Solved Math Exams

Lesson

Taylor Series Error Bounds

When we approximate a function with a Taylor polynomial, we're leaving off the "tail" of the infinite series. Taylor's Inequality tells us exactly how big that error can be.

Taylor's Inequality

If $|f^{(n+1)}(t)| \leq M$ for all $t$ between $a$ and $x$, then the remainder (error) of the $n$th-degree Taylor polynomial satisfies:

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

Where:

$R_n(x) = f(x) - T_n(x)$ is the error (actual value minus approximation)
$M$ is an upper bound for the $(n+1)$th derivative on the interval
$a$ is the center of the Taylor series
$n$ is the degree of your Taylor polynomial

Why This Makes Sense

The error bound has three key ingredients:

$M$ (derivative bound): If the function's higher derivatives are large, the function is "wiggly" and harder to approximate with polynomials. Larger $M$ → larger potential error.
$(n+1)!$ in the denominator: Factorials grow incredibly fast. This is why Taylor polynomials work so well—each additional term you include dramatically shrinks the error bound.
$|x - a|^{n+1}$: The farther $x$ is from the center $a$, the worse your approximation gets. Taylor polynomials are local approximations.

How to Use Taylor's Inequality

Step 1: Identify $f(x)$, the center $a$, the degree $n$, and the $x$-value you're approximating.

Step 2: Find the $(n+1)$th derivative: $f^{(n+1)}(t)$.

Step 3: Find $M$—the maximum of $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$.

Step 4: Plug into the formula: $|R_n(x)| \leq \frac{M}{(n+1)!}|x - a|^{n+1}$

Example 1: Error Bound for $e^x$

Problem: Use a 4th-degree Taylor polynomial centered at $a = 0$ to approximate $e^{0.1}$. Find an upper bound for the error.

Step 1: Identify the setup.

$f(x) = e^x$
$a = 0$ (Maclaurin series)
$n = 4$ (4th-degree polynomial)
$x = 0.1$

Step 2: Find $f^{(5)}(t)$.

All derivatives of $e^x$ are $e^x$, so $f^{(5)}(t) = e^t$.

Step 3: Find $M$ on $[0, 0.1]$.

Since $e^t$ is increasing, its maximum on $[0, 0.1]$ is at $t = 0.1$: $$M = e^{0.1} < e^1 = e < 3$$

(We use $M = 3$ as a convenient overestimate.)

Step 4: Apply Taylor's Inequality.

$$|R_4(0.1)| \leq \frac{3}{5!}|0.1 - 0|^5 = \frac{3}{120}(0.00001) = \frac{0.00003}{120} = 0.00000025$$

Answer: The error is at most $2.5 \times 10^{-7}$. Our 4th-degree approximation is accurate to at least 6 decimal places!

Example 2: How Many Terms for Desired Accuracy?

Problem: How many terms of the Maclaurin series for $\cos x$ are needed to compute $\cos 1$ accurate to within $0.0001$?

Step 1: Set up the inequality.

We need $|R_n(1)| \leq 0.0001$.

Step 2: Analyze the derivatives of $\cos x$.

The derivatives cycle: $\cos x, -\sin x, -\cos x, \sin x, \cos x, ...$

All have absolute value $\leq 1$, so $M = 1$ for any $n$.

Step 3: Apply Taylor's Inequality.

$$|R_n(1)| \leq \frac{1}{(n+1)!}|1|^{n+1} = \frac{1}{(n+1)!}$$

Step 4: Find the smallest $n$ where $\frac{1}{(n+1)!} < 0.0001$.

Test values:

$n = 6$: $\frac{1}{7!} = \frac{1}{5040} \approx 0.000198$ ❌
$n = 7$: $\frac{1}{8!} = \frac{1}{40320} \approx 0.0000248$ ✓

Answer: We need the 7th-degree polynomial, which means terms through $x^7$.

Since $\cos x$ only has even powers, this means: $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}$ (4 non-zero terms).

Example 3: Approximating a Cube Root

Problem: Use Taylor's Inequality to estimate the accuracy of the approximation $\sqrt[3]{1.03} \approx 1 + \frac{1}{3}(0.03) = 1.01$.

Step 1: Identify the setup.

$f(x) = \sqrt[3]{1+x} = (1+x)^{1/3}$
$a = 0$
$n = 1$ (linear approximation)
$x = 0.03$

Step 2: Find $f''(t)$.

$f(x) = (1+x)^{1/3}$

$f'(x) = \frac{1}{3}(1+x)^{-2/3}$

$f''(x) = \frac{1}{3} \cdot (-\frac{2}{3})(1+x)^{-5/3} = -\frac{2}{9}(1+x)^{-5/3}$

Step 3: Find $M$ on $[0, 0.03]$.

$$|f''(t)| = \frac{2}{9}(1+t)^{-5/3} = \frac{2}{9(1+t)^{5/3}}$$

This is decreasing in $t$ (larger denominator as $t$ increases), so the maximum is at $t = 0$:

$$M = \frac{2}{9(1)^{5/3}} = \frac{2}{9}$$

Step 4: Apply Taylor's Inequality.

$$|R_1(0.03)| \leq \frac{2/9}{2!}|0.03|^2 = \frac{2}{9 \cdot 2}(0.0009) = \frac{0.0009}{9} = 0.0001$$

Answer: The approximation $\sqrt[3]{1.03} \approx 1.01$ is accurate to within $0.0001$.

Shortcut: Alternating Series Error Bound

If your Taylor series is an alternating series (signs alternate $+, -, +, -$), there's an easier method:

$$|R_n| \leq |a_{n+1}|$$

The error is at most the absolute value of the first omitted term.

When to use this: For series like $e^{-x}$, $\cos x$, $\sin x$, $\ln(1+x)$ when $x$ makes the series alternate.

Example: For $\cos 1 = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + ...$

If we stop at $\frac{1}{4!}$, the error is at most $\frac{1}{6!} = \frac{1}{720} \approx 0.00139$.

Common Mistakes

❌ Mistake: Using the wrong derivative

You need the $(n+1)$th derivative, not the $n$th. If using a 4th-degree polynomial, find $f^{(5)}$.

❌ Mistake: Forgetting to find the maximum

$M$ must bound $|f^{(n+1)}(t)|$ for ALL $t$ in the interval, not just at the endpoints.

❌ Mistake: Using $|x|$ instead of $|x - a|$

When the series is centered at $a \neq 0$, use the distance from center: $|x - a|$.

Formulas & Reference

Taylor's Inequality (Error Bound)

|R_n(x)| \leq \frac{M}{(n+1)!}|x - a|^{n+1}

Gives an upper bound for the error when approximating f(x) with the nth-degree Taylor polynomial centered at a. Valid when |f^(n+1)(t)| ≤ M for all t between a and x.

Variables:

$R_n(x)$:: the remainder (error) = f(x) - T_n(x)
$M$:: upper bound for |f^(n+1)(t)| on the interval between a and x
$n$:: degree of the Taylor polynomial
$a$:: center of the Taylor series
$x$:: point where you are approximating
$(n+1)!$:: factorial of (n+1)

Alternating Series Error Bound

|R_n| \leq |a_{n+1}|

For alternating series satisfying the Alternating Series Test, the error is at most the absolute value of the first omitted term. Often simpler than Taylor's Inequality.