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Lagrange
Error Bound

Quantify the maximum error when a Taylor polynomial approximates a function: bound the remainder with one clean inequality.

AP Calculus BC

What Is a Taylor Polynomial?

A Taylor polynomial of degree \( n \) centered at \( x = c \) approximates a function \( f(x) \) using its derivatives at \( c \):

This polynomial matches \( f \) and its first \( n \) derivatives exactly at \( x = c \). Away from \( c \), there is an error: the difference \( f(x) - P_n(x) \).

The Remainder \( R_n(x) \)

The remainder (or error) is the gap between the true function and the polynomial approximation:

We usually cannot compute \( R_n(x) \) exactly because we do not know \( f(x) \) precisely (otherwise we would not need the approximation). Instead, we bound its absolute value.

Visualizing the Error

x y c Error f(x) P₂(x)

The Lagrange Error Bound

\( M \)
Max of \( |f^{(n+1)}| \) on the interval between \( c \) and \( x \)
\( n+1 \)
One degree higher than the polynomial used
\( |x - c| \)
Distance from the center of approximation to the target value

Why Does the Bound Work?

Taylor's Theorem with remainder says there exists some \( z \) between \( c \) and \( x \) such that:

We do not know \( z \) exactly: it could be anywhere on the interval. So we take the worst case: replace \( |f^{(n+1)}(z)| \) with its maximum \( M \) on that interval. This turns an exact (but unknown) equality into a known inequality.

Error Bound vs. Actual Error

Lagrange Error Bound Actual Error
Formula \( \frac{M}{(n+1)!}|x-c|^{n+1} \) \( |f(x) - P_n(x)| \)
Requires knowing \( f(x) \)? No Yes
Exact or over-estimate? Over-estimate (upper bound) Exact
When used Guaranteeing accuracy Checking after the fact

How to Apply the Bound

1
Identify the setup: What is \( f(x) \), the center \( c \), the degree \( n \), and the target \( x \)?
2
Compute \( f^{(n+1)}(x) \): Take one more derivative than the degree of the polynomial.
3
Find \( M \): Determine the maximum of \( |f^{(n+1)}(z)| \) for all \( z \) between \( c \) and \( x \).
4
Plug into the formula: \( |R_n(x)| \leq \dfrac{M}{(n+1)!}\,|x - c|^{n+1} \)
5
Simplify: The result is the worst-case error guarantee.

Example 1: Bounding Error for \( e^x \)

Use the 3rd-degree Taylor polynomial for \( e^x \) centered at \( c = 0 \) to approximate \( e^{0.5} \). Bound the error.

1
Setup: \( f(x) = e^x \), \( c = 0 \), \( n = 3 \), \( x = 0.5 \).
2
Next derivative: \( f^{(4)}(x) = e^x \).
3
Find \( M \): On \( [0, 0.5] \), \( e^x \) is increasing, so \( M = e^{0.5} \leq 2 \).
4
Plug in: \( |R_3(0.5)| \leq \dfrac{2}{4!}(0.5)^4 = \dfrac{2}{24}\cdot\dfrac{1}{16} = \dfrac{1}{192} \approx 0.0052 \).

Example 2: \( \sin x \) Approximation

Use the 5th-degree Maclaurin polynomial for \( \sin x \) to approximate \( \sin(0.1) \). Bound the error.

1
Setup: \( f(x) = \sin x \), \( c = 0 \), \( n = 5 \), \( x = 0.1 \).
2
Next derivative: \( f^{(6)}(x) = -\sin x \), so \( |f^{(6)}(x)| = |\sin x| \leq 1 \).
3
Find \( M \): \( M = 1 \) (since \( |\sin z| \leq 1 \) everywhere).
4
Plug in: \( |R_5(0.1)| \leq \dfrac{1}{6!}(0.1)^6 = \dfrac{10^{-6}}{720} \approx 1.39 \times 10^{-9} \).

Example 3: Finding the Minimum Degree

What is the least degree \( n \) of a Maclaurin polynomial for \( \cos x \) so that the error at \( x = 1 \) is less than \( 0.001 \)?

1
All derivatives of \( \cos x \) satisfy \( |f^{(n+1)}(z)| \leq 1 \), so \( M = 1 \) always.
2
We need: \( \dfrac{1}{(n+1)!}(1)^{n+1} = \dfrac{1}{(n+1)!} < 0.001 \).
3
Test values: \( 6! = 720 \Rightarrow \frac{1}{720} \approx 0.00139 \) (too big). \( 7! = 5040 \Rightarrow \frac{1}{5040} \approx 0.000198 \) (small enough).
4
So \( n + 1 = 7 \Rightarrow n = 6 \). The 6th-degree Maclaurin polynomial suffices.

Error Shrinks with Higher Degree

Degree n Error Bound 1.000 n=0 0.500 n=1 0.167 n=2 0.042 n=3 0.008 n=4

Special Case: Alternating Series Estimation

When a Taylor series is alternating (signs alternate) and the terms decrease in absolute value, the Alternating Series Estimation Theorem (ASET) provides a different bound:

This is often tighter than the Lagrange bound. But the Lagrange bound applies to all Taylor series, not just alternating ones.

Lagrange vs. ASET

Lagrange Bound ASET Bound
Applies to Any Taylor polynomial Alternating series only
Requires Max of \( |f^{(n+1)}| \) Decreasing terms
Bound quality Often over-estimates Usually tighter
AP exam use Most common When series alternates

Common Mistakes

✕ Wrong derivative order
Using \( f^{(n)} \) instead of \( f^{(n+1)} \). The bound requires one degree beyond the polynomial.
✕ Wrong \( M \) value
Evaluating \( f^{(n+1)} \) only at \( c \) instead of finding its maximum on the entire interval between \( c \) and \( x \).
✕ Forgetting the factorial
Writing \( M \cdot |x-c|^{n+1} \) without dividing by \( (n+1)! \). The factorial is critical.

Quick Reference Summary

The Formula
\( |R_n(x)| \leq \dfrac{M}{(n+1)!}|x-c|^{n+1} \)
\( M \) Means
Max of \( |f^{(n+1)}(z)| \) on the interval between \( c \) and \( x \)
Use When
You need a guaranteed worst-case error for any Taylor approximation
Key Insight
Higher \( n \) or smaller \( |x-c| \) means smaller error bound

Question 1 of 5

When applying the Lagrange Error Bound with a 4th-degree Taylor polynomial, which derivative do you need to bound?

Question 2 of 5

For \( f(x) = \sin x \) with a 3rd-degree Maclaurin polynomial approximating \( \sin(0.2) \), what is \( M \)?

Question 3 of 5

The 2nd-degree Taylor polynomial for \( e^x \) centered at \( 0 \) is used to approximate \( e^{0.3} \). Using \( M = e^{0.3} < 1.35 \), what is the Lagrange error bound?

Question 4 of 5

What is the least degree \( n \) of a Maclaurin polynomial for \( e^x \) so the error at \( x = 1 \) is less than \( 0.001 \)? (Use \( M = e < 3 \).)

Question 5 of 5

Which of the following will decrease the Lagrange error bound?