Lagrange Error Bound

A Taylor polynomial gives an approximation, but an approximation is only useful if you know how far off it might be. The alternating series error bound handles alternating series. The Lagrange error bound works for any Taylor polynomial, alternating or not, as long as you can put a ceiling on the next derivative. It’s a favourite on the AP BC exam.

Key ideas

The remainder

The remainder (or error) of the $n$th-degree Taylor polynomial is

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

It’s what you’d still need to add to the polynomial to get the exact value.

The Lagrange error bound

If $P_n$ is the $n$th-degree Taylor polynomial for $f$ about $x = a$, then

$$|R_n(x)| = |f(x) - P_n(x)| \le \frac{M}{(n+1)!}\,|x - a|^{n+1}$$

where $M$ is any number with $\left|f^{(n+1)}(z)\right| \le M$ for every $z$ between $a$ and $x$.

Notice the pattern: the bound looks just like the next term of the Taylor polynomial, $\dfrac{f^{(n+1)}(a)}{(n+1)!}(x - a)^{n+1}$, except that $f^{(n+1)}(a)$ is replaced by $M$, the largest the $(n+1)$th derivative can be on the interval.

Choosing M

$M$ only has to be an upper bound for $\left|f^{(n+1)}\right|$ on the interval from $a$ to $x$; it doesn’t have to be the smallest one. A slightly bigger $M$ gives a slightly weaker, but still true, bound. Common choices:

Situation	Good choice of M
$f^{(n+1)}$ is $\pm\sin$ or $\pm\cos$	$M = 1$
$\lvert f^{(n+1)} \rvert$ is increasing or decreasing on the interval	its value at the endpoint where it’s largest
$f^{(n+1)}(x) = e^x$ on $[0, b]$	$e^b$, rounded up (e.g. $e \lt 3$)
The problem gives "$\lvert f^{(n+1)}(x) \rvert \le K$"	$M = K$

Never round $M$ down: a bound that’s too small isn’t a bound.

Finding the degree you need

To guarantee an error less than some tolerance, increase $n$ until $\dfrac{M}{(n+1)!}|x - a|^{n+1}$ drops below it. Because of the factorial, this usually happens quickly.

Which error bound should I use?

• If the series you’re using is alternating, with terms decreasing to $0$, the alternating series error bound is usually easier.
• Otherwise (for example, $e^x$ at a positive $x$, or $\ln x$ for $x \lt 1$, where all terms have the same sign), use the Lagrange error bound.
• If a question says “Lagrange error bound,” use it, even if the series happens to alternate.

On the AP exam, show the setup: state $M$ and why it works, then write the bound with numbers substituted. You don’t need to simplify the final number.

Worked examples

Example 1: sin x near 0

Use $P_3(x) = x - \dfrac{x^3}{6}$ to approximate $\sin 0.5$ (radians), and use the Lagrange error bound to show the error is less than $0.003$.

Solution.

$$\sin 0.5 \approx 0.5 - \frac{0.125}{6} \approx 0.479167$$

Here $n = 3$, so you need the fourth derivative: $f^{(4)}(x) = \sin x$. Since $|\sin z| \le 1$ for every $z$, take $M = 1$.

$$|R_3(0.5)| \le \frac{1}{4!}\,|0.5 - 0|^4 = \frac{0.0625}{24} \approx 0.0026 \lt 0.003$$

(Check: $\sin 0.5 \approx 0.479426$, so the actual error is about $0.00026$, well within the bound. Since the $x^4$ coefficient of sine is $0$, this is also $P_4$, and using $n = 4$ would give the even smaller bound $\dfrac{0.5^5}{5!} \approx 0.00026$.)

Example 2: ln x, choosing M at an endpoint

Use the second-degree Taylor polynomial for $f(x) = \ln x$ about $x = 1$ to approximate $\ln 1.2$, and find a Lagrange error bound.

Solution. $f(1) = 0$, $f'(x) = \dfrac{1}{x}$ gives $f'(1) = 1$, and $f''(x) = -\dfrac{1}{x^2}$ gives $f''(1) = -1$:

$$P_2(x) = (x - 1) - \frac{(x - 1)^2}{2} \quad\Rightarrow\quad \ln 1.2 \approx 0.2 - 0.02 = 0.18$$

The next derivative is $f'''(x) = \dfrac{2}{x^3}$. On $1 \le z \le 1.2$ it is positive and decreasing, so its largest value is at the left endpoint: $M = f'''(1) = 2$.

$$|R_2(1.2)| \le \frac{2}{3!}\,(0.2)^3 = \frac{0.016}{6} \approx 0.00267$$

(Check: $\ln 1.2 \approx 0.182322$, so the actual error is about $0.00232$.)

Example 3: A bound given in the problem

A function $f$ has a third-degree Taylor polynomial about $x = 1$ called $P_3$. It is known that $\left|f^{(4)}(x)\right| \le 6$ for $1 \le x \le 1.5$. Show that $P_3(1.4)$ approximates $f(1.4)$ with an error less than $0.01$.

Solution. Take $M = 6$, $n = 3$, $|x - a| = 0.4$:

$$|R_3(1.4)| \le \frac{6}{4!}\,(0.4)^4 = \frac{6(0.0256)}{24} = 0.0064 \lt 0.01$$

Example 4: How many terms for e?

Using $M = 3$, find the smallest $n$ for which the Lagrange error bound guarantees that the Maclaurin polynomial $P_n(1)$ approximates $e$ within $0.001$.

Solution. Every derivative of $e^x$ is $e^x$, which is increasing, so on $0 \le z \le 1$ the largest value is $e^1 \lt 3$. So $M = 3$ works, and $|x - a| = 1$:

$$|R_n(1)| \le \frac{3}{(n+1)!}\,(1)^{n+1} = \frac{3}{(n+1)!}$$

You need $(n+1)! \gt 3000$. Since $6! = 720$ and $7! = 5040$, you need $n + 1 = 7$, so $n = 6$.

Common mistakes

Using the wrong derivative. For $P_n$, the bound uses the $(n+1)$th derivative and $(n+1)!$, one step past the polynomial. For $P_2$, use $f'''$ and $3!$.

Evaluating M at the centre instead of finding the maximum. $M$ must bound $\left|f^{(n+1)}\right|$ on the whole interval from $a$ to $x$. In Example 2 the maximum happened to be at the centre, but for $e^x$ on $[0, 0.5]$ it’s at $0.5$.

Rounding M down. $e^{0.2} \approx 1.2214$, so $M = 1.22$ is too small. Use $1.23$, $1.3$, or even $3$.

Forgetting the absolute value or using x instead of x − a. The bound uses $|x - a|^{n+1}$. For a polynomial about $x = 1$ evaluated at $0.9$, that’s $|-0.1|^{n+1}$, not $0.9^{n+1}$.

Thinking the bound is the error. The bound says the error is no more than this amount. The actual error is often much smaller.

Using the alternating series bound when the terms don’t alternate. For $\ln 0.9$, the Taylor terms about $1$ are all negative, so only the Lagrange error bound applies.

Practice

1. (Warm-up) The approximation $\cos 0.3 \approx 1 - \dfrac{0.3^2}{2} = 0.955$ uses $P_2$ for $\cos x$ about $0$. Find a Lagrange error bound.

Solution

$n = 2$, so use $f'''(x) = \sin x$, with $|\sin z| \le 1$, so $M = 1$:

$$|R_2(0.3)| \le \frac{1}{3!}\,(0.3)^3 = \frac{0.027}{6} = 0.0045$$

2. (Warm-up) A function has $\left|f^{(4)}(x)\right| \le 10$ for all $x$. Its third-degree Maclaurin polynomial is used to approximate $f(0.5)$. Find an upper bound for the error.

Solution

$$|R_3(0.5)| \le \frac{10}{4!}\,(0.5)^4 = \frac{10(0.0625)}{24} \approx 0.026$$

3. (Warm-up) $P_2(x) = 1 + x + \dfrac{x^2}{2}$ is used to approximate $e^{0.5}$. Explain why $M = 2$ is a valid choice, and find the error bound.

Solution

$f'''(x) = e^x$ is increasing, so on $0 \le z \le 0.5$ its largest value is $e^{0.5} \approx 1.649 \lt 2$. So $M = 2$ works.

$$|R_2(0.5)| \le \frac{2}{3!}\,(0.5)^3 = \frac{0.25}{6} \approx 0.0417$$

(Actual: $P_2(0.5) = 1.625$ and $e^{0.5} \approx 1.6487$, an error of about $0.0237$.)

4. (Core) In the Taylor polynomials lesson, $P_3(0.2) \approx 1.221333$ was used to approximate $e^{0.2}$. Use the Lagrange error bound with $M = 1.3$ to show the error is less than $0.0001$.

Solution

$f^{(4)}(x) = e^x$ is increasing, so on $0 \le z \le 0.2$ its maximum is $e^{0.2} \approx 1.2214 \lt 1.3$.

$$|R_3(0.2)| \le \frac{1.3}{4!}\,(0.2)^4 = \frac{1.3(0.0016)}{24} \approx 0.0000867 \lt 0.0001$$

5. (Core) The tangent line to $f(x) = \sqrt{x}$ at $x = 4$ gives $\sqrt{4.2} \approx 2.05$. Find a Lagrange error bound for this approximation.

Solution

The tangent line is $P_1$, so use $f''(x) = -\dfrac{1}{4}x^{-3/2}$. Its absolute value $\dfrac{1}{4x^{3/2}}$ is decreasing on $4 \le z \le 4.2$, so the maximum is at $z = 4$:

$$M = \frac{1}{4 \cdot 8} = \frac{1}{32}$$$$|R_1(4.2)| \le \frac{1/32}{2!}\,(0.2)^2 = \frac{0.04}{64} = 0.000625$$

(Actual error: $2.05 - 2.04939 \approx 0.00061$.)

6. (Core) A function $f$ has $f(1) = 2$, $f'(1) = -1$ and $f''(1) = 3$, and $\left|f'''(x)\right| \le 12$ for $1 \le x \le 1.3$.

• (a) Use the second-degree Taylor polynomial about $x = 1$ to approximate $f(1.3)$.
• (b) Find a Lagrange error bound for your approximation.
• (c) Can $f(1.3)$ equal $1.9$? Explain.

Solution

(a)

$$P_2(x) = 2 - (x - 1) + \frac{3}{2}(x - 1)^2 \quad\Rightarrow\quad f(1.3) \approx 2 - 0.3 + 1.5(0.09) = 1.835$$

(b)

$$|R_2(1.3)| \le \frac{12}{3!}\,(0.3)^3 = 2(0.027) = 0.054$$

(c) No. $f(1.3)$ must be within $0.054$ of $1.835$, that is, between $1.781$ and $1.889$. Since $|1.9 - 1.835| = 0.065 \gt 0.054$, $f(1.3)$ cannot be $1.9$.

7. (Core) Using $M = 1$, what is the smallest degree $n$ for which the Lagrange error bound guarantees that the Maclaurin polynomial for $\sin x$ approximates $\sin 1$ within $0.001$?

Solution

Every derivative of $\sin x$ is $\pm\sin x$ or $\pm\cos x$, so $M = 1$, and $|x - a| = 1$:

$$|R_n(1)| \le \frac{1}{(n+1)!}$$

You need $(n+1)! \gt 1000$. Since $6! = 720$ and $7! = 5040$, $n + 1 = 7$, so $n = 6$. (The degree 6 polynomial is the same as the degree 5 one, $x - \dfrac{x^3}{6} + \dfrac{x^5}{120}$, because sine has no $x^6$ term.)

8. (Challenge) Use the third-degree Taylor polynomial for $\ln x$ about $x = 1$ to approximate $\ln 0.9$, and find a Lagrange error bound. Why can’t you use the alternating series error bound here?

Solution

The derivatives of $\ln x$ at $1$ are $f'(1) = 1$, $f''(1) = -1$, $f'''(1) = 2$, so

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

With $x - 1 = -0.1$:

$$\ln 0.9 \approx -0.1 - 0.005 - 0.000333 \approx -0.105333$$

Next, $f^{(4)}(x) = -\dfrac{6}{x^4}$, and $\dfrac{6}{x^4}$ is decreasing, so on $0.9 \le z \le 1$ its maximum is at $z = 0.9$: $M = \dfrac{6}{0.9^4} \approx 9.145$ (use $M = 9.2$ to be safe).

$$|R_3(0.9)| \le \frac{9.2}{4!}\,|-0.1|^4 = \frac{9.2(0.0001)}{24} \approx 0.0000383$$

At $x = 0.9$ every term $-0.1$, $-0.005$, $-0.000333$, … is negative, so the series isn’t alternating and the alternating series bound doesn’t apply. (Check: $\ln 0.9 \approx -0.105361$, an error of about $0.0000272$.)

9. (Challenge) Use $P_6(1) = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!} + \dfrac{1}{6!}$ and the Lagrange error bound with $M = 3$ to show that $2.718 \lt e \lt 2.719$.

Solution

$$P_6(1) = \frac{1957}{720} \approx 2.718056$$

As in Example 4, $M = 3$ works on $0 \le z \le 1$:

$$|R_6(1)| \le \frac{3}{7!} = \frac{3}{5040} \approx 0.000595$$

So $e \le 2.718056 + 0.000595 = 2.718651 \lt 2.719$.

Also, every term of the Maclaurin series for $e^1$ is positive, so the remainder (the sum of the terms left out) is positive: $e \gt P_6(1) \approx 2.718056 \gt 2.718$.

Together, $2.718 \lt e \lt 2.719$.