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

The Alternating Series Remainder applies only to convergent alternating series and says the error is bounded by the absolute value of the first omitted term. The Lagrange Remainder applies to any Taylor polynomial (not just alternating series) and requires bounding the $(n+1)$th derivative. When both apply, the alternating series bound is often tighter and easier to compute.

Lagrange Remainder — Definition, Formula & Examples

The Lagrange Remainder is the error term that tells you the maximum possible difference between a function and its Taylor polynomial approximation of degree

n

. It provides a concrete upper bound so you can guarantee how accurate your polynomial approximation actually is.

If $f$ is a function with continuous derivatives through order $n+1$ on an interval containing $a$ and $x$ , then the remainder $R_n(x) = f(x) - T_n(x)$ , where $T_n(x)$ is the $n$ th-degree Taylor polynomial centered at $a$ , can be expressed as $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$ for some $c$ between $a$ and $x$ . This is also known as Taylor's theorem with the Lagrange form of the remainder.

Key Formula

|R_n(x)| = \left|\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\right| \leq \frac{M}{(n+1)!}|x-a|^{n+1}

Where:

$R_n(x)$ = The remainder (error) between f(x) and the nth-degree Taylor polynomial
$f^{(n+1)}(c)$ = The (n+1)th derivative of f evaluated at some unknown c between a and x
$M$ = An upper bound for |f^{(n+1)}(t)| on the interval between a and x
$a$ = The center of the Taylor expansion
$n$ = The degree of the Taylor polynomial
$x$ = The point at which you are approximating f

How It Works

You use the Lagrange Remainder to determine how many terms of a Taylor series you need to achieve a desired accuracy. Since the exact value of

c

is unknown, you find the maximum of

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

for

t

between

a

and

x

, then plug that into the formula to get an upper bound on

|R_n(x)|

. If this upper bound is smaller than your error tolerance, you know the

n

th-degree Taylor polynomial is accurate enough. This technique turns a vague approximation into a rigorous guarantee.

Worked Example

Problem: Find an upper bound for the error when approximating

e^{0.5}

using the 3rd-degree Taylor polynomial for

e^x

centered at

a = 0

Step 1: Write the Lagrange Remainder formula with

n = 3

a = 0

, and

x = 0.5

|R_3(0.5)| \leq \frac{M}{4!}|0.5 - 0|^4 = \frac{M}{24}(0.5)^4

Step 2: Find

M

, the maximum of

|f^{(4)}(t)|

[0, 0.5]

. Since

f(x) = e^x

, every derivative is

e^x

. On

[0, 0.5]

, the maximum occurs at

t = 0.5

, so

M = e^{0.5}

. We can use the safe overestimate

M \leq e^1 = e \approx 2.72

M \leq e \approx 2.72

Step 3: Substitute into the bound and compute.

|R_3(0.5)| \leq \frac{2.72}{24}(0.0625) = \frac{2.72 \times 0.0625}{24} \approx \frac{0.17}{24} \approx 0.00708

Answer: The error in approximating

e^{0.5}

with the 3rd-degree Taylor polynomial is at most approximately

0.0071

. (The actual error is about

0.0052

, confirming the bound works.)

Another Example

Problem: How many terms of the Maclaurin series for

\sin(x)

are needed to approximate

\sin(1)

with error less than

0.001

Step 1: All derivatives of

\sin(x)

satisfy

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

, so

M = 1

for any

n

|R_n(1)| \leq \frac{1}{(n+1)!}|1|^{n+1} = \frac{1}{(n+1)!}

Step 2: Find the smallest

n

such that

\frac{1}{(n+1)!} < 0.001

. Test values:

\frac{1}{5!} = \frac{1}{120} \approx 0.0083

(too large),

\frac{1}{6!} = \frac{1}{720} \approx 0.00139

(still too large),

\frac{1}{7!} = \frac{1}{5040} \approx 0.000198

(small enough).

\frac{1}{7!} = \frac{1}{5040} \approx 0.000198 < 0.001

Step 3: Since

\frac{1}{(n+1)!} < 0.001

when

n+1 = 7

, we need

n = 6

. The 6th-degree Taylor polynomial (which has terms up to

x^5

for sine, since the

x^6

coefficient is zero) suffices.

T_6(x) = x - \frac{x^3}{6} + \frac{x^5}{120}

Answer: Three nonzero terms of the Maclaurin series (up to

x^5

) approximate

\sin(1)

with error less than

0.001

Why It Matters

In Calculus II and III courses, the Lagrange Remainder is the standard tool for proving that a Taylor series converges to a specific function and for certifying numerical accuracy. Engineers and scientists use it when replacing complicated functions with polynomial approximations in computational algorithms, ensuring the approximation error stays within design tolerances. It also appears in numerical analysis when analyzing the accuracy of interpolation and quadrature methods.

Common Mistakes

Mistake: Using

f^{(n)}

instead of

f^{(n+1)}

in the remainder formula.

Correction: The Lagrange Remainder involves the

(n+1)

th derivative, one order higher than the degree of the Taylor polynomial. The polynomial already accounts for derivatives through order

n

Mistake: Choosing

M

as the value of

|f^{(n+1)}|

at the center

a

rather than its maximum on the entire interval.

Correction: You must bound

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

over all

t

between

a

and

x

. Taking the value only at

a

may underestimate the true maximum, making your error bound invalid.

Related Terms

Alternating Series Remainder — Alternative error bound for alternating series
Alternating Series — Series type with its own simpler remainder estimate
Derivative of a Power Series — Derivatives of Taylor series used in remainder analysis
Convergence Tests — Tests that determine if the remainder goes to zero
Integral of a Power Series — Integration of series approximations with error control
Absolute Convergence — Convergence property related to bounding remainder terms
Comparison Test — Often used alongside remainder bounds for series