10.12 Lagrange Error Bound

The Lagrange error bound tells you the largest possible error when you use a Taylor or Maclaurin polynomial to approximate a function. You find it by computing the next term after your polynomial, using the maximum value of the next derivative on the interval between your center and your input. For AP Calculus BC, identify the next derivative and justify the maximum value you use.

What Is the Lagrange Error Bound?

The Lagrange error bound gives a maximum possible error for a Taylor polynomial approximation. If $P_n(x)$ is centered at $x=a$, then the error satisfies $|R_n(x)| \leq \frac{M|x-a|^{n+1}}{(n+1)!}$, where $M$ is the maximum value of $|f^{(n+1)}|$ on the interval between $a$ and $x$.

For AP Calculus BC, the key is choosing the correct next derivative and justifying $M$. A degree $n$ polynomial uses the $(n+1)$th derivative in the bound, not the derivative that made the last included term.

Why This Matters for the AP Calculus Exam

This is a BC only topic, so AB students can skip it. On the BC exam, error bounds show up when you need to justify how accurate a Taylor polynomial approximation is. You may be asked to set up the remainder term, plug in the maximum derivative value, and show the error stays below a target value. Clear setup and correct notation matter for showing your reasoning on free response work.

Two tools appear in this part of the course:

• The Lagrange error bound, which works for any Taylor polynomial.
• The alternating series error bound, which sometimes applies when the Taylor series alternates.

Key Takeaways

• The remainder is the difference between the true function and your polynomial: $R_n(x) = f(x) - P_n(x)$.
• The Lagrange form of the remainder is $R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$, where $c$ is some value between $a$ and $x$.
• To bound the error, replace $f^{(n+1)}(c)$ with $M$, the maximum of $\left|f^{(n+1)}\right|$ on the interval from $a$ to $x$, giving $|R_n(x)| \le \dfrac{M|x-a|^{n+1}}{(n+1)!}$.
• The error bound uses the $(n+1)$th derivative, one degree past your polynomial.
• When the Taylor series alternates and meets the conditions, the error is at most the size of the first omitted term.
• You can choose $n$ large enough to force the error below a required value.

Review: Taylor Polynomials

A Taylor polynomial approximates a function using a polynomial built from the function's derivatives at a center point $a$. Here is the general form:

$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n$$

A few common series centered at 0 (Maclaurin series) are worth memorizing, but any Taylor series can be built by taking derivatives of the function at the center.

To review building these polynomials, check out 10.11 Finding Taylor Polynomial Approximations of Functions.

Lagrange Error Bound

Think of a Taylor polynomial as an approximation. The true function equals your polynomial plus a remainder:

$$f(x)=P_n(x)+R_n(x)$$ $$R_n(x)=f(x)-P_n(x)$$

Using Taylor's theorem, the remainder has this form:

$$R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$

In plain terms, the remainder looks like the next term after your polynomial, the one of degree $n+1$. The catch is that you do not know the exact $c$. To get a usable bound, you replace $f^{(n+1)}(c)$ with $M$, the largest value $\left|f^{(n+1)}\right|$ can take on the interval between $a$ and $x$. That gives:

$$|R_n(x)| \le \dfrac{M|x-a|^{n+1}}{(n+1)!}$$

Example 1: Lagrange Error Bound

Using the 3rd degree Maclaurin polynomial of $e^x$ to approximate $\frac{1}{e}$, find the Lagrange error bound.

First write the 3rd degree Maclaurin polynomial of $e^x$:

$$P_3(x)= 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$$

Since $\frac{1}{e} = e^{-1}$, plug in $x = -1$:

$$P_3(-1)=1-1+\frac{1}{2}-\frac{1}{6}=\frac{1}{3}≈0.33333333333$$

Now find the error bound using the $(n+1) = 4$th term:

$$R_3(-1)=\frac{f^{(4)}(z)(-1)^4}{4!}$$

To bound $f^{(4)}(z)$, find the maximum of the 4th derivative on the interval $[-1,0]$. The 4th derivative of $e^x$ is $e^x$, which is largest at $x = 0$, giving a value of 1. So set $M = 1$:

$$R_3(-1)=\frac{1}{24}≈0.04166666666$$

The true error is at most about $0.0417$.

How to Use This on the AP Calculus Exam

Free Response

A common task gives you a table of derivative values or tells you a derivative is bounded on an interval, then asks you to show the error of a Taylor polynomial is below some target. Your job is to set up the remainder term, choose the right maximum derivative value, and compute.

Problem Solving

• Identify your polynomial degree $n$. The error bound uses the $(n+1)$th derivative.
• Find $M$, the maximum of $\left|f^{(n+1)}\right|$ on the interval between the center $a$ and the input $x$. Use monotonicity to pick the endpoint that gives the largest value.
• Plug into $\dfrac{M|x-a|^{n+1}}{(n+1)!}$ and compare to the required error.

Practice AP FRQ 2008 #3

This is Question 3 from the 2008 AP Calculus BC examination administered by College Board. All credit to College Board.

Let $h$ be a function having derivatives of all orders for $x>0$. Selected values of $h$ and its first four derivatives are shown in the table above. The function $h$ and these four derivatives are increasing on the interval $1≤x≤3$.

b) Write the third-degree Taylor polynomial for $h$ about $x=2$ and use it to approximate $h(1.9)$.

Build the polynomial using the values at $x = 2$:

$$P_3(x)=80+128(x-2)+\frac{488(x-2)^2}{6}+\frac{448(x-2)^3}{18}$$ $$P_3(1.9)=80+128(-0.1)+\frac{488(-0.1)^2}{6}+\frac{448(-0.1)^3}{18}=67.988$$

c) Use the Lagrange error bound to show that the third-degree Taylor polynomial for $h$ about $x=2$ approximates $h(1.9)$ with error less than $3 \times 10^{-4}$.

Because the fourth derivative is increasing on $1≤x≤3$, its maximum on that interval is $\frac{584}{9}$, the value at $x = 3$. Use that as $M$:

$$R_3(1.9)=\frac{584(-0.1)^4}{9 \cdot 4!}=2.7 \times 10^{-4}<3 \times 10^{-4}$$

The error stays below the target.

Common Misconceptions

• The error bound is an upper limit, not the exact error. The true error is at most the bound, not equal to it.
• Use the $(n+1)$th derivative, not the $n$th. A degree 3 polynomial uses the 4th derivative in its error bound.
• $M$ is the maximum of the absolute value of the next derivative on the interval, not just its value at the center. If the derivative is increasing, the maximum is at the higher endpoint.
• The interval for finding $M$ runs between the center $a$ and the input $x$, not the full radius of convergence.
• The alternating series error bound only applies when the series actually alternates and meets its conditions. It does not replace the Lagrange bound in every case.
• Do not forget the factorial $(n+1)!$ in the denominator. Leaving it out gives a much larger, wrong bound.

Related AP Calculus Guides

• Unit 10 Overview: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.4 Integral Test for Convergence