Lagrange Error Bound — AP Calculus BC

AP Calculus BC · Infinite Sequences and Series · 14 min read

1. What is Lagrange Error Bound? ★★☆☆☆ ⏱ 3 min

Lagrange error bound (also called the Lagrange remainder bound) is a general technique for finding the maximum possible absolute error when approximating a function with its nth-degree Taylor polynomial. It is a core topic in Unit 10 of the AP Calculus BC CED, accounting for ~2-4% of total exam score, appearing in both multiple-choice and free-response sections.

Unlike alternating series error bound, which only applies to alternating series meeting Alternating Series Test conditions, Lagrange error bound works for any Taylor polynomial approximation of a sufficiently differentiable function, making it a universal error estimation tool. On the AP exam, common prompts ask you to find maximum approximation error, verify error is below a given tolerance, or prove a full Taylor series converges to the original function. The key skill tested in most questions is correctly bounding the (n+1)th derivative of the original function.

2. The Lagrange Error Bound Formula ★★★☆☆ ⏱ 4 min

For a function $f(x)$ that is $(n+1)$ times differentiable on an interval containing the Taylor center $c$ and approximation point $x$, the error (or remainder) of the nth-degree Taylor approximation $P_n(x)$ is defined as $R_n(x) = f(x) - P_n(x)$. Lagrange's form of the remainder, derived from the generalized Mean Value Theorem, states there exists some $z$ strictly between $c$ and $x$ such that:

R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x - c)^{n+1}

The Lagrange error bound is the maximum possible absolute value of this remainder. We find the maximum value of $|f^{(n+1)}(z)|$ for all $z$ between $c$ and $x$, call this maximum $M$, giving the bound:

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

Intuitively, the formula matches observable behavior of Taylor polynomials: the further the approximation point is from the center $c$, the larger the maximum error, and functions with more extreme higher-order derivatives have larger approximation error.

Exam tip: Any valid overestimate of M is acceptable on the AP exam; you do not need the exact maximum M. Using a simple overestimate like $M=3$ for $e^z$ on $z<1$ saves time and avoids calculation errors.

3. Bounding the (n+1)th Derivative ★★★☆☆ ⏱ 4 min

Finding a valid $M$, the maximum of $|f^{(n+1)}(z)|$ on the interval between $c$ and $x$, is the most commonly tested step of Lagrange error bound problems on the AP exam. For most standard functions tested on the exam, there are predictable shortcuts to find M without complicated optimization:

For $f(x) = \sin(kx)$ or $f(x) = \cos(kx)$: All derivatives are $\pm k^{n+1} \sin(kz)$ or $\pm k^{n+1} \cos(kz)$. Since $|\sin(kz)| \leq 1$ and $|\cos(kz)| \leq 1$ for any $z$, $M = |k|^{n+1}$ is always a valid bound, regardless of interval.
For $f(x) = e^{kx}$ (with $k>0$): $f^{(n+1)}(z) = k^{n+1} e^{kz}$, which is increasing, so maximum at the largest $z$ in the interval. If the upper bound of $z$ is less than 1, $M < 3k^{n+1}$ is a safe overestimate.
For logarithmic functions like $f(x) = \ln(1+x)$: All derivatives of $\ln(1+x)$ have decreasing absolute value as $x$ increases, so maximum at the left endpoint of the interval.

Exam tip: For sine and cosine, you will almost never need to search for a maximum M beyond the $k^{n+1}$ shortcut; this is a common AP exam time-saver that is always valid.

4. Proving Taylor Series Convergence ★★★★☆ ⏱ 5 min

A full Taylor series centered at $c$ converges to $f(x)$ at a point $x$ if and only if $\lim_{n \to \infty} R_n(x) = 0$, meaning the error approaches zero as we add more terms to the polynomial. Lagrange error bound gives a straightforward way to prove this convergence for all $x$ in the interval of convergence. The logic is: if we can bound $|f^{(n+1)}(z)|$ by a constant $M$ that does not depend on $n$, then the error bound $\frac{M |x-c|^{n+1}}{(n+1)!}$ will go to zero as $n$ approaches infinity, because factorial growth outpaces exponential growth of any fixed base. This is a common free-response question that requires explicit justification.

Exam tip: Explicitly state that $M$ is independent of $n$ and that factorial growth dominates $|x|^{n+1}$ for full credit on convergence proof FRQs; these are required reasoning steps.

5. AP-Style Worked Practice Problems ★★★★☆ ⏱ 6 min

📐 Worked Example

Let $f(x) = e^{-x}$, and let $P_3(x)$ be the 3rd-degree Taylor polynomial for $f(x)$ centered at $c=0$.<br>(a) Find the Lagrange error bound for $|f(0.4) - P_3(0.4)|$.<br>(b) State whether your bound from (a) guarantees that the approximation is accurate to within 0.001 of the true value of $e^{-0.4}$. Justify your answer.<br>(c) Prove that the Maclaurin series for $e^{-x}$ converges to $e^{-x}$ for all real $x$.

Part (a): For $n=3$, the 4th derivative of $f(x) = e^{-x}$ is $f^{(4)}(z) = e^{-z}$, so $|f^{(4)}(z)| = e^{-z}$. $z$ is between 0 and 0.4, so $e^{-z}$ is decreasing, maximum at $z=0$, so $M=e^0 = 1$. The error bound is:
$|R_3(0.4)| \leq \frac{1 \cdot (0.4)^4}{4!} = \frac{0.0256}{24} \approx 0.00107$
Part (b): No, the bound is approximately 0.00107, which is larger than 0.001. Because the Lagrange bound is an upper bound on the true error, we cannot guarantee the error is smaller than 0.001.
Part (c): For any fixed real $x$, $z$ is between 0 and $x$. If $x \geq 0$, then $|f^{(n+1)}(z)| = e^{-z} \leq 1 = M$, a constant independent of $n$. If $x < 0$, then $|f^{(n+1)}(z)| = e^{-z} \leq e^{-x} = M$, also a constant independent of $n$. The error bound is:
$|R_n(x)| \leq \frac{M |x|^{n+1}}{(n+1)!}$
For any fixed $x$, $\lim_{n \to \infty} \frac{|x|^{n+1}}{(n+1)!} = 0$. By Squeeze Theorem, $\lim_{n \to \infty} |R_n(x)| = 0$, so the series converges to $e^{-x}$ for all real $x$.

📐 Worked Example

A physicist approximates the velocity of a falling object near Earth's surface after $t$ seconds, when accounting for air resistance, as $v(t) = v_t (1 - e^{-kt})$, where $v_t = 50$ m/s is terminal velocity and $k = 0.2$ s⁻¹. The physicist uses a 4th-degree Maclaurin polynomial to approximate $v(1)$. What is the maximum possible absolute error in the velocity approximation, in m/s? Interpret your result.

We have $v(t) = 50(1 - e^{-0.2t})$, so for a 4th-degree polynomial, $n=4$, meaning we need the 5th derivative. The 5th derivative is $v^{(5)}(z) = 50 \cdot (0.2)^5 e^{-0.2z}$, where $z$ is between 0 and 1. $|v^{(5)}(z)|$ is decreasing, so maximum at $z=0$, giving $M = 50(0.2)^5 = 50(0.00032) = 0.0016$.
Substitute into the error bound formula:
$|R_4(1)| \leq \frac{M |1 - 0|^5}{5!} = \frac{0.0016}{120} \approx 1.33 \times 10^{-5}$
Interpretation: The maximum possible error in the velocity approximation is less than $1.4 \times 10^{-5}$ m/s, which is negligible for any practical physics measurement, so the 4th-degree approximation is extremely accurate for this scenario.

Common Pitfalls

Why: Students confuse the degree of the polynomial with the order of the derivative needed for the remainder term.

Why: Students default to evaluating derivatives at the center, since that is how Taylor polynomials are built.

Why: Students confuse the Taylor polynomial term formula (which uses n! for the nth term) with the error bound formula.

Why: Students want a 'tighter' bound and use a value smaller than the actual maximum derivative.

Why: Students think they need an exact maximum, leading to unnecessary work and calculation errors.

Why: Students skip this justification step to save time.