# Lagrange Error Bound

> AP Calculus BC · Infinite Sequences and Series
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u10-lagrange-error-bound/

This module covers Lagrange error bound for nth-degree Taylor polynomials, maximum error estimation, bounding the (n+1)th derivative, and proving Taylor series convergence for AP Calculus BC exam questions.

**Prerequisites:** [Taylor polynomial and Maclaurin series construction](https://www.owlsprep.com/study/ap-calculus-bc-u10-taylor-maclaurin-series/); Derivatives of common transcendental functions; Basic limit properties of infinite series

## Learning objectives

- Calculate the Lagrange error bound for an nth-degree Taylor polynomial approximation
- Find a valid upper bound M for the (n+1)th derivative of common functions
- Prove convergence of a Taylor series to the original function using Lagrange error bound
- Solve AP-exam style Lagrange error bound problems

## 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 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.

**Lagrange Error Bound** — An upper bound on the maximum absolute error when approximating a function $f(x)$ with its nth-degree Taylor polynomial $P_n(x)$ centered at $c$, valid for sufficiently differentiable functions.

*Notation:* $|R_n(x)| \leq \frac{M |x-c|^{n+1}}{(n+1)!}$

## 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 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.

**Worked example:** 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 approximation point $x=0.2$, so $0 \leq z \leq 0.2$. 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:
4. $$|R_3(0.2)| \leq \frac{M |x - c|^4}{4!} = \frac{3 \cdot (0.2)^4}{24}$$
5. 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 the exact maximum M. Using a simple overestimate like $M=3$ for $e^z$ on $z<1$ saves time and avoids calculation errors.

## Bounding the (n+1)th Derivative

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.

**Worked example:** $f(x) = \sin(2x)$, 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 $0.5$, $|\cos(2z)| \leq 1$, so $|f^{(5)}(z)| \leq 32 \cdot 1 = 32$. Thus $M=32$.
3. Substitute into the error bound formula:
4. $$|R_4(0.5)| \leq \frac{M |x - c|^5}{5!} = \frac{32 \cdot (0.5)^5}{120}$$
5. Calculate the result: $(0.5)^5 = 1/32$, so $32 \cdot (1/32) = 1$, so $|R_4(0.5)| \leq 1/120 \approx 0.0083$.

> **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.

## Proving Taylor Series Convergence

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.

**Worked example:** 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 $|f^{(n+1)}(z)| \leq 1$ for any $n$ and any $z$. Thus $M=1$, a constant independent of $n$.
2. The Lagrange error bound for any $n$ is:
3. $$|R_n(x)| \leq \frac{1 \cdot |x|^{n+1}}{(n+1)!}$$
4. 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 $x$.
5. By the Squeeze Theorem, $0 \leq \lim_{n \to \infty} |R_n(x)| \leq 0$, 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.

## AP-Style Worked Practice Problems

**Worked example:** 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 $\ln(3)$?<br>A) $\frac{1}{24}$<br>B) $\frac{1}{12}$<br>C) $\frac{1}{3}$<br>D) $\frac{1}{2}$

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

**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 $x$.

1. 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:
2. $$|R_3(0.4)| \leq \frac{1 \cdot (0.4)^4}{4!} = \frac{0.0256}{24} \approx 0.00107$$
3. 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.
4. 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:
5. $$|R_n(x)| \leq \frac{M |x|^{n+1}}{(n+1)!}$$
6. 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.

1. 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 $M = 50(0.2)^5 = 50(0.00032) = 0.0016$.
2. Substitute into the error bound formula:
3. $$|R_4(1)| \leq \frac{M |1 - 0|^5}{5!} = \frac{0.0016}{120} \approx 1.33 \times 10^{-5}$$
4. 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

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

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Error Definition | $R_n(x) = f(x) - P_n(x)$ | $P_n(x)$ = nth-degree Taylor polynomial centered at $c$ |
| Lagrange Remainder Exact Form | $R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$ | $z$ is strictly between $c$ and $x$ |
| Lagrange Error Bound | $\|R_n(x)\| \leq \frac{M \|x - c\|^{n+1}}{(n+1)!}$ | $M$ = maximum of $\|f^{(n+1)}(z)\|$ for $z$ between $c$ and $x$ |
| M for $\sin(kx), \cos(kx)$ | $M = \|k\|^{n+1}$ | Valid for any interval, since $\|\sin\|, \|\cos\| \leq 1$ |
| M for $e^{kx}$ ($k>0$, $z \in [0,a]$) | $M = k^{n+1} e^{ka}$ | If $a < 1$, use $M < 3k^{n+1}$ as a safe overestimate |
| M for $\ln(1+x)$ | Maximum at left endpoint of the interval | Derivatives of $\ln(1+x)$ have decreasing absolute value |
| Taylor Series Convergence Condition | $\lim_{n \to \infty} \|R_n(x)\| = 0$ | If true, the Taylor series converges to $f(x)$ at $x$ |
| Proving Convergence for All x | $\lim_{n \to \infty} \frac{M \|x\|^{n+1}}{(n+1)!} = 0$ | Factorial growth outpaces any fixed $\|x\|^k$, so limit is always 0 |

## 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 common transcendental functions are valid for all x in their interval of convergence, and to compare the 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.

- [Alternating series error bound](https://www.owlsprep.com/study/ap-calculus-bc-u10-alternating-series-error-bound/)
- [Radius and interval of convergence of power series](https://www.owlsprep.com/study/ap-calculus-bc-u10-radius-and-interval-of-convergence/)
- [Finding Taylor or Maclaurin series for a function](https://www.owlsprep.com/study/ap-calculus-bc-u10-finding-taylor-or-maclaurin-series/)

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