What is the difference between the Lagrange Error Bound and the Alternating Series Remainder?

The Alternating Series Remainder applies only to convergent alternating series and says the error is at most the absolute value of the first omitted term. The Lagrange Error Bound works for any Taylor polynomial (alternating or not) but requires you to find an upper bound $M$ on the $(n+1)$th derivative. When a Taylor series is alternating, the Alternating Series Remainder is often easier to use.

Lagrange Error Bound — Definition, Formula & Examples

The Lagrange Error Bound is the maximum possible error when you use a Taylor polynomial of degree

n

to approximate a function. It tells you how far off your polynomial approximation can be from the true value of the function.

If $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$ , then the remainder $R_n(x) = f(x) - P_n(x)$ satisfies $|R_n(x)| \leq \frac{M}{(n+1)!}|x - a|^{n+1}$ , where $M = \max|f^{(n+1)}(c)|$ for all $c$ between $a$ and $x$ . This is also called the Taylor Remainder Theorem or Lagrange Remainder.

Key Formula

|R_n(x)| \leq \frac{M}{(n+1)!}\,|x - a|^{n+1}

Where:

$R_n(x)$ = The error (remainder) when approximating f(x) by its degree-n Taylor polynomial
$M$ = An upper bound for |f^{(n+1)}(c)| on the interval between a and x
$n$ = The degree of the Taylor polynomial
$a$ = The center of the Taylor polynomial
$x$ = The point where you are approximating f

How It Works

To use the Lagrange Error Bound, you need three things: the degree

n

of your Taylor polynomial, the center

a

, and the value

x

where you are approximating. Then you find the maximum value

M

of the absolute value of the

(n+1)

th derivative on the interval between

a

and

x

. Plug these into the formula to get an upper bound on the error. On the AP Calculus BC exam, you are often asked either to compute this bound for a given

n

, or to find the smallest

n

that guarantees the error is below a specified tolerance. Finding

M

is usually the trickiest part — for functions like

\sin x

\cos x

, and

e^x

, the derivatives cycle or grow predictably, which simplifies the bound.

Worked Example

Problem: Use the Lagrange Error Bound to find the maximum error when approximating

e^1

using the 4th-degree Taylor polynomial for

e^x

centered at

a = 0

Identify the values: We have

n = 4

a = 0

, and

x = 1

. We need the 5th derivative of

e^x

f^{(5)}(x) = e^x

Find M: We need the maximum of

|e^c|

for

c

between 0 and 1. Since

e^x

is increasing, the maximum occurs at

c = 1

. We use

M = e^1 \approx 3

(a common AP technique is to overestimate with

M = 3

M = e^1 \leq 3

Apply the formula: Substitute into the Lagrange Error Bound formula.

|R_4(1)| \leq \frac{3}{5!}\cdot|1-0|^5 = \frac{3}{120} = 0.025

Interpret: The 4th-degree Taylor polynomial for

e^x

a=0

approximates

e^1

with an error of at most

0.025

Answer: The maximum error is at most

0.025

Another Example

Problem: Find the smallest

n

so that the

n

th-degree Maclaurin polynomial for

\sin x

approximates

\sin(0.5)

with error less than

0.0001

Set up the bound: All derivatives of

\sin x

satisfy

|f^{(n+1)}(c)| \leq 1

, so

M = 1

. With

a = 0

and

x = 0.5

, we need:

\frac{1}{(n+1)!}\,(0.5)^{n+1} < 0.0001

Test n = 3: Substitute

n = 3

\frac{(0.5)^4}{4!} = \frac{0.0625}{24} \approx 0.0026

Test n = 5: Substitute

n = 5

\frac{(0.5)^6}{6!} = \frac{0.015625}{720} \approx 0.0000217

Conclusion: Since

0.0000217 < 0.0001

, a 5th-degree polynomial is sufficient.

Answer: The smallest

n

5

Why It Matters

The Lagrange Error Bound appears on nearly every AP Calculus BC exam, typically in the free-response section on Taylor series. Engineers and scientists rely on the same idea when they replace complicated functions with polynomial approximations and need to guarantee their result is accurate to a certain number of decimal places. Mastering this bound also builds intuition for how quickly Taylor series converge — a foundation for numerical methods courses in college.

Common Mistakes

Mistake: Using the

(n)

th derivative instead of the

(n+1)

th derivative to find

M

Correction: The error bound uses one derivative beyond the degree of the polynomial. For a degree-

n

polynomial, evaluate the

(n+1)

th derivative.

Mistake: Choosing

M

at the center

a

instead of maximizing

|f^{(n+1)}(c)|

over the entire interval between

a

and

x

Correction: You must find the maximum of

|f^{(n+1)}|

on the whole interval

[a, x]

(or

[x, a]

). Evaluating only at

a

can underestimate the true bound.

Related Terms

Alternating Series Remainder — Simpler error bound for alternating Taylor series
Alternating Series — Series type where a different error bound applies
Alternating Series Test — Convergence test often paired with remainder estimates
Binomial Series — Special Taylor series where the bound can be applied
Alternating Harmonic Series — Classic series used in error-bound examples
Arithmetic Series — Contrasting finite series with no approximation error