Lagrange Error Bound | AP Calc BC Unit 10 Study Guide

The Core Idea: Lagrange Error Bound

When we use a Taylor polynomial, $P_{n} (x)$ , to approximate the value of a function, $f (x)$ , we are almost always introducing some amount of error. The approximation is useful only if we have a sense of how large this error might be. The core idea of the Lagrange error bound is to provide a "worst-case scenario" or an upper bound for the absolute value of this error. It doesn't tell us the exact error, but it guarantees that the true error is no larger than a specific, calculable value.

The error itself is defined as the absolute difference between the actual function value and the value given by the Taylor polynomial approximation, $∣ f (x) - P_{n} (x) ∣$ . The Lagrange error bound provides a formula that calculates a maximum possible value for this error. This is crucial for determining the accuracy of our Taylor polynomial approximation on a given interval.

Key Formulas

The topic is governed by two fundamental pieces of essential knowledge.

The Definition of Error
The error, sometimes denoted as the remainder $R_{n} (x)$ , in using an $n$th-degree Taylor polynomial $P_n(x)$ to approximate a function $f (x)$ is the absolute difference between the two values.
$E rror = ∣ R_{n} (x) ∣ = ∣ f (x) - P_{n} (x) ∣$
The Lagrange Error Bound Formula
For an $n$th-degree Taylor polynomial $P_n(x)$ for a function $f$ centered at $x = c$ , the error is bounded by the following inequality:
$∣ R_{n} (x) ∣ \leq \frac{f ^{(n + 1)} ( z )}{( n + 1 )!} (x - c)^{n + 1}$
Where:
- $n$ is the degree of the Taylor polynomial.
- $c$ is the center of the Taylor polynomial.
- $x$ is the value at which the function is being approximated.
- $f^{(n + 1)} (z)$ represents the maximum value of the $(n+1)$th derivative of $f$ on the closed interval between $c$ and $x$ . The value $z$ is some number in that interval where this maximum occurs.

Understanding The Maximum Value of the Derivative

The most critical and often most challenging part of applying the Lagrange error bound formula is correctly identifying the term $f^{(n + 1)} (z)$ . This term does not mean you should plug $c$ or $x$ into the $(n+1)$th derivative. Instead, $f^{(n+1)}(z)$ stands for the maximum absolute value that the $(n+1)$th derivative can attain anywhere on the interval between the center $c$ and the point of approximation $x$ . For the purpose of the formula, we can define a value $M$ such that:

$M = z \in [c, x] max ∣ f^{(n + 1)} (z) ∣$

The formula is then typically written as:

$∣ R_{n} (x) ∣ \leq \frac{M}{( n + 1 )!} ∣ x - c ∣^{n + 1}$

To find $M$ , you must analyze the behavior of the $(n+1)$th derivative function, $f^{(n+1)}(x)$ , but only on the specific interval between $c$ and $x$ . For example, if you are approximating $f (1.2)$ with a polynomial centered at $c = 1$ , you must find the maximum value of $∣ f^{(n + 1)} (x) ∣$ on the interval $[1, 1.2]$ . In many AP exam problems, this value $M$ is either given directly in the problem statement or can be easily determined from a known function like $s in (x)$ or $e^{x}$ .

Core Concepts & Rules

A Taylor polynomial $P_{n} (x)$ is an approximation of a function $f (x)$ . The difference between them is the error.
The error of an $n$th-degree Taylor polynomial approximation is given by $Error = |f(x) - P_n(x)|$ .
The Lagrange error bound provides an upper limit, not the exact value, of the approximation error.
The formula for the bound requires the $(n+1)$th derivative, one degree higher than the polynomial used for the approximation. - The key to the formula is finding $M$ , the maximum absolute value of the $(n+1)$th derivative on the interval between the center of the approximation, $c$ , and the point of approximation, $x$ .

Step-by-Step Example 1: Finding the Bound

Problem: A third-degree Taylor polynomial, $P_{3} (x)$ , centered at $c = 0$ is used to approximate $f (x) = cos (x)$ at $x = 0.4$ . Find the Lagrange error bound for this approximation.

Solution:

Step 1: Identify all components.

The function is $f (x) = cos (x)$ .
The degree of the polynomial is $n = 3$ .
The center of the polynomial is $c = 0$ .
The point of approximation is $x = 0.4$ .
We need the Formula89 = 4$th derivative.

**Step 2: Find the $(n+1)$th derivative.** - $f'(x) = -\sin(x)$

$f^{''} (x) = - cos (x)$
$f^{'''} (x) = sin (x)$
$f^{(4)} (x) = cos (x)$

Step 3: Find $M$ , the maximum value of $∣ f^{(4)} (x) ∣$ on the interval between $c$ and $x$ .

The interval is $[0, 0.4]$ .
We need to find the maximum value of $∣ f^{(4)} (x) ∣ = ∣ cos (x) ∣$ on $[0, 0.4]$ .
On the interval $[0, 0.4]$ (which is in the first quadrant), $cos (x)$ is a positive and decreasing function. Its maximum value will occur at the left endpoint, $x = 0$ .
$max ∣ cos (x) ∣$ on $[0, 0.4]$ is $cos (0) = 1$ .
Therefore, M=1.

Step 4: Substitute all values into the Lagrange error bound formula.

$∣ R_{3} (0.4) ∣ \leq \frac{M}{( 3 + 1 )!} ∣0.4 - 0 ∣^{3 + 1}$

$∣ R_{3} (0.4) ∣ \leq \frac{1}{4 !} ∣0.4 ∣^{4}$

$∣ R_{3} (0.4) ∣ \leq \frac{1}{24} (0.0256)$

Step 5: Calculate the final bound.

$∣ R_{3} (0.4) ∣ \leq \frac{0.0256}{24} \approx 0.001067$

The error in approximating $cos (0.4)$ with its third-degree Maclaurin polynomial is guaranteed to be less than or equal to approximately $0.001067$ .

Step-by-Step Example 2: Exam-Style Application

Problem: Let $f$ be a function having derivatives of all orders for all real numbers. A fourth-degree Taylor polynomial for $f$ about $c = 1$ is used to approximate $f (1.3)$ . The fifth derivative of $f$ satisfies the inequality $∣ f^{(5)} (x) ∣ \leq 20$ for all $x$ in the interval $[1, 1.3]$ . Find the Lagrange error bound for the approximation of $f (1.3)$ .

Solution:

Step 1: Identify all components from the problem statement.

The degree of the polynomial is $n = 4$ .
The center of the polynomial is $c = 1$ .
The point of approximation is $x = 1.3$ .
The problem directly gives us the bound for the $(n+1) = 5$th derivative. **Step 2: Find $M$ , the maximum value of $∣ f^{(n + 1)} (x) ∣$ on the interval.**
The interval is between $c = 1$ and $x = 1.3$ , which is $[1, 1.3]$ .
The problem states that $∣ f^{(5)} (x) ∣ \leq 20$ on this interval.
This means the maximum possible value for $∣ f^{(5)} (x) ∣$ is 20.
Therefore, M=20.

Step 3: Substitute all values into the Lagrange error bound formula.

$∣ R_{4} (1.3) ∣ \leq \frac{M}{( 4 + 1 )!} ∣1.3 - 1 ∣^{4 + 1}$

$∣ R_{4} (1.3) ∣ \leq \frac{20}{5 !} ∣0.3 ∣^{5}$

Step 4: Calculate the final bound.

$∣ R_{4} (1.3) ∣ \leq \frac{20}{120} (0.3)^{5}$

$∣ R_{4} (1.3) ∣ \leq \frac{1}{6} (0.00243)$

$∣ R_{4} (1.3) ∣ \leq 0.000405$

The maximum possible error for this approximation is $0.000405$ .

Using Your Calculator

The Lagrange error bound is an analytical tool, meaning its setup is not dependent on a calculator. You must determine the derivative, find the maximum value $M$ , and set up the inequality by hand.

A calculator is only useful for the final step: performing the arithmetic to find the numerical value of the bound.

For Example 2 above, after setting up the expression $\frac{20}{120} (0.3)^{5}$ , you would use your calculator to compute the final decimal value:

$(20/120) * (0.3)^{5}$
$(1/6) * 0.00243$
The calculator would return $0.000405$ .

Do not rely on the calculator to find $M$ or any derivatives. These steps must be done through analytical reasoning based on the information given in the problem.

AP Exam Quick Hit

Common Question Types

Given Maximum Value: The problem provides a function $f$ , the degree $n$ of the Taylor polynomial, the center $c$ , the point of approximation $x$ , and explicitly states an upper bound for the $(n+1)$th derivative on the relevant interval (e.g.,$ |f^{(n+1)}(x)| \le K$). You are asked to find the Lagrange error bound. This is the most common type on the FRQ section.
Known Function: The problem gives a common function like $f (x) = e^{x}$ , $f (x) = sin (x)$ , or $f (x) = ln (x)$ . You must first calculate the $(n+1)$th derivative and then determine its maximum absolute value on the interval between $c$ and $x$ yourself before plugging into the formula.
Finding $n$ for a Given Accuracy: The problem asks for the minimum degree $n$ of a Taylor polynomial required to approximate a function value to within a certain tolerance (e.g., an error less than $0.001$ ). You must set up the Lagrange error bound inequality $\frac{M}{( n + 1 )!} ∣ x - c ∣^{n + 1} \leq 0.001$ and find the smallest integer $n$ that satisfies it, often by testing values.

Common Mistakes

Off-by-One Error ( $n$ vs. $n + 1$ ): Using the degree of the polynomial, $n$ , in the formula instead of $n + 1$ . The formula always uses the Formula24!Formula23^{n+1} $. - **Incorrectly Finding $M$ :** Using the value of the derivative at an endpoint, $f^{(n + 1)} (c)$ or $f^{(n + 1)} (x)$ , instead of finding the absolute maximum of $∣ f^{(n + 1)} (z) ∣$ over the entire interval between $c$ and $x$ .
Wrong Interval for $M$ : Searching for the maximum value of the derivative on an interval other than the one specified by the center $c$ and the point of approximation $x$ .
Ignoring Absolute Value: Forgetting to take the absolute value of the $(n+1)$th derivative when finding its maximum, which can lead to an incorrect value for $M$ if the derivative is negative on the interval.

Lagrange Error Bound - AP Calculus BC Study Guide