How do you define $e^x$? One approach: declare $e = \lim_{n \to \infty} (1 + 1/n)^n$ and define $e^x$ by continuity and the laws of exponents. But this requires proving that $e^{p/q} = (e^{1/q})^p$ makes sense for rationals, extending to irrationals by density of $\mathbb{Q}$, and verifying differentiability — a laborious chain of arguments that obscures the function's essential character. A far cleaner approach: *define* $e^x$ by the power series $\sum_{n=0}^\infty x^n / n!$ and *derive* all its properties — the functional equation $e^{x+y} = e^x e^y$, the differential equation $f' = f$, the value $e = e^1$ — from the [series](/page/Series) alone. The power series does not merely *represent* the exponential; it *is* the exponential, in the most logically economical sense. This is the central idea of power series: they provide a unified framework for defining and manipulating transcendental [functions](/page/Function). A power series $\sum a_n (x - c)^n$ is an "infinite polynomial" that converges on some interval and diverges outside it. Within its interval of convergence, it inherits the best properties of polynomials — [continuity](/page/Continuity%20(Real%20Analysis)), infinite differentiability, term-by-term calculus — while reaching functions (exponentials, trigonometric functions, logarithms) that no finite polynomial can represent exactly. The [radius of convergence](/theorems/273) $R$ governs the domain, the Cauchy-Hadamard formula computes $R$ from the coefficients, and the identity theorem imposes a remarkable rigidity: if two power series agree on a sequence with a [limit](/page/Limit) point, they agree everywhere. ## Definition and the Radius of Convergence [definition:Power Series] A **power series** centred at $c \in \mathbb{R}$ is a formal expression \begin{align*} \sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x - c) + a_2(x - c)^2 + \cdots \end{align*} where $(a_n)_{n=0}^\infty$ is a sequence of real numbers called the **coefficients**. For each fixed $x$, the series becomes a numerical series whose convergence may depend on $x$. [/definition] At $x = c$ the series always converges (to $a_0$). The fundamental question is: for which other $x$ does it converge? The answer has a strikingly clean geometric structure — convergence always occurs on an interval symmetric about $c$. [quotetheorem:202] The proof reveals *why* the convergence region is an interval. If the series converges at some $x_0$, then $|a_n (x_0 - c)^n| \leq M$ for some bound $M$ (because convergent [sequences](/page/Sequence) are bounded). For any $x$ closer to $c$ than $x_0$, the ratio $\rho = |x - c| / |x_0 - c| < 1$ gives $|a_n(x - c)^n| \leq M\rho^n$, and comparison with the geometric series $\sum M\rho^n$ yields absolute convergence. Convergence at one point *dominates* all nearer points — and this domination forces the convergence set to be an interval. ### The Cauchy-Hadamard Formula The radius $R$ can be computed directly from the coefficients, without testing convergence at individual points. [definition:Radius Of Convergence] The **radius of convergence** of $\sum a_n (x - c)^n$ is the quantity \begin{align*} \frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}, \end{align*} with the conventions $1/0 = \infty$ and $1/\infty = 0$. The series converges absolutely for $|x - c| < R$ and diverges for $|x - c| > R$. [/definition] The formula uses the $\limsup$ rather than the $\lim$ because the limit may not exist (the $n$th root of $|a_n|$ may oscillate). The $\limsup$ captures the "worst-case" growth rate of the coefficients, which is what determines convergence. When the ordinary limit exists, the ratio test gives an alternative formula: $1/R = \lim_{n \to \infty} |a_{n+1}/a_n|$ (provided the limit exists). This is often easier to compute in practice. [example:Three Fundamental Radii] **$R = 1$ (the geometric series).** $\sum x^n$ has $a_n = 1$, so $|a_n|^{1/n} = 1$ and $R = 1$. The series converges to $1/(1-x)$ for $|x| < 1$ and diverges for $|x| \geq 1$. This is the prototype: every power series locally resembles a geometric series with ratio $|x - c|/R$. **$R = \infty$ (the exponential).** $\sum x^n/n!$ has $|a_{n+1}/a_n| = 1/(n+1) \to 0$, so $R = \infty$. Factorial decay of the coefficients overwhelms any polynomial growth of $|x|^n$, forcing convergence everywhere. **$R = 0$ (a pathological case).** $\sum n! \, x^n$ has $|a_{n+1}/a_n| = n + 1 \to \infty$, so $R = 0$. The series converges only at $x = 0$. Factorial *growth* of the coefficients overwhelms any decay from $|x|^n$ for $x \neq 0$. [/example] ### Behaviour at the [Boundary](/page/Boundary) The Cauchy-Hadamard formula determines convergence for $|x - c| < R$ (absolute convergence) and $|x - c| > R$ (divergence), but says nothing about the boundary points $x = c \pm R$. Boundary behaviour must be analysed case by case and can differ at the two endpoints. [example:Boundary Diversity] The three series $\sum x^n/n^2$, $\sum (-1)^{n-1} x^n/n$, and $\sum x^n/n$ all have radius $R = 1$, but: - $\sum 1/n^2$ converges (at $x = 1$) and $\sum (-1)^n/n^2$ converges (at $x = -1$): convergence at *both* endpoints. - $\sum (-1)^{n-1}/n$ converges (at $x = 1$, alternating harmonic) but $\sum (-1)^{n-1}(-1)^n/n = -\sum 1/n$ diverges (at $x = -1$): convergence at one endpoint only. - $\sum 1/n$ diverges (at $x = 1$) and $\sum (-1)^n/n$ converges (at $x = -1$): the opposite endpoint converges. No formula determines boundary convergence from the coefficients alone — it depends on the precise interplay between coefficient decay and sign cancellation. [/example] ## Term-Wise Calculus The deepest property of power series is that calculus operations — differentiation and integration — commute with infinite summation within the interval of convergence. This is far from obvious: term-wise differentiation fails for many uniformly convergent series of functions. Power series are special because the derived series has the *same* radius of convergence. ### Term-Wise Differentiation [quotetheorem:206] The key fact is that differentiation does not change the radius. The Cauchy-Hadamard formula for the derived series gives $\limsup |n a_n|^{1/n} = \limsup n^{1/n} |a_n|^{1/n} = \limsup |a_n|^{1/n}$, since $n^{1/n} \to 1$. The factor $n$ introduced by differentiation grows only polynomially, while the $n$th root strips this to a factor of $n^{1/n} \to 1$ — invisible to the Cauchy-Hadamard formula. By induction, a power series with $R > 0$ is *infinitely differentiable* on $(c - R, c + R)$, and the $k$th derivative is obtained by differentiating term by term $k$ times: \begin{align*} f^{(k)}(x) = \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1) \, a_n (x - c)^{n-k}. \end{align*} Setting $x = c$ gives $f^{(k)}(c) = k! \, a_k$, so the coefficients are determined by the derivatives: $a_k = f^{(k)}(c)/k!$. A power series is the [Taylor series](/page/Derivative) of its own sum — and the Taylor series converges back to the function. ### Term-Wise Integration [quotetheorem:207] Integration also preserves the radius of convergence (by the same $n^{1/n} \to 1$ argument applied to the factors $1/(n+1)$). Moreover, integration can *improve* boundary behaviour: the series $\sum x^n/n$ diverges at $x = 1$, but its term-wise integral $\sum x^{n+1}/(n(n+1))$ converges at $x = 1$. Integration smooths out boundary divergence, while differentiation may worsen it. ### Continuity [quotetheorem:205] Continuity follows from the Weierstrass $M$-test: on any closed subinterval $[c - r, c + r]$ with $r < R$, the terms are bounded by $|a_n| r^n$ (a convergent series of constants), so the power series converges [uniformly](/page/Uniform%20Convergence). The [uniform limit theorem](/page/Uniform%20Convergence) then gives continuity of the sum. ## Defining Transcendental Functions The term-wise calculus results justify a powerful strategy: *define* transcendental functions by their power series and *derive* their properties from the series. This avoids circular definitions and provides immediate analyticity. ### The Exponential Function [definition:Exponential Function Via Power Series] The **exponential function** is defined by \begin{align*} \exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \end{align*} which has radius of convergence $R = \infty$. [/definition] Term-wise differentiation gives $\exp'(x) = \sum_{n=1}^{\infty} x^{n-1}/(n-1)! = \exp(x)$. So $\exp$ is the unique solution to $f' = f$ with $f(0) = 1$. The functional equation $\exp(x + y) = \exp(x) \exp(y)$ follows from the Cauchy product of absolutely convergent series. We define $e = \exp(1)$ and write $e^x = \exp(x)$. ### Sine and Cosine [definition:Sine And Cosine Via Power Series] \begin{align*} \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \qquad \cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}. \end{align*} Both have $R = \infty$. [/definition] Term-wise differentiation yields $\sin' x = \cos x$ and $\cos' x = -\sin x$, with $\sin 0 = 0$ and $\cos 0 = 1$. The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ follows by differentiating the left side (getting $2\sin x \cos x - 2\cos x \sin x = 0$) and evaluating at $x = 0$. Euler's formula $e^{ix} = \cos x + i \sin x$ emerges by substituting $ix$ into the exponential series and separating real and imaginary parts — a computation that unifies the exponential and trigonometric functions within the complex numbers. ### The Natural Logarithm [definition:Natural Logarithm Via Power Series] For $|x| < 1$: \begin{align*} \log(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \end{align*} with $R = 1$. The series also converges at $x = 1$ (alternating harmonic series), giving $\log 2 = 1 - 1/2 + 1/3 - 1/4 + \cdots$. [/definition] Term-wise differentiation gives $\frac{d}{dx} \log(1 + x) = \sum_{n=0}^{\infty} (-x)^n = 1/(1 + x)$ for $|x| < 1$, confirming the standard derivative. The full logarithm on $(0, \infty)$ is constructed by analytic continuation using the functional equation $\log(ab) = \log a + \log b$. ## Analyticity and the Identity Theorem A function $f: I \to \mathbb{R}$ on an open interval $I$ is **analytic** at $c \in I$ if it equals a convergent power series in some neighbourhood of $c$: $f(x) = \sum a_n (x - c)^n$ for $|x - c| < \rho$. It is **analytic on $I$** if analytic at every point. Every polynomial, every rational function (away from poles), and the functions $\exp$, $\sin$, $\cos$, $\log$ are analytic on their natural domains. Analyticity is strictly stronger than infinite differentiability — every analytic function is $C^\infty$, but the converse fails. [example:Smooth But Not Analytic] Define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = e^{-1/x^2}$ for $x \neq 0$ and $f(0) = 0$. This function is $C^\infty$ with $f^{(k)}(0) = 0$ for all $k$ (because $e^{-1/x^2}$ decays faster than any power of $|x|$ as $x \to 0$). Its Taylor series at $0$ is identically zero — yet $f(x) > 0$ for all $x \neq 0$. The Taylor series converges everywhere but does not converge to $f$: it "misses" the function's nonzero values entirely. This example shows that having all derivatives at a point is not enough for analyticity. The Taylor series captures the *jet* (the sequence of derivatives) perfectly, but the function may have behaviour invisible to any finite-order jet — a phenomenon impossible for analytic functions. [/example] The rigidity of analytic functions is captured by the identity theorem, which asserts that local agreement forces global agreement. [quotetheorem:208] The proof has two ingredients. First, if $f$ and $g$ agree on a sequence converging to $p$, then all their derivatives agree at $p$ — because any nonzero Taylor coefficient would make $f - g$ behave like $(x - p)^m$ near $p$ for some $m$, contradicting the infinitely many zeros. Second, the [set](/page/Set) where all derivatives agree is both open (power series convergence provides a neighbourhood) and closed (continuity of derivatives) in the connected interval $I$, so it equals all of $I$ — a [topological](/page/Topology) "domino effect" that propagates local agreement globally. The identity theorem has a striking consequence: the coefficients of a power series are **unique**. If $\sum a_n (x - c)^n = \sum b_n (x - c)^n$ on any open interval containing $c$, then $a_n = b_n$ for all $n$. This justifies formal manipulations: multiplying, composing, or inverting power series term by term produces the *only possible* coefficients. ## The Complex Perspective Although this page begins with real power series, the deepest understanding of convergence, coefficients, and singularities comes from complex analysis. In $\mathbb{C}$, a power series $\sum a_n (z - c)^n$ converges in a *disk* $|z - c| < R$ rather than an interval, and the radius of convergence $R$ equals the **distance from $c$ to the nearest singularity** of the complex function defined by the series — a geometric fact invisible from the real line alone. ### Why The Radius Is Geometric [example:Why The Geometric Series Has $R = 1$] The series $\sum z^n$ converges to $1/(1 - z)$. The function $1/(1 - z)$ has a pole at $z = 1$, distance $1$ from the centre $c = 0$. Hence $R = 1$. The radius "knows" about the singularity — even if we restrict to the real line, where $1/(1-x)$ looks innocuous on $(-1, 1)$. [/example] [example:A Real-Analytic Mystery Explained] The function $f(x) = 1/(1 + x^2)$ is smooth on all of $\mathbb{R}$ with no apparent obstruction. Yet its Taylor series at $c = 0$ has radius $R = 1$. Why? In $\mathbb{C}$, $f(z) = 1/(1 + z^2)$ has poles at $z = \pm i$, distance $1$ from the origin. The real function is smooth, but the complex extension reveals hidden singularities that limit the radius. Without complex analysis, $R = 1$ would seem arbitrary; with it, the radius is geometrically inevitable. [/example] Functions with $R = \infty$ (like $\exp$, $\sin$, $\cos$) are **entire**: they have no singularities anywhere in $\mathbb{C}$. Functions with finite $R$ have at least one singularity on the circle $|z - c| = R$. ### Cauchy's Integral Formula for Coefficients The most profound result connecting power series to complex integration is Cauchy's formula: if $f(z) = \sum a_n (z - c)^n$ converges in $|z - c| < R$, then the coefficients are given by contour [integrals](/page/Integral): \begin{align*} a_n = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - c)^{n+1}} \, dz, \end{align*} where $\gamma$ is any positively oriented simple closed contour encircling $c$ within the disk of convergence. This formula reveals that the coefficients are determined by the *global* behaviour of $f$ on a contour — not just by derivatives at a single point. The derivation is elegant: substitute the series $f(z) = \sum a_k (z - c)^k$ into the integral and exchange summation and integration (justified by uniform convergence on compact subsets): \begin{align*} \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z-c)^{n+1}} \, dz = \sum_{k=0}^{\infty} a_k \cdot \frac{1}{2\pi i} \oint_\gamma (z - c)^{k - n - 1} \, dz. \end{align*} The integral $\frac{1}{2\pi i} \oint_\gamma (z - c)^m \, dz$ equals $1$ when $m = -1$ (i.e., $k = n$) and $0$ otherwise — this is the fundamental orthogonality of the powers $(z - c)^m$ on a circle. So the sum collapses to $a_n$. Combined with $a_n = f^{(n)}(c)/n!$ (from term-wise differentiation), Cauchy's formula gives the remarkable identity \begin{align*} f^{(n)}(c) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - c)^{n+1}} \, dz, \end{align*} which computes derivatives of *any* order from a single contour integral. This is why holomorphic functions are infinitely differentiable — a single complex derivative implies all of them, because the integral formula provides them all at once. ### Cauchy's Inequality and the Cauchy-Hadamard Formula Taking $\gamma$ to be the circle $|z - c| = r$ with $0 < r < R$ and bounding the integral: \begin{align*} |a_n| = \left|\frac{1}{2\pi i} \oint_{|z-c|=r} \frac{f(z)}{(z-c)^{n+1}} \, dz\right| \leq \frac{1}{2\pi} \cdot \frac{M(r)}{r^{n+1}} \cdot 2\pi r = \frac{M(r)}{r^n}, \end{align*} where $M(r) = \max_{|z-c|=r} |f(z)|$. This is **Cauchy's inequality**: $|a_n| \leq M(r)/r^n$ for every $r < R$. Cauchy's inequality is the *source* of the Cauchy-Hadamard formula. It shows that the coefficients cannot grow faster than $M(r)/r^n$ for any $r < R$, which forces $\limsup |a_n|^{1/n} \leq 1/r$ for every $r < R$, giving $\limsup |a_n|^{1/n} \leq 1/R$. The reverse inequality (divergence for $|z - c| > R$) completes the proof. Cauchy's inequality also gives **Liouville's theorem** as an immediate corollary: if $f$ is entire ($R = \infty$) and bounded ($|f(z)| \leq M$ for all $z$), then $|a_n| \leq M/r^n$ for *every* $r > 0$, so $a_n = 0$ for $n \geq 1$, and $f$ is constant. A bounded entire function is constant — one of the most striking rigidity results in all of mathematics, from which the [Fundamental Theorem of Algebra](/theorems/347) follows. ### Laurent Series A power series $\sum_{n=0}^{\infty} a_n (z - c)^n$ represents a function that is holomorphic in a disk — it has no singularities inside its region of convergence. But many important functions *have* singularities: $1/z$ at $z = 0$, $\cot z$ at $z = k\pi$, $e^{1/z}$ at $z = 0$. To represent such functions by series, we need **negative powers** of $(z - c)$. [definition:Laurent Series] A **Laurent series** centred at $c$ is a doubly infinite series \begin{align*} \sum_{n=-\infty}^{\infty} a_n (z - c)^n = \cdots + \frac{a_{-2}}{(z-c)^2} + \frac{a_{-1}}{z - c} + a_0 + a_1(z-c) + a_2(z-c)^2 + \cdots \end{align*} The series converges in an **annulus** $r < |z - c| < R$, where $0 \leq r < R \leq \infty$. The non-negative powers $\sum_{n \geq 0} a_n (z - c)^n$ converge for $|z - c| < R$ (the "analytic part") and the negative powers $\sum_{n < 0} a_n (z - c)^n$ converge for $|z - c| > r$ (the "principal part"). [/definition] The Laurent series is unique: if $f$ is holomorphic in the annulus $r < |z - c| < R$, there is exactly one Laurent series converging to $f$ there, with coefficients given by \begin{align*} a_n = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - c)^{n+1}} \, dz \end{align*} for any positively oriented circle $\gamma$ in the annulus — the same formula as for Taylor coefficients, now valid for *all* integers $n$. [example:Laurent Series Of $1/z(1-z)$] The function $f(z) = 1/(z(1-z))$ has singularities at $z = 0$ and $z = 1$. In the annulus $0 < |z| < 1$ (between the two singularities): \begin{align*} f(z) = \frac{1}{z} \cdot \frac{1}{1 - z} = \frac{1}{z} \sum_{n=0}^{\infty} z^n = \sum_{n=0}^{\infty} z^{n-1} = \frac{1}{z} + 1 + z + z^2 + \cdots \end{align*} The principal part is $1/z$ (a single negative-power term); the analytic part is the geometric series. In the annulus $|z| > 1$, a different Laurent expansion holds: \begin{align*} f(z) = \frac{1}{z} \cdot \frac{-1}{z(1 - 1/z)} = \frac{-1}{z^2} \sum_{n=0}^{\infty} z^{-n} = -\frac{1}{z^2} - \frac{1}{z^3} - \frac{1}{z^4} - \cdots \end{align*} The Laurent expansion depends on the annulus — the same function can have different Laurent series in different regions. [/example] ### The Residue The coefficient $a_{-1}$ in a Laurent series has a distinguished role: it is the only coefficient that contributes to a contour integral of $f$ around the singularity. [definition:Residue] If $f$ has an isolated singularity at $c$ with Laurent series $\sum a_n (z - c)^n$ in some punctured disk $0 < |z - c| < R$, the **residue** of $f$ at $c$ is \begin{align*} \operatorname{Res}(f, c) = a_{-1} = \frac{1}{2\pi i} \oint_\gamma f(z) \, dz, \end{align*} where $\gamma$ is any small positively oriented circle around $c$. [/definition] The residue is the only Laurent coefficient that "survives" integration: $\oint (z - c)^n \, dz = 0$ for $n \neq -1$ (since $(z - c)^n$ has an antiderivative $(z - c)^{n+1}/(n+1)$ for $n \neq -1$), but $\oint (z - c)^{-1} \, dz = 2\pi i$ (the winding number integral). This is why the residue controls all contour integrals — and why the **Residue Theorem** reduces contour integration to the purely algebraic problem of extracting $a_{-1}$ from a Laurent expansion. ### Classification of Singularities The Laurent series classifies isolated singularities into three types, each with sharply different behaviour: **Removable singularities** have no principal part ($a_n = 0$ for all $n < 0$). The function extends to a holomorphic function at $c$ by defining $f(c) = a_0$. Example: $\sin(z)/z$ at $z = 0$ has Laurent series $1 - z^2/6 + \cdots$ with no negative powers. **Poles** have a finite principal part ($a_{-m} \neq 0$ for some $m \geq 1$, $a_n = 0$ for $n < -m$). The integer $m$ is the **order** of the pole. Near a pole of order $m$, $|f(z)| \to \infty$ like $|z - c|^{-m}$. Example: $1/z^2$ at $z = 0$ is a pole of order $2$ with residue $0$. **Essential singularities** have an infinite principal part ($a_n \neq 0$ for infinitely many $n < 0$). Near an essential singularity, $f$ takes every complex value (with at most one exception) infinitely often in any punctured neighbourhood — this is the **Great Picard Theorem**, the most extreme possible behaviour. Example: $e^{1/z} = \sum_{n=0}^{\infty} z^{-n}/n!$ at $z = 0$ has an infinite principal part. The type of singularity is encoded entirely in the Laurent series: removable (no negative powers), pole (finitely many), essential (infinitely many). This classification, together with the residue theorem, makes Laurent series the central computational tool of complex analysis.