The Lagrange Error Bound — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

A Taylor polynomial gives an estimate. Engineers and the exam both ask the next question: how far off can it be? The Lagrange error bound answers with a single number, using the largest the next derivative can be.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the third-degree Taylor polynomial for $e^x$ about $x = 0$ ?

$1 + x + \frac{x^2}{2} + \frac{x^3}{6}$
$1 + x + x^2 + x^3$
$x + \frac{x^2}{2} + \frac{x^3}{3}$
$1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

Show the answer

Every derivative at 0 is 1, so each coefficient is $\frac{1}{n!}$ ; a polynomial stops at its last term.

Correct: $1 + x + \frac{x^2}{2} + \frac{x^3}{6}$ :
$1 + x + x^2 + x^3$ :
$x + \frac{x^2}{2} + \frac{x^3}{3}$ :
$1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$ :

2. On $[0, 2]$ , where does $g(x) = \frac{1}{1 + x}$ take its largest value?

At $x = 0$
At $x = 2$
At $x = 1$
It has no largest value

Show the answer

$g$ is decreasing, so its largest value is at the left end.

Correct: At $x = 0$ :
At $x = 2$ :
At $x = 1$ :
It has no largest value:

3. For an alternating series that meets the test's conditions, the error of $S_n$ is at most

$a_{n+1}$ , the first term left out
$a_n$ , the last term used
$S_{n+1}$
$a_1$

Show the answer

The alternating series error bound.

Correct: $a_{n+1}$ , the first term left out:
$a_n$ , the last term used:
$S_{n+1}$ :
$a_1$ :

Part 4 · See it

See it first

On -1.6 to 1.6, a dashed curve x to the fourth over 24, the Lagrange error bound for cos x and its second-degree Taylor polynomial about 0, with the region under it lightly shaded, and a solid curve, the actual error |cos x minus (1 minus x squared over 2)|, which stays below it. At x = 1 the bound is about 0.0417 and the actual error about 0.0403. — For $\cos x$ and $P_2(x) = 1 - \frac{x^2}{2}$ , the actual error stays under the Lagrange bound $\frac{x^4}{4!}$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$P_n$ matches $f$ and its first $n$ derivatives at $a$ the difference $f - P_n$ grows only through the derivative after those, $f^{(n+1)}$
Bound $\left|f^{(n+1)}(z)\right| \le M$ for every $z$ between $a$ and $x$ the error can be no bigger than $\frac{M}{(n+1)!}|x - a|^{n+1}$
Compare that bound with the accuracy asked for"error $\le$ bound $<$ given number" shows the estimate is good enough

Part 6 · Key ideas

Key ideas

$\left|f(x) - P_n(x)\right| \le \frac{\max\left|f^{(n+1)}(z)\right|}{(n+1)!}|x - a|^{n+1}$ , the maximum taken for $z$ between $a$ and $x$ .
Say where the maximum is and why: " $e^z$ is increasing, so on $[0, 0.5]$ its maximum is $e^{0.5} < 2$ ."
The bound is a ceiling: write "error $\le$ ", never "error $=$ ".
If the values form an alternating sum with sizes decreasing to 0, the alternating series error bound may be quicker.

Part 7 · Misconception

A common mistake

The wrong idea: For the bound on $P_3$ at $x = 0.5$ , use $f^{(4)}(0)$ , the fourth derivative at the center.

What actually happens: Use the largest $\left|f^{(4)}(z)\right|$ for $z$ anywhere between the center and $0.5$ , and say why that value is the largest.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Derivatives at the center and a bound on the next one

Values of $f$ and its derivatives at $x = 0$ ; on $0 \le x \le 0.5$ , $\left|f^{(4)}(x)\right| \le 10$
$f(0)$	$f'(0)$	$f''(0)$	$f'''(0)$	$\left\|f^{(4)}(x)\right\|$ on $[0, 0.5]$
$1$	$2$	$-3$	$6$	at most $10$

1. Use the third-degree Taylor polynomial for $f$ about $x = 0$ to estimate $f(0.5)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$P_3(x)$ $= 1 + 2x - \frac{3}{2}x^2 + x^3$ , so $P_3(0.5)$ $= 1$ $+ 1 - 0.375$ $+ 0.125$ $= 1.75$ $= \frac{7}{4}$ .

Answer: $\frac{7}{4}$ ≈ 1.750

2. Use the Lagrange error bound to find a number $E$ with $\left|f(0.5) - P_3(0.5)\right| \le E$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The Lagrange error bound for $P_3$ is $\frac{\max\left|f^{(4)}(z)\right|}{4!}|x - 0|^4 \le \frac{10}{24}(0.5)^4$ $= \frac{10}{384}$ $= \frac{5}{192}$ $\approx 0.026$ .

Answer: $\frac{5}{192}$ ≈ 0.026

3. Which statement shows that $P_3(0.5)$ approximates $f(0.5)$ with an error less than $0.03$ ?

$\left|f(0.5) - P_3(0.5)\right|$ $= \frac{5}{192} < 0.03$
$\left|f(0.5) - P_3(0.5)\right| \le \frac{10}{3!}(0.5)^3 < 0.03$
$\left|f(0.5) - P_3(0.5)\right| \le \frac{6}{4!}(0.5)^4 < 0.03$
$\left|f(0.5) - P_3(0.5)\right| \le \frac{5}{192} < 0.03$

Show the answer

The Lagrange error bound gives $\left|f(0.5) - P_3(0.5)\right| \le \frac{10}{4!}(0.5)^4$ $= \frac{5}{192}$ $\approx 0.026$ , and $\frac{5}{192} < 0.03$ .

$\left|f(0.5) - P_3(0.5)\right|$ $= \frac{5}{192} < 0.03$ : The bound gives "at most"; the error itself is not known to equal $\frac{5}{192}$ .
$\left|f(0.5) - P_3(0.5)\right| \le \frac{10}{3!}(0.5)^3 < 0.03$ : That uses the wrong derivative and power, and $\frac{10}{6}(0.125) \approx 0.21$ is not below 0.03.
$\left|f(0.5) - P_3(0.5)\right| \le \frac{6}{4!}(0.5)^4 < 0.03$ : $6 = f'''(0)$ is a value at the center; the bound needs the largest size of $f^{(4)}$ on the interval, 10.
Correct: $\left|f(0.5) - P_3(0.5)\right| \le \frac{5}{192} < 0.03$ : Right: the bound with its inputs, then the comparison.

4. Which is the Lagrange error bound for $P_n(x)$ , the $n$ th-degree Taylor polynomial for $f$ about $x = a$ ?

$\frac{\max\left|f^{(n)}(z)\right|}{n!}|x - a|^n$
$\frac{\max\left|f^{(n+1)}(z)\right|}{(n+1)!}|x - a|^{n+1}$
$\frac{\left|f^{(n+1)}(a)\right|}{(n+1)!}\left|x - a\right|^{n+1}$
$\max\left|f^{(n+1)}(z)\right| \cdot |x - a|^{n+1}$

Show the answer

$\left|f(x) - P_n(x)\right| \le \frac{\max\left|f^{(n+1)}(z)\right|}{(n+1)!}|x - a|^{n+1}$ , where the maximum is over the $z$ between $a$ and $x$ .

$\frac{\max\left|f^{(n)}(z)\right|}{n!}|x - a|^n$ : The bound uses the derivative one past the degree: $f^{(n+1)}$ .
Correct: $\frac{\max\left|f^{(n+1)}(z)\right|}{(n+1)!}|x - a|^{n+1}$ : Right: the next derivative, the next factorial, the next power.
$\frac{\left|f^{(n+1)}(a)\right|}{(n+1)!}\left|x - a\right|^{n+1}$ : The derivative is maximized over the interval between $a$ and $x$ , not taken at the center.
$\max\left|f^{(n+1)}(z)\right| \cdot |x - a|^{n+1}$ : Divide by $(n+1)!$ .

5. To bound the error of the second-degree Taylor polynomial for $f$ about $x = 1$ at $x = 1.3$ , over which values of $z$ do you find the largest $\left|f'''(z)\right|$ ?

$z = 1$ alone, the center
$z = 1.3$ alone, the point being estimated
$0 \le z \le 1.3$ , from zero to the point
$1 \le z \le 1.3$ , from the center to the point

Show the answer

The Lagrange error bound uses the largest $\left|f'''(z)\right|$ for $z$ between the center $a = 1$ and $x = 1.3$ . Say where that maximum is and why (for example, because $|f'''|$ is increasing there).

$z = 1$ alone, the center: One value does not cover the interval; the largest size between 1 and 1.3 is needed.
$z = 1.3$ alone, the point being estimated: The largest size may occur anywhere between 1 and 1.3.
$0 \le z \le 1.3$ , from zero to the point: The interval runs from the center, 1, not from 0.
Correct: $1 \le z \le 1.3$ , from the center to the point: Right: the $z$ in the bound lies between the center and $x$ .

6. The Lagrange error bound for an estimate is $0.004$ . Which sentence is correct?

The error is $0.004$
The estimate is $0.004$ too large
The error is at most $0.004$
The error is at least $0.004$

Show the answer

The Lagrange error bound says $\left|f(x) - P_n(x)\right| \le 0.004$ : the error is at most $0.004$ , in either direction.

The error is $0.004$ : The actual error is less than or equal to the bound, usually less.
The estimate is $0.004$ too large: The bound gives no direction and no exact size.
Correct: The error is at most $0.004$ : Right: the bound caps the error.
The error is at least $0.004$ : The bound is an upper limit, not a lower one.

Part 9 · Summary

Summary

The Lagrange error bound caps the error of a Taylor polynomial:

\left|f(x) - P_n(x)\right| \le \frac{M}{(n+1)!}|x - a|^{n+1}

, where

M

is the largest value of

\left|f^{(n+1)}\right|

between

a

and

x

. Justify

M

, compute the bound, and compare it with the accuracy required.

Part 10 · Up next

What comes next

Part 11 · Connections