Taylor's Remainder Theorem — Mathematics Knowledge System

Overview

Taylor's theorem quantifies the error made when a function f is approximated by its n-th degree Taylor polynomial Pₙ at a point a: the difference f(x) − Pₙ(x), called the remainder Rₙ(x), can be written explicitly in several equivalent forms. The Lagrange form expresses Rₙ(x) using the (n+1)-th derivative evaluated at some unknown point between a and x, directly generalizing the Mean Value Theorem (which is exactly the n=0 case). This remainder formula converts an approximation into a rigorous inequality: bounding the (n+1)-th derivative on the relevant interval bounds the error, which is what justifies using truncated Taylor series for numerical computation with guaranteed precision.

Intuition

A Taylor polynomial is the 'best local imitation' of f near a using only polynomial building blocks matched in value and derivatives up to order n. The remainder Rₙ(x) measures exactly how much that imitation misses by. The Lagrange form says this leftover error behaves like the NEXT term you would have added to the polynomial (order n+1), except the derivative is evaluated not at a itself but at some unknown, unspecified point c in between — this is what makes the formula exact rather than merely approximate, at the cost of not knowing precisely where c is. In practice you don't pin down c exactly; instead you bound |f⁽ⁿ⁺¹⁾| by some constant M on the whole interval, which bounds the error by M|x-a|ⁿ⁺¹/(n+1)!, and that's usually all that's needed to certify a desired accuracy.

Formal Definition

Definition

Let f be (n+1)-times differentiable on an interval containing a and x. Define the n-th Taylor polynomial and remainder:

P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k, \qquad f(x) = P_n(x) + R_n(x)

Taylor polynomial and remainder

R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \quad \text{for some } c \text{ strictly between } a \text{ and } x

Lagrange form of the remainder

n = 0: \quad f(x) = f(a) + f'(c)(x-a) \quad \text{(exactly the Mean Value Theorem)}

MVT as the n=0 special case

Worked Examples

The remainder uses the 3rd derivative; for f(x)=eˣ, f'''(x)=eˣ for all x.
$R_2(x) = \frac{e^c}{3!}x^3 \text{ for some } c \in (0,x)$
On [0,1], since eˣ is increasing, e^c ≤ e^1 = e < 3.
$e^c < 3 \text{ for } c \in (0,1)$
Bound the remainder: |R₂(x)| < (3/6)·1³ = 1/2 for x ∈ [0,1].
$|R_2(x)| < \frac{3}{3!}\cdot 1^3 = \frac{1}{2}$

Answer: The error of the degree-2 Taylor approximation to eˣ on [0,1] is bounded by 1/2 (a loose but valid, rigorous bound).

Practice Problems

Difficulty 4/10

Write the Lagrange remainder R₃(x) for f(x) = cos(x) expanded at a=0.

Difficulty 3/10

Taylor's theorem with n=0 (the Lagrange remainder for the 0th degree polynomial) is equivalent to:

The Intermediate Value TheoremThe Mean Value TheoremRolle's theorem onlyL'Hôpital's rule

Difficulty 7/10

Show that if |f⁽ⁿ⁺¹⁾(t)| ≤ M for all t between a and x, then |Rₙ(x)| ≤ M|x-a|ⁿ⁺¹/(n+1)!.

Quiz

The Lagrange form of the Taylor remainder Rₙ(x) involves:

The n-th derivative evaluated at aThe (n+1)-th derivative evaluated at some unknown point c between a and xAn integral that cannot be simplifiedOnly the function values, not derivatives

Taylor's theorem with n=0 reduces to which classical theorem?

Rolle's theoremThe Mean Value TheoremThe Fundamental Theorem of CalculusThe Squeeze Theorem

Summary

Taylor's theorem writes f(x) = Pₙ(x) + Rₙ(x), where Pₙ is the n-th Taylor polynomial and Rₙ is the remainder (error).
The Lagrange form gives Rₙ(x) = f⁽ⁿ⁺¹⁾(c)(x-a)ⁿ⁺¹/(n+1)! for some unknown c between a and x, generalizing the MVT (its n=0 case).
Bounding |f⁽ⁿ⁺¹⁾| on the interval by a constant M yields the practical error bound |Rₙ(x)| ≤ M|x-a|ⁿ⁺¹/(n+1)!, used to certify numerical approximation accuracy.

References

WebsiteWikipedia — Taylor's theorem