Math Core

A Taylor polynomial gives an approximation, but an approximation is only useful if you know how far off it might be. The Lagrange error bound gives a guaranteed upper limit on the error, using only a bound on the next derivative. It works for any Taylor polynomial, not just alternating ones.

The remainder

When you approximate $f(x)$ by its $n$ th-degree Taylor polynomial $P_n(x)$ centered at $a$ , the error is the remainder

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

You usually can't compute $R_n(x)$ exactly (if you could, you wouldn't need the approximation). But you can bound it.

Where the bound comes from

Look at the pattern in the Taylor polynomial: the term after $\dfrac{f^{(n)}(a)}{n!}(x-a)^n$ would be $\dfrac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}$ . Lagrange showed that the remainder is exactly this next term, except with the derivative evaluated at some unknown point $c$ between $a$ and $x$ rather than at $a$ :

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

You don't know $c$ , but if you know the largest that $|f^{(n+1)}|$ can be on the interval, you can replace $f^{(n+1)}(c)$ by that maximum.

Lagrange error bound

If $|f^{(n+1)}(t)| \le M$ for every $t$ between $a$ and $x$ , then

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

Read the formula as "next term, with the derivative replaced by its maximum." Three ingredients: $M$ (the max of the $(n+1)$ st derivative), the factorial $(n+1)!$ , and the distance from the center raised to the $(n+1)$ power.

sin x and its cubic Taylor polynomial x − x³/6. The error is tiny near 0 and grows as |x − a| grows, just as the factor |x − a|⁴ in the bound predicts.Open in grapher →

Finding M

$M$ must bound $|f^{(n+1)}(t)|$ on the whole interval between the center $a$ and the point $x$ , not just at the endpoints you happen to plug in.

For $\sin x$ and $\cos x$ , every derivative is a sine or cosine, so $M = 1$ always works.
For $e^x$ on $[0, x]$ , the derivative $e^t$ is increasing, so $M = e^x$ . Since the goal is to avoid using the value you're approximating, a convenient overestimate is often used, like $e^{0.5} < 2$ or $e^1 < 3$ .
For other functions, find where $|f^{(n+1)}|$ is largest on the interval; for a monotonic derivative, that's at an endpoint.
On the AP exam, $M$ is often given to you: " $|f^{(4)}(x)| \le 6$ for all $x$ in the interval."

Common mistake

The most common mistakes are using the wrong derivative and the wrong factorial. For $P_n$ , the bound uses the $(n+1)$ st derivative and $(n+1)!$ . For a third-degree polynomial, that's the fourth derivative and $4! = 24$ . Also, the answer is a bound on the error, so write " $\text{error} \le 0.0026$ ", not "error $= 0.0026$ ."

Worked examples

Worked example: Sine near zero

The polynomial $P_3(x) = x - \dfrac{x^3}{6}$ is used to approximate $\sin(0.5)$ . Use the Lagrange error bound to bound the error.

Solution. Here $n = 3$ , $a = 0$ , $x = 0.5$ . The fourth derivative of $\sin x$ is $\sin x$ , and $|\sin t| \le 1$ , so take $M = 1$ :

|\sin(0.5) - P_3(0.5)| \le \frac{1}{4!}(0.5)^4 = \frac{0.0625}{24} \approx 0.0026.

(The actual error is about $0.00026$ , well within the bound.)

Worked example: A given derivative bound

The function $f$ has derivatives of all orders, and $|f^{(4)}(x)| \le 6$ for $1 \le x \le 1.5$ . The third-degree Taylor polynomial for $f$ about $x = 1$ is used to approximate $f(1.4)$ . Show that the error is less than $0.01$ .

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

|f(1.4) - P_3(1.4)| \le \frac{6}{4!}(0.4)^4 = \frac{6(0.0256)}{24} = 0.0064 < 0.01.

Worked example: Estimating e

The Maclaurin polynomial $P_n(x) = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}$ is used to approximate $e = e^1$ . Using $M = 3$ , find the smallest $n$ for which the Lagrange error bound is less than $0.001$ .

Solution. Every derivative of $e^x$ is $e^x$ , and on $[0, 1]$ it is at most $e < 3$ . So

|e - P_n(1)| \le \frac{3}{(n+1)!}(1)^{n+1} = \frac{3}{(n+1)!}.

We need $(n+1)! > 3000$ . Since $6! = 720$ and $7! = 5040$ , the smallest choice is $n + 1 = 7$ , so $n = 6$ .

Worked example: Finding M yourself

The second-degree Taylor polynomial for $\ln x$ about $x = 1$ is $P_2(x) = (x-1) - \dfrac{(x-1)^2}{2}$ . Bound the error when $P_2(1.2)$ is used to approximate $\ln(1.2)$ .

Solution. We need the third derivative: $f'''(x) = \dfrac{2}{x^3}$ . On $[1, 1.2]$ this is decreasing, so its maximum is at $x = 1$ : $M = 2$ . Then

|\ln(1.2) - P_2(1.2)| \le \frac{2}{3!}(0.2)^3 = \frac{2(0.008)}{6} \approx 0.00267.

Tip

When a Taylor series is alternating (like the ones for $\sin x$ and $\cos x$ at a positive $x$ ), the alternating series error bound is often simpler and tighter. The Lagrange bound is the tool to use when the series isn't alternating, or when the question says "Lagrange."

Practice

Practice 1

The polynomial $P_3(x) = x - \dfrac{x^3}{6}$ approximates $\sin x$ . Use the Lagrange error bound with $M = 1$ to bound $|\sin(0.5) - P_3(0.5)|$ . Round to four decimal places.