Taylor Series

⚡ Key Takeaways

Taylor series represents a function as an infinite sum of polynomial terms built from its derivatives at a single point. Key result: if $f$ is smooth and well-behaved near $a$ , then $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$ . Applies to approximation of functions, numerical methods, solving ODEs, physics expansions, and proving identities. Who/when: named for Brook Taylor (1715), with key contributions from Maclaurin, Gregory, and rigorous foundations by Lagrange and Cauchy.

💡 The Core Insight

A function's behavior near a point is encoded entirely in its derivatives at that point. If you know $f(a)$ , $f'(a)$ , $f''(a)$ , and so on, you have captured the value, slope, curvature, and all higher-order bending of the curve at $a$ . Reassembling these into a polynomial $\sum \frac{f^{(n)}(a)}{n!}(x-a)^n$ recovers the function — provided the "error" left after finitely many terms shrinks to zero. The factorial denominators are exactly what makes the $n$ -th derivative of the series at $a$ reproduce $f^{(n)}(a)$ .

DefinitionTaylor Series and Taylor Polynomial

Let $f$ be a function that is $n$ times differentiable at a point $a$ . The $n$ -th Taylor polynomial of $f$ centered at $a$ is

1

\quad P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n.

Derivation hint: demand that $P_n$ match $f$ in value and in all derivatives up to order $n$ at $a$ ; differentiating $P_n$ $k$ times and evaluating at $a$ kills all but the $k$ -th term, forcing coefficient $c_k = f^{(k)}(a)/k!$ . If $f$ is infinitely differentiable at $a$ , the Taylor series is the formal infinite sum

2

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

Derivation hint: take the limit $n \to \infty$ in [1]; convergence and equality with $f$ are separate questions addressed below.

Symbols: $f$ is the function; $a$ the center (or base point); $f^{(k)}$ the $k$ -th derivative; $k!$ the factorial; $(x-a)$ the displacement from the center. When $a=0$ the series is called a Maclaurin series:

3

\quad f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n.

Derivation hint: set $a=0$ in [2].

TheoremTaylor's Theorem with Remainder

Let $f$ be $(n+1)$ times differentiable on an open interval containing $a$ and $x$ . Then

4

\quad f(x) = P_n(x) + R_n(x),

where the remainder $R_n(x)$ admits the Lagrange form

5

\quad R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}

for some $\xi$ strictly between $a$ and $x$ . The Taylor series converges to $f(x)$ if and only if $R_n(x) \to 0$ as $n \to \infty$ . Derivation hint: apply the Cauchy Mean Value Theorem repeatedly, or use the integral form of the remainder and the Mean Value Theorem for integrals; $\xi$ is the unattained intermediate point.

ProofTaylor's Theorem via Repeated Integration by Parts

We establish [4]–[6]. 1. Base case (FTC). For $f$ continuously differentiable,

f(x) = f(a) + \int_a^x f'(t)\,dt.

This is the $n=0$ statement with $P_0(x)=f(a)$ and $R_0(x)=\int_a^x f'(t)\,dt$ . 2. Integrate by parts. In $\int_a^x f'(t)\,dt$ , choose $u=f'(t)$ , $dv=dt$ , but with the clever antiderivative $v=-(x-t)$ (constant of integration chosen to vanish at $t=x$ ). Then

\int_a^x f'(t)\,dt = \big[-(x-t)f'(t)\big]_a^x + \int_a^x (x-t)f''(t)\,dt = (x-a)f'(a) + \int_a^x (x-t)f''(t)\,dt.

3. Iterate. Repeating with $v = -\frac{(x-t)^{k+1}}{(k+1)}$ at each stage produces, after $n$ integrations by parts,

f(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{1}{n!}\int_a^x (x-t)^n f^{(n+1)}(t)\,dt,

which is exactly $P_n(x) + R_n(x)$ with $R_n$ in the integral form [6]. 4. From integral to Lagrange form. Since $(x-t)^n$ does not change sign on $[a,x]$ , the Mean Value Theorem for integrals gives a $\xi$ between $a$ and $x$ with

R_n(x) = \frac{f^{(n+1)}(\xi)}{n!}\int_a^x (x-t)^n\,dt = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1},

recovering [5]. $\blacksquare$

∎

mathematician · Brook Taylor | 1685–1731

An English mathematician and a Fellow of the Royal Society, Taylor published the general expansion in his 1715 Methodus Incrementorum Directa et Inversa, the founding text of the calculus of finite differences. He derived the series essentially as a limit of Newton's interpolation formula. Taylor did not address convergence — that rigor came a century later from Cauchy and Lagrange. He also made early contributions to perspective in art and to the physics of vibrating strings. Despite the eponym, special cases were known earlier to James Gregory, Newton, and the Indian Kerala school (Madhava) centuries before.

🧠 Intuition

Think of building a clone of $f$ that agrees with the original ever more closely. The constant term $f(a)$ matches height. Adding $f'(a)(x-a)$ matches slope — now your clone is the tangent line. Adding $\frac{f''(a)}{2}(x-a)^2$ matches curvature — the clone now bends the right way. Each new term pins down one more derivative, so the polynomial "hugs" $f$ over a wider neighborhood. The factorial $1/n!$ is the price of forcing the $n$ -th derivative to come out right: differentiating $(x-a)^n$ exactly $n$ times yields $n!$ , which the denominator cancels. Convergence is the question of whether infinitely many corrections ever catch up to the whole function — and the answer depends on how fast the higher derivatives grow.

Figure

Successive Taylor polynomials of sin(x) about 0 hugging the curve over a widening interval

ExampleMaclaurin series of

e^x

and a numerical estimate

Given: $f(x)=e^x$ , center $a=0$ . Estimate $e^{0.5}$ using $P_4$ and bound the error.

Steps: 1. All derivatives satisfy $f^{(n)}(x)=e^x$ , so $f^{(n)}(0)=1$ for every $n$ . 2. The Maclaurin series is $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots$ 3. Evaluate $P_4$ at $x=0.5$ :

P_4(0.5) = 1 + 0.5 + \tfrac{0.25}{2} + \tfrac{0.125}{6} + \tfrac{0.0625}{24} = 1 + 0.5 + 0.125 + 0.0208\overline{3} + 0.0026042 = 1.6484375.

4. Error bound via Lagrange [5]: $|R_4| = \frac{e^{\xi}}{5!}(0.5)^5$ with $0<\xi<0.5$ , so $e^{\xi}<e^{0.5}<2$ . Hence

|R_4| < \frac{2}{120}(0.03125) = 0.00052.

Answer: $\boxed{e^{0.5}\approx 1.64844,\quad \text{error}<5.2\times10^{-4}}$ (true value $1.648721\ldots$ , units: dimensionless). ✓

ExampleTaylor expansion of

\ln x

about

a=1

and its interval of validity

Given: $f(x)=\ln x$ , center $a=1$ . Find the series and the radius within which it converges.

Steps: 1. Derivatives: $f'(x)=x^{-1}$ , $f''(x)=-x^{-2}$ , $f'''(x)=2x^{-3}$ , generally $f^{(n)}(x)=(-1)^{n-1}(n-1)!\,x^{-n}$ for $n\ge 1$ . 2. At $a=1$ : $f(1)=0$ , $f^{(n)}(1)=(-1)^{n-1}(n-1)!$ . 3. Coefficient: $\frac{f^{(n)}(1)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n!} = \frac{(-1)^{n-1}}{n}$ . Thus

\ln x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots

4. Ratio test: $\lim_{n\to\infty}\left|\frac{(x-1)^{n+1}/(n+1)}{(x-1)^n/n}\right| = |x-1|<1$ , giving radius $R=1$ , i.e. $0<x<2$ . At $x=2$ it converges (alternating harmonic $\to \ln 2$ ); at $x=0$ it diverges.

Answer: $\boxed{\ln x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n,\ \ 0<x\le 2}$ (dimensionless). ✓

⚠️ Common mistake

A function being $C^\infty$ (infinitely differentiable) does NOT guarantee it equals its Taylor series. Convergence of the series is necessary, and even then the limit need not be $f$ . Always check that the remainder $R_n(x)\to 0$ . Also: the Taylor series may converge only on a small interval (its radius of convergence) even when $f$ is defined everywhere — see $\arctan x$ or $\frac{1}{1+x^2}$ , whose Maclaurin series diverges for $|x|>1$ despite the function being perfectly smooth on all of $\mathbb{R}$ . The cause is a complex-plane singularity at $x=\pm i$ , invisible on the real line.

CounterexampleThe flat function whose Taylor series is identically zero

Define

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

This function is $C^\infty$ on all of $\mathbb{R}$ , and one can show by induction that every derivative at the origin vanishes: $f^{(n)}(0)=0$ for all $n$ . Therefore its Maclaurin series is

7

\quad T(x) = \sum_{n=0}^{\infty}\frac{0}{n!}x^n = 0.

Derivation hint: each derivative is a rational function times $e^{-1/x^2}$ ; since $e^{-1/x^2}$ decays faster than any power of $x$ grows, the limit at $0$ is $0$ . The series converges everywhere (to $0$ ) but equals $f$ only at $x=0$ , since $f(x)>0$ for all $x\neq 0$ . This is the canonical example of a smooth, non-analytic function: the remainder $R_n(x)$ never tends to zero away from the origin.

connection

Taylor series are a special case of Power Series, inheriting their radius of convergence and term-by-term differentiation/integration. A function equal to its Taylor series on an open set is called real analytic, the central object of Real Analysis. Extending the center and variable into the complex plane links Taylor series to Complex Analysis, where Cauchy's theory shows every holomorphic function is automatically analytic — explaining the mysterious radii of convergence via nearest singularities. The discrete analogue is the Newton forward-difference formula of the Calculus of Finite Differences, from which Taylor originally derived his series. Truncated Taylor polynomials are the foundation of the Big-O / little-o asymptotics used in Asymptotic Analysis, and the multivariable generalization uses the gradient and Hessian Matrix.

🔬 Where this appears

⏳ History

- c. 1350 — Madhava of Sangamagrama (Kerala school) derives series for

\sin

,

\cos

, and

\arctan

, including the Madhava–Leibniz series for

\pi

— the earliest known instances of such expansions.

- 1668 — Nicholas Mercator and James Gregory publish series; Gregory finds Taylor-type expansions decades before Taylor.

- 1671 — Isaac Newton develops general series methods in De Methodis, expanding functions via his binomial theorem and reversion of series.

- 1715 — Brook Taylor states the general theorem in Methodus Incrementorum, derived from finite differences; no discussion of convergence.

- 1742 — Colin Maclaurin popularizes the

a=0

case in his Treatise of Fluxions, lending his name to that special form.

- 1772 — Joseph-Louis Lagrange introduces the remainder term and proposes building all of calculus on Taylor series (the "calculus of analytic functions").

- 1821 — Augustin-Louis Cauchy provides rigorous treatment of convergence and the remainder in his Cours d'analyse, and gives the counterexample

e^{-1/x^2}

showing smoothness ≠ analyticity.

- 1823 — Cauchy establishes the integral form of the remainder and the complex-analytic foundation linking convergence radius to singularities.

- 1840s–1850s — Weierstrass and Riemann found the modern theory of analytic functions, embedding Taylor series in Complex Analysis.

open-problem

The gap between smooth ( $C^\infty$ ) and analytic functions is profound: analytic functions form a meager subset, yet they are the ones for which Taylor series fully determine global behavior. Borel's theorem (1895) shows the contrast is total — any sequence of real numbers can serve as the derivatives $f^{(n)}(a)$ of some smooth function, so the Taylor coefficients carry essentially no constraint in the smooth category. Modern questions include: characterizing the precise growth conditions on derivatives (Denjoy–Carleman classes) that force analyticity; understanding divergent asymptotic series (e.g. in quantum field theory and the Stirling expansion) which approximate functions superbly for finitely many terms yet diverge — managed through Borel summation and resurgence theory; and the practical question of optimal polynomial approximation (Chebyshev, minimax) versus Taylor truncation, since Taylor series concentrate all accuracy at the center and may be far from optimal across an interval.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also