — pole, branch point, or essential singularity alike.

Case	Result	Meaning
Entire function	$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!},\quad R=\infty$	No finite singularity, so the series converges on all of $\mathbb{C}$ ; the same holds for $\sin z$ and $\cos z$ .
Removable singularity	$\frac{\sin z}{z}=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n+1)!},\quad R=\infty$	The apparent singularity at $0$ cancels; the function extends to an entire one, so nothing limits the radius.
Pole-limited radius	$\frac{1}{1+z^2}=\sum_{n=0}^{\infty}(-1)^n z^{2n},\quad R=1$	Poles at $z=\pm i$ sit at distance $1$ , capping the radius despite perfect smoothness on the real axis.
Branch point	$\log(1+z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}z^{n},\quad R=1$	The branch point at $z=-1$ , distance $1$ away, limits the expansion about the origin.

Common trapA real function's Taylor series diverges only where the function misbehaves on the real line.

This is false, and it is the single most common misconception about Taylor series in physics. The radius is set by the nearest singularity anywhere in the complex plane, not by anything visible on $\mathbb{R}$ . The textbook example is $1/(1+x^2)$ : it is smooth, bounded, and monotone-tailed across the entire real line, with no hint of trouble — yet its Taylor series about the origin has radius exactly $1$ , because the analytic continuation $1/(1+z^2)$ has poles at $z=\pm i$ , each at distance $1$ from $0$ . If you only ever look at the real axis, the divergence at $|x|=1$ is inexplicable; step into the complex plane and it is forced. The same trap explains why a perturbation series in physics can diverge even in a regime where the physical quantity it approximates is perfectly finite — the divergence is announcing a complex singularity in the coupling, not a breakdown of the physics.

Worked examples

1 · Geometric expansion of a rational function, twice (confidence builder)

Expand $f(z)=1/(3-z)$ as a Taylor series about $z_0=0$ and find its radius; then re-expand about $z_0=1$ and find the new radius. Verify the first series numerically.

Step 1. About the origin, factor out the constant to hit the geometric seed (3): $\dfrac{1}{3-z}=\dfrac{1}{3}\cdot\dfrac{1}{1-z/3}=\dfrac{1}{3}\sum_{n=0}^{\infty}\left(\dfrac{z}{3}\right)^n=\sum_{n=0}^{\infty}\dfrac{z^n}{3^{n+1}}$ . The coefficients are $a_n=3^{-(n+1)}$ .

Step 2. The only singularity is the simple pole at $z=3$ , at distance $3$ from the origin, so $R=3$ . Cross-check by Cauchy–Hadamard: $|a_n|^{1/n}=\big(3^{-(n+1)}\big)^{1/n}=3^{-(n+1)/n}\to 3^{-1}$ , hence $1/R=1/3$ and $\mathbf{R=3}$ — both routes agree.

Step 3. Numerical check at $z=1$ (inside the disc, since $1<3$ ): the closed form gives $f(1)=1/(3-1)=1/2$ , while the series gives $\sum_{n=0}^{\infty}3^{-(n+1)}=\dfrac{1/3}{1-1/3}=\dfrac{1/3}{2/3}=\mathbf{1/2}$ . The series reproduces the function exactly.

Step 4. Re-expand about $z_0=1$ . The nearest (and only) singularity is still the pole at $z=3$ , now at distance $|3-1|=2$ , so the new radius is $\mathbf{R=2}$ . Concretely, $\dfrac{1}{3-z}=\dfrac{1}{2-(z-1)}=\dfrac{1}{2}\sum_{n=0}^{\infty}\left(\dfrac{z-1}{2}\right)^n$ , with coefficients $2^{-(n+1)}$ and Cauchy–Hadamard radius $2$ , confirming the geometric reading: moving the centre changes the radius by exactly how the distance to the pole changes.

2 · The real-line paradox with numbers

Take the real function $f(x)=1/(1+x^2)$ . Find the radius of its Taylor series about $x_0=0$ , show numerically that the series diverges just past that radius, and then find the radius about $x_0=2$ .

Step 1. About the origin the series is $\sum_{n=0}^{\infty}(-1)^n x^{2n}$ , obtained by putting $w=-x^2$ into the geometric seed (3). Its analytic continuation $1/(1+z^2)$ has poles at $z=\pm i$ , each at distance $|{\pm i}-0|=1$ from the origin, so $\mathbf{R=1}$ — even though $f$ is smooth and bounded for every real $x$ .

Step 2. Watch the boundary numerically. At $x=0.9$ the ratio $x^2=0.81<1$ , so the terms $(0.81)^n$ shrink geometrically and the series converges (to $1/1.81=0.5525$ ). At $x=1.1$ the ratio $x^2=1.21>1$ , so the terms $(1.21)^n$ grow without bound and the series diverges — even though $f(1.1)=1/2.21=0.4525$ is a perfectly ordinary number. The divergence is a property of the series, not of the function.

Step 3. See it in the coefficients. The degree- $10$ partial sum $S(x)=1-x^2+x^4-x^6+x^8-x^{10}$ gives $S(0.5)=0.7998$ against the true $0.8000$ — excellent — but $S(1.2)=-3.244$ against the true $0.4525$ , a wild miss. Beyond $|x|=1$ the leading term $-x^{10}$ dominates and the polynomial plunges, exactly the amber curve of the second figure.

Step 4. Re-centre at $x_0=2$ . The nearest singularities are still $z=\pm i$ , now at distance $|2-i|=\sqrt{2^2+1^2}=\sqrt{5}\approx 2.236$ from the centre, so the Taylor series about $x_0=2$ has $\mathbf{R=\sqrt5\approx2.236}$ . Because $2.236>2$ , that disc reaches back past the origin — a single re-centring more than doubles the interval of convergence, and the whole change is dictated by the geometry of two invisible complex poles.

Physics of the portrait

The theory of Taylor series is really the theory of Cauchy's integral formula wearing a different hat. It was Augustin-Louis Cauchy who turned the eighteenth-century formal series of Brook Taylor into a theorem with a proof, a domain of validity, and a radius — and who supplied, in the same stroke, the coefficient estimates that still carry the deepest consequences of the subject.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · complex analysis and the integral formula

Augustin-Louis Cauchy is, more than any single figure, the architect of rigorous analysis, and the Taylor series of a holomorphic function is one of his monuments. Brook Taylor had written the series bearing his name in 1715 as a formal manipulation, and Lagrange had tried to found all of calculus on the assumption that functions equal their Taylor series — an assumption that is simply false for real functions (the standard counterexample $e^{-1/x^2}$ , smooth with every derivative zero at the origin, has the zero series and equals nothing of the sort). Cauchy's revolution, beginning with his 1821 Cours d'analyse, was to insist on convergence and limits rather than formal algebra. In the complex domain he then discovered why the miracle that fails on $\mathbb{R}$ succeeds on $\mathbb{C}$ : his integral formula expresses a holomorphic function's interior values through a boundary integral, and expanding its kernel as a geometric series yields the Taylor theorem with the coefficients $a_n=f^{(n)}(z_0)/n!$ delivered exactly. From the same integral he extracted the estimates $|a_n|\le M(r)/r^n$ , the compact inequalities from which Liouville's theorem and the fundamental theorem of algebra fall out almost immediately. Cauchy's output was staggering — some seven hundred papers, and his name attaches to the integral formula, the residue theorem, the Cauchy–Riemann equations, and Cauchy sequences alike. The final piece of the story, the coefficient formula $1/R=\limsup|a_n|^{1/n}$ for the radius, was supplied by Jacques Hadamard in 1888, completing the dictionary between a function's singularities and the arithmetic of its Taylor coefficients that Cauchy had opened.

History

1715

— Brook Taylor publishes the general series in Methodus Incrementorum Directa et Inversa

1821

— Cauchy's Cours d'analyse puts convergence and limits on a rigorous footing

1831

— Cauchy states the integral formula and derives his coefficient estimates

1841

— Weierstrass founds the theory of analytic functions on convergent power series

1888

— Hadamard gives the radius formula 1/R = limsup of the nth root of the coefficient modulus

Going further

Texts. Stone & Goldbart, Mathematics for Physics, complex-analysis chapter (Taylor and Laurent series) · Arfken & Weber, Mathematical Methods for Physicists, Taylor expansion of analytic functions and Cauchy's estimates · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, Ch. 24–25 · Ahlfors, Complex Analysis, Ch. 4 (Taylor's theorem, Cauchy estimates, Liouville) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 6. The key results to keep in mind: Taylor's theorem via geometric expansion of the Cauchy kernel; the coefficient identity $a_n=f^{(n)}(z_0)/n!$ and its integral form (2); the radius theorem $R=\operatorname{dist}(z_0,\text{singularity})$ and its Cauchy–Hadamard companion $1/R=\limsup|a_n|^{1/n}$ ; the worked series $1/(3-z)=\sum z^n/3^{n+1}$ ( $R=3$ , sum $1/2$ at $z=1$ ; $R=2$ about $z_0=1$ ), $1/(1+z^2)$ ( $R=1$ about $0$ , $R=\sqrt5$ about $2$ ), $e^z$ and $\sin z/z$ ( $R=\infty$ ), $\log(1+z)$ ( $R=1$ , branch point at $-1$ ); and the estimate $|a_n|\le M(r)/r^n$ leading to Liouville.

Prerequisites. Analytic Functions and the Cauchy-Riemann Equations · Complex Integration and Contour Integrals · Cauchy's Integral Formula and Derivatives · Functions of a Complex Variable and Limits. Leads to. Laurent Series · Zeros, Poles, and Isolated Singularities · Consequences: Liouville, Maximum Modulus, Morera · Residues and the Residue Theorem.

Ask the tutor: Ask me to expand any function you like about any centre and I will find its radius of convergence two independent ways — by hunting the nearest complex singularity and by the Cauchy–Hadamard growth of its coefficients — and show you why they agree.

Is being infinitely differentiable the same as being analytic?

On the complex plane, yes; on the real line, emphatically no. A complex function that is differentiable even once on an open set is automatically infinitely differentiable and equal to its Taylor series — the three notions collapse into one. On $\mathbb{R}$ they come apart badly: the function $e^{-1/x^2}$ (defined as $0$ at the origin) is infinitely differentiable there with every derivative equal to zero, so its Taylor series about $0$ is identically zero and represents the function nowhere except at the single point $x=0$ . Such a function is smooth but not analytic. The gap is exactly what complex differentiability closes, which is why analyticity is a complex-analytic idea at heart.

How can a function be perfectly smooth on the real line yet have a finite radius of convergence?

Because the radius is decided in the complex plane, where the function's analytic continuation may have singularities the real axis never sees. The canonical case is $1/(1+x^2)$ : on $\mathbb{R}$ it is a smooth, bounded bell with no bad behaviour anywhere, but its continuation $1/(1+z^2)$ has poles at $z=\pm i$ . The radius of the Taylor series about the origin equals the distance to the nearest of these, namely $1$ , so the series converges only on $(-1,1)$ regardless of how tame the real function looks. The lesson is that real convergence questions are secretly complex-analytic, and you cannot answer them by staring at the real graph.

What is the practical difference between the two formulas for the radius?

They answer the same question from opposite ends. The geometric formula $R=\operatorname{dist}(z_0,\text{nearest singularity})$ is fastest when you know the function in closed form — you locate its poles and branch points and measure. The Cauchy–Hadamard formula $1/R=\limsup|a_n|^{1/n}$ is what you use when you only know the coefficients, for instance a series generated by a recurrence or defined by its power-series expansion in the first place. In practice you check one against the other: compute $R$ from the singularities, then confirm that the coefficients grow like $R^{-n}$ . When both agree you can be confident you have found every relevant singularity.

Why is uniform convergence the crucial hypothesis in the derivation?

Because the whole proof hinges on interchanging an infinite sum with a contour integral, and that swap is only valid when the series converges uniformly on the contour. In the kernel expansion the ratio $|(z-z_0)/(\zeta-z_0)|$ equals $\rho/r$ , a single constant strictly below $1$ that does not depend on where $\zeta$ sits on the circle — that uniform bound is exactly what makes the geometric series converge uniformly, and hence what licenses term-by-term integration. If convergence were merely pointwise you could not move the sum past the integral, and the coefficients would not come out as clean Cauchy integrals. Uniformity is not a technicality; it is the hinge of the argument.

How do Cauchy's estimates lead to Liouville's theorem and the fundamental theorem of algebra?

The estimate $|a_n|\le M(r)/r^n$ says a coefficient cannot exceed the size of $f$ on a circle of radius $r$ , divided by $r^n$ . If $f$ is entire (holomorphic on all of $\mathbb{C}$ ) and bounded by a constant $M$ , then for every $n\ge1$ the bound $M/r^n$ can be made as small as you like by taking $r\to\infty$ , forcing $a_n=0$ . Only $a_0$ survives, so $f$ is constant — that is Liouville's theorem. Applying it to $1/P(z)$ for a non-constant polynomial $P$ with no root produces a bounded entire function that would have to be constant, a contradiction; hence every non-constant polynomial has a complex root, the fundamental theorem of algebra.

Why can I compute Taylor coefficients by any trick I like?

Because the coefficients are unique. If $\sum a_n(z-z_0)^n=\sum b_n(z-z_0)^n$ on a disc, then $a_n=b_n$ for every $n$ — a power-series representation is one of a kind. So you may obtain the series by whatever route is easiest: differentiate repeatedly, multiply two known series (as for $e^z\cos z$ ), integrate a known series term by term (as for $\log(1+z)$ from $1/(1+z)$ ), or split a rational function by partial fractions and expand each piece geometrically. Any legitimate method returns the Taylor series, valid on the smaller of the relevant discs.