Taylor's Theorem and Error Bounds

⚡ Key Takeaways

Taylor's Theorem states that a sufficiently differentiable function can be approximated near a point by a polynomial built from its derivatives, with an explicit, controllable error term (the remainder). The key result: $f(x) = P_n(x) + R_n(x)$ , where $P_n$ is the degree- $n$ Taylor polynomial and $R_n$ is bounded by the size of the $(n{+}1)$ -th derivative. It applies wherever local polynomial approximation, error estimation, or series expansion is needed — numerical analysis, physics, engineering. Named for Brook Taylor (1715), with the remainder forms due to Lagrange, Cauchy, and Schlömilch (18th–19th c.).

💡 The Core Insight

The value, slope, curvature, and all higher derivatives of $f$ at a single point $a$ encode enough information to reconstruct $f$ approximately throughout a neighborhood. Matching the first $n+1$ derivatives at $a$ forces the approximating polynomial to "hug" the curve, and the leftover error is governed entirely by the next derivative — the one we did not match. The remainder is not a vague fudge factor; it is a genuine equality of the form $R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ for some unknown $\xi$ between $a$ and $x$ .

DefinitionTaylor Polynomial

Let $f$ be $n$ times differentiable at $a$ . The Taylor polynomial of degree $n$ 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) + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n.

Derivation hint: it is the unique degree- $\le n$ polynomial $P$ satisfying $P^{(k)}(a)=f^{(k)}(a)$ for $k=0,\dots,n$ ; differentiate the sum $k$ times and evaluate at $x=a$ — only the $k$ -th term survives. Here $f^{(k)}(a)$ is the $k$ -th derivative of $f$ at $a$ , $k!$ is the factorial, and $(x-a)^k$ measures displacement from the base point $a$ . When $a=0$ this is called the Maclaurin polynomial. The remainder is defined by $R_n(x) = f(x) - P_n(x)$ .

TheoremTaylor's Theorem with Lagrange Remainder

Let $f:[a,x]\to\mathbb{R}$ (or $[x,a]$ ) be such that $f, f', \dots, f^{(n)}$ are continuous on the closed interval and $f^{(n+1)}$ exists on the open interval. Then there exists a point $\xi$ strictly between $a$ and $x$ such that

2

\quad f(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}.

Derivation hint: apply the Cauchy Mean Value Theorem (or repeated Rolle's Theorem) to an auxiliary function chosen so its zeros pin down $\xi$ ; see the proof below. The final term is the Lagrange form of the remainder, $R_n(x) = \dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ .

DefinitionOther Forms of the Remainder

For $f^{(n+1)}$ continuous, the integral (exact) form is

3

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

Derivation hint: start from $f(x)-f(a)=\int_a^x f'(t)\,dt$ and integrate by parts repeatedly, each time differentiating the $(x-t)$ power and integrating the derivative factor. The Cauchy form (for some $\theta\in(0,1)$ , with $\xi=a+\theta(x-a)$ ) is

4

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

Derivation hint: apply the ordinary Mean Value Theorem (not Cauchy's generalized one) to the integral form, or use the auxiliary function with a linear weight instead of a power weight.

DefinitionPeano Remainder — Qualitative Form

If only $f^{(n)}(a)$ exists (no $(n{+}1)$ -th derivative required), then

5

\quad f(x) = P_n(x) + o\big((x-a)^n\big) \quad\text{as } x\to a,

Derivation hint: subtract $P_n$ , apply L'Hôpital's rule $n$ times to $R_n(x)/(x-a)^n$ , using the matching of all derivatives up to order $n$ . meaning $\lim_{x\to a} R_n(x)/(x-a)^n = 0$ . This is the weakest, purely local statement and underlies asymptotic analysis.

ProofLagrange Remainder via Auxiliary Function

Goal: Find $\xi$ between $a$ and $x$ with $R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ . Fix $x$ and treat $a$ (renamed $t$ ) as the variable. 1. Define an auxiliary function. For $t$ in the closed interval, set

g(t) = f(x) - \sum_{k=0}^{n}\frac{f^{(k)}(t)}{k!}(x-t)^k.

Then $g(x) = 0$ (only the $k=0$ term survives, giving $f(x)-f(x)$ ) and $g(a) = R_n(x)$ . 2. Differentiate $g$ . Each term's derivative telescopes by the product rule:

g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x-t)^n.

All intermediate terms cancel pairwise — this is the engine of the proof. 3. Define a comparison function. Let $h(t) = \left(\dfrac{x-t}{x-a}\right)^{n+1}$ , so $h(a)=1$ , $h(x)=0$ , and $h'(t) = -\dfrac{(n+1)(x-t)^n}{(x-a)^{n+1}}$ . 4. Apply Cauchy's Mean Value Theorem to $g$ and $h$ on $[a,x]$ : there is $\xi$ between $a$ and $x$ with

\frac{g(x)-g(a)}{h(x)-h(a)} = \frac{g'(\xi)}{h'(\xi)}.

5. Substitute the endpoint values. The left side is $\dfrac{0 - R_n(x)}{0 - 1} = R_n(x)$ . The right side simplifies:

\frac{g'(\xi)}{h'(\xi)} = \frac{-\frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^n}{-\frac{(n+1)(x-\xi)^n}{(x-a)^{n+1}}} = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}.

6. Conclude. Equating both sides gives the Lagrange remainder. $\blacksquare$

∎

mathematician · Brook Taylor | 1685–1731

An English mathematician and member of the Royal Society, Taylor published the formula bearing his name in his 1715 treatise Methodus Incrementorum Directa et Inversa. He derived it as a limit of Newton's interpolation formula and used it to study vibrating strings and the motion of projectiles — though without any notion of convergence or remainder. The crucial error term came a generation later from Joseph-Louis Lagrange (1797), and the rigorous treatment with explicit remainders is largely the work of Cauchy in the 1820s. Taylor also founded the mathematical theory of linear perspective and contributed to the calculus of finite differences.

🧠 Intuition

Think of building an approximation one derivative at a time. Start with a flat horizontal line at height $f(a)$ — it gets the value right but ignores slope. Add a tilt $f'(a)(x-a)$ — now the tangent line matches direction. Add curvature $\frac{f''(a)}{2}(x-a)^2$ — now you bend the right way. Each new term corrects the highest unmatched derivative, and the error you have left is dominated by the first derivative you stopped matching, scaled by $(x-a)^{n+1}/(n+1)!$ . Two forces fight in the remainder: the factorial in the denominator crushes the error (it grows astronomically), while $(x-a)^{n+1}$ explodes if $|x-a|>1$ . Near the base point, the factorial wins decisively, which is why Taylor polynomials are local tools — superb close to $a$ , often useless far away.

Figure

Degree-1, -2, and -3 Taylor approximations of a curve at base point

a

, with the error (remainder) shown as the vertical gap.

ExampleEstimating

e^{0.1}

with a Maclaurin Polynomial

Given: Approximate $e^{0.1}$ using the degree-2 Maclaurin polynomial of $f(x)=e^x$ , and bound the error.

Setup. Here $a=0$ , $x=0.1$ . Since $f^{(k)}(x)=e^x$ , all derivatives at $0$ equal $1$ :

P_2(x) = 1 + x + \frac{x^2}{2}.

Compute the approximation.

P_2(0.1) = 1 + 0.1 + \frac{0.01}{2} = 1.105.

Bound the error (Lagrange). With $n=2$ ,

R_2(0.1) = \frac{f^{(3)}(\xi)}{3!}(0.1)^3 = \frac{e^{\xi}}{6}(0.001), \quad 0<\xi<0.1.

Since $e^\xi < e^{0.1} < 1.2$ ,

|R_2(0.1)| < \frac{1.2}{6}(0.001) = 0.0002.

Answer.

\boxed{e^{0.1} \approx 1.105, \quad \text{error} < 2\times 10^{-4}}

(The true value $e^{0.1}=1.10517\ldots$ , so the actual error is $\approx 1.7\times10^{-4}$ — consistent with the bound.) Units are dimensionless.

ExampleHow Many Terms for

\sin(0.5)

to

10^{-6}

?

Given: Find the degree $n$ of the Maclaurin polynomial of $\sin x$ needed to compute $\sin(0.5)$ with error below $10^{-6}$ .

Setup. For $\sin x$ , every derivative satisfies $|f^{(n+1)}(\xi)|\le 1$ . The Lagrange bound is

|R_n(0.5)| \le \frac{|0.5|^{\,n+1}}{(n+1)!}.

Test values.

$n+1$	$(0.5)^{n+1}/(n+1)!$
4	$0.0625/24 \approx 2.6\times10^{-3}$
6	$0.015625/720 \approx 2.2\times10^{-5}$
8	$0.00390625/40320 \approx 9.7\times10^{-8}$

So $n+1=8$ , i.e. $n=7$ , gives error $<10^{-6}$ . (Because $\sin$ has only odd-power terms, this means using terms up to $x^7$ .)

Compute.

P_7(0.5) = 0.5 - \frac{0.5^3}{6} + \frac{0.5^5}{120} - \frac{0.5^7}{5040} \approx 0.479425539.

Answer.

\boxed{n=7 \text{ suffices}; \quad \sin(0.5)\approx 0.4794255, \quad \text{error} < 10^{-6}}

(True value $\sin 0.5 = 0.4794255386\ldots$ .)

⚠️ Common mistake

"Infinitely differentiable" does NOT imply "equals its Taylor series." Many students assume that if all derivatives exist, the Taylor series converges to the function. Convergence of the series and the series summing to $f$ are separate questions — the remainder $R_n(x)$ must tend to $0$ as $n\to\infty$ , and this can fail even when the series converges to a different value. Also beware: the point $\xi$ in the Lagrange remainder is unknown — you do not get to choose it. You can only bound $|f^{(n+1)}|$ over the whole interval to get a usable error estimate.

CounterexampleA Smooth Function Whose Taylor Series Is Useless

Consider

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

This function is $C^\infty$ (infinitely differentiable) everywhere, and one can show by induction that $f^{(k)}(0) = 0$ for every $k$ . Therefore its Maclaurin series is identically $0$ :

\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k = 0 + 0\cdot x + 0\cdot x^2 + \cdots = 0.

The series converges everywhere — but to the zero function, not to $f$ , which is positive for all $x\neq 0$ . The remainder $R_n(x) = f(x)$ never shrinks. This shows smoothness is not enough; such non-analytic functions are central to the study of Analytic Functions.

connection

Taylor's Theorem sits at the junction of several core ideas. It generalizes the Mean Value Theorem — indeed the case $n=0$ of equation [2] is the Mean Value Theorem $f(x)=f(a)+f'(\xi)(x-a)$ . Letting $n\to\infty$ leads to Taylor Series and the theory of Power Series, where the radius of convergence governs validity. The integral remainder [3] is a direct consequence of Integration by Parts and the Fundamental Theorem of Calculus. In several variables it generalizes to expansions involving the Hessian Matrix and higher tensors, foundational to Optimization. The Peano form [5] is the bridge to Asymptotic Expansions and the theory of Big-O Notation used throughout numerical analysis.

🔬 Where this appears

⏳ History

- 1671 — James Gregory discovers special cases of Taylor series (for

\arctan

,

\tan

,

\sec

) decades before Taylor, unpublished in full.

- 1715 — Brook Taylor publishes the general formula in Methodus Incrementorum, derived via finite differences, with no remainder or convergence theory.

- 1742 — Colin Maclaurin popularizes the special case

a=0

in his Treatise of Fluxions; the name "Maclaurin series" sticks despite his crediting Taylor.

- 1772 — Lagrange begins promoting Taylor expansions as the foundation of calculus, hoping to eliminate limits and infinitesimals.

- 1797 — Lagrange introduces the explicit remainder term in Théorie des fonctions analytiques, giving the Lagrange form.

- 1821 — Augustin-Louis Cauchy, in Cours d'analyse and later works, supplies rigorous definitions of limits, derives the Cauchy form of the remainder, and exposes that the series need not equal the function.

- 1823 — Cauchy publishes the

e^{-1/x^2}

counterexample, demolishing Lagrange's program of founding calculus on Taylor series alone.

- 1880s — Oskar Schlömilch & Roché give the general (parametrized) remainder form unifying the Lagrange and Cauchy versions.

- 20th c. — Giuseppe Peano formalizes the qualitative

o((x-a)^n)

remainder, central to modern asymptotic analysis.

open-problem

The boundary between "has a Taylor series" and "equals its Taylor series" defines the class of real-analytic functions, and characterizing this class intrinsically is subtle. A deep result (Pringsheim, Borel) shows that for any prescribed sequence of real numbers $\{a_k\}$ — no matter how wildly it grows — there exists a $C^\infty$ function with $f^{(k)}(0)=a_k$ (Borel's theorem). Thus the Taylor coefficients place essentially no constraint on a smooth function. Quantifying how a smooth-but-non-analytic function's remainder behaves leads to the theory of Gevrey classes and quasi-analytic functions (Denjoy–Carleman theorem), which gives sharp conditions on the growth of $|f^{(n)}|$ guaranteeing that a function is determined by its derivatives at a point. These questions remain active in PDE regularity theory and the study of divergent asymptotic series, where optimal truncation ("superasymptotics") squeezes useful accuracy out of series that never converge.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also