Lagrange Error Bound — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Lagrange error bound formula for nth-degree Taylor polynomials, maximum error estimation, bounding the (n+1)th derivative, justifying Taylor series convergence, and solving common AP exam error-bound questions.

You should already know: Taylor polynomial and Maclaurin series construction, derivatives of common transcendental functions, basic limit properties of infinite series.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Lagrange error bound?

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 (Infinite Sequences and Series) of the AP Calculus BC CED, accounting for approximately 2-4% of total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Unlike the alternating series error bound, which only applies to alternating series meeting the alternating series test conditions, Lagrange error bound works for any Taylor polynomial approximation of a sufficiently differentiable function, making it a universal tool for error estimation. On the AP exam, common prompts ask you to find the maximum error of an approximation, verify the error is less than a given tolerance, or prove that a full Taylor series converges to the original function on its interval. The method relies on bounding the (n+1)th derivative of the original function, which is the key skill most exam questions test.

2. The Lagrange Error Bound Formula

For a function $f(x)$ that is $(n+1)$ times differentiable on an interval containing the Taylor center $c$ and the 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 that there exists some number $z$ 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. Because we only need an upper bound (not the exact error), we find the maximum value of $|f^{(n+1)}(z)|$ for all $z$ between $c$ and $x$, call this maximum $M$, then the bound becomes: $$|f(x)-P_n(x)|=|R_n(x)|\le \frac{M|x-c{|}^{n+1}}{(n+1)!}$$ Intuitively, the formula matches what we observe about Taylor polynomials: the further your approximation point is from the center $c$, the larger the maximum error, and functions with more extreme higher-order derivatives will have larger approximation error.

Worked Example

Problem: Find the Lagrange error bound for the 3rd-degree Maclaurin polynomial of $f(x)=e^x$ when approximating $e^{0.2}$.

1. For a 3rd-degree polynomial, $n=3$, so we need the 4th derivative of $f(x)$. All derivatives of $e^x$ are $e^x$, so $f^{(4)}(z)=e^z$.
2. $z$ lies between the center $c=0$ and the approximation point $x=0.2$, so $0\le z\le 0.2$. Since $e^z$ is increasing, its maximum on this interval is at $z=0.2$. We can safely overestimate $M=e^{0.2}<e^1<3$, which is a valid upper bound.
3. Substitute into the error bound formula: $|R_3(0.2)|\le \frac{M|x-c{|}^4}{4!}=\frac{3\cdot (0.2{)}^4}{24}$.
4. Calculate the result: $(0.2{)}^4=0.0016$, so $\frac{3\cdot 0.0016}{24}=0.0002$. The maximum error is at most 0.0002.

Exam tip: Any valid overestimate of M is acceptable on the AP exam; you do not need to calculate 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

Finding a valid value of $M$, 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 calculus.

• For $f(x)=\sin (kx)$ or $f(x)=\cos (kx)$: All derivatives are of the form $\pm k^{n+1}\sin (kz)$ or $\pm k^{n+1}\cos (kz)$. Since $|\sin (kz)|\le 1$ and $|\cos (kz)|\le 1$ for any $z$, $M=|k{|}^{n+1}$ is always a valid bound, no matter the interval.
• For $f(x)=e^{kx}$: If $k>0$, $f^{(n+1)}(z)=k^{n+1}{e}^{kz}$, 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.

Worked Example

Problem: $f(x)=\sin (2x)$, centered at $c=0$. What is the Lagrange error bound when using a 4th-degree Taylor polynomial to approximate $\sin (1)$?

1. $n=4$, so we need the 5th derivative of $f(x)$. Calculating derivatives: $f^{(1)}(x)=2\cos (2x)$, $f^{(2)}=-4\sin (2x)$, $f^{(3)}=-8\cos (2x)$, $f^{(4)}=16\sin (2x)$, $f^{(5)}(z)=32\cos (2z)$.
2. For any z between 0 and 0.5 (the approximation point), $|\cos (2z)|\le 1$, so $|f^{(5)}(z)|\le 32\cdot 1=32$. Thus $M=32$.
3. Substitute into the error bound formula: $|R_4(0.5)|\le \frac{M|x-c{|}^5}{5!}=\frac{32\cdot (0.5{)}^5}{120}$.
4. Calculate the result: $(0.5{)}^5=1/32$, so $32\cdot (1/32)=1$, so $|R_4(0.5)|\le 1/120\approx 0.0083$.

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 with Lagrange Error Bound

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 goes to 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 FRQ question that requires explicit justification.

Worked Example

Problem: Prove that the Maclaurin series for $f(x)=\sin (x)$ converges to $\sin (x)$ for all real x.

1. For any fixed real x, z lies between 0 and x. The (n+1)th derivative of $\sin (x)$ is either $\pm \sin z$ or $\pm \cos z$, so $|f^{(n+1)}(z)|\le 1$ for any n and any z. Thus $M=1$, a constant independent of n.
2. The Lagrange error bound for any n is $|R_n(x)|\le \frac{1\cdot |x{|}^{n+1}}{(n+1)!}$.
3. For any fixed real x, ${\lim }_{n\to \infty }\frac{|x{|}^{n+1}}{(n+1)!}=0$, because factorial growth always outpaces the growth of any fixed power of x.
4. By the Squeeze Theorem, $0\le {\lim }_{n\to \infty }|R_n(x)|\le 0$, so ${\lim }_{n\to \infty }{R}_n(x)=0$, which means the Maclaurin series for $\sin (x)$ converges to $\sin (x)$ for all real x.

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. Common Pitfalls (and how to avoid them)

• Wrong move: Using the nth derivative instead of the (n+1)th derivative when calculating M for an nth-degree Taylor polynomial. Why: Students confuse the degree of the polynomial with the order of the derivative needed for the remainder term. Correct move: Always add 1 to the degree of the polynomial to get the derivative order: for $P_n(x)$, use $f^{(n+1)}(z)$.
• Wrong move: Evaluating the derivative at the center c to get M, instead of finding the maximum over the entire interval between c and x. Why: Students default to evaluating derivatives at the center, since that is how Taylor polynomials are built. Correct move: Always check the entire interval between c and x to find the maximum of $|f^{(n+1)}(z)|$.
• Wrong move: Using n! instead of (n+1)! in the denominator of the error bound. Why: Students confuse the Taylor polynomial term formula (which uses n! for the nth term) with the error bound formula. Correct move: Write the full error bound formula before plugging in values, double-check that both the exponent of $(x-c)$ and the denominator factorial are $(n+1)$.
• Wrong move: Using an underestimate of M to get a smaller error bound than the true maximum possible error. Why: Students want a "tighter" bound and use a value smaller than the actual maximum derivative. Correct move: Always overestimate M; any valid upper bound is acceptable, and underestimating M leads to incorrect conclusions about error tolerance.
• Wrong move: Trying to find the exact value of M via critical point analysis when a simple overestimate is sufficient. Why: Students think they need an exact maximum, leading to unnecessary work and calculation errors. Correct move: Use the standard shortcut for M that matches your function (e.g., $M=k^{n+1}$ for sine/cosine) to save time.
• Wrong move: Forgetting to state that M is bounded independently of n when proving Taylor series convergence. Why: Students skip this justification step to save time. Correct move: Always explicitly note that M is a constant that does not depend on n before taking the limit.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Let $P_2(x)$ be the 2nd-degree Taylor polynomial for $f(x)=\ln (1+x)$ centered at $c=1$. What is the Lagrange error bound when using $P_2(2)$ to approximate $\ln (3)$? A) $\frac{1}{24}$ B) $\frac{1}{12}$ C) $\frac{1}{3}$ D) $\frac{1}{2}$

Worked Solution: First, for $n=2$, we calculate the 3rd derivative of $f(x)=\ln (1+x)$: $f^{\prime }(x)=(1+x{)}^{-1}$, $f^{\prime \prime }(x)=-(1+x{)}^{-2}$, $f^{\prime \prime \prime }(z)=2(1+z{)}^{-3}$. z lies between 1 and 2, so $|f^{\prime \prime \prime }(z)|=2/(1+z{)}^3$, which is decreasing, so maximum at $z=1$, giving $M=2/(2{)}^3=1/4$. Substitute into the error bound formula: $|R_2(2)|\le \frac{(1/4)|2-1{|}^3}{3!}=1/(4\cdot 6)=1/24$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=e^{-x}$, and let $P_3(x)$ be the 3rd-degree Taylor polynomial for f(x) centered at $c=0$. (a) Find the Lagrange error bound for $|f(0.4)-P_3(0.4)|$. (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. (c) Prove that the Maclaurin series for $e^{-x}$ converges to $e^{-x}$ for all real x.

Worked Solution: (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: $$|R_3(0.4)|\le \frac{1\cdot (0.4{)}^4}{4!}=\frac{0.0256}{24}\approx 0.00107$$ (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. (c) For any fixed real x, z is between 0 and x. If $x\ge 0$, then $|f^{(n+1)}(z)|=e^{-z}\le 1=M$, a constant independent of n. If $x<0$, then $|f^{(n+1)}(z)|=e^{-z}\le e^{-x}=M$, also a constant independent of n. The error bound is $|R_n(x)|\le \frac{M|x{|}^{n+1}}{(n+1)!}$, and 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.

Question 3 (Application / Real-World Style)

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 in context.

Worked Solution: 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 $M=50(0.2{)}^5=50(0.00032)=0.0016$. The Lagrange error bound is: $$|R_4(1)|\le \frac{M|1-0{|}^5}{5!}=\frac{0.0016}{120}\approx 1.33\times 10^{-5}\text{ m/s}$$ 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.

7. Quick Reference Cheatsheet

8. What's Next

Lagrange error bound is a foundational tool for all work with Taylor series, the core advanced series topic on the AP Calculus BC exam. Next, you will apply error bound techniques to find the minimum degree of a Taylor polynomial needed to meet a given error tolerance, a common AP FRQ question that relies entirely on the skills you learned here. Lagrange error bound also prepares you to confirm that power series representations of transcendental functions are valid for all x in their interval of convergence, and to compare approximation accuracy of Taylor polynomials centered at different points. Without mastering the process of bounding the (n+1)th derivative and correctly applying the error bound formula, you will not be able to earn full credit on series FRQs that require error justification.

• Power series radius and interval of convergence
• Taylor and Maclaurin series construction
• Alternating series error bound
• Representing functions as power series