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Taylor's Inequality

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Reference: Stewart §11.10

Textbook Reference

Primary source OpenStax Calculus Volume 2, Section 6.3: “Taylor and Maclaurin Series”
Direct link https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series
Textbook used in class Stewart, Calculus, Section 11.10: “Taylor and Maclaurin Series” (Examples 5, 6, 7)

Opening Scenario

Knowing that $f(x) = T_n(x) + R_n(x)$ is not enough: you need to know how large the remainder $R_n(x)$ can be. Taylor’s Inequality gives a concrete bound. It is also the bridge between the Taylor polynomial (a finite approximation) and the Taylor series (an infinite sum that equals $f$): the series equals $f$ if and only if $R_n(x) \to 0$ as $n \to \infty$.


Quick Reference

Taylor’s Inequality. If $|f^{(n+1)}(t)| \leq M$ for all $t$ between $a$ and $x$, then \[ |R_n(x)| \leq \frac{M}{(n+1)!}|x - a|^{n+1}. \]

Proof that $e^x$ equals its Maclaurin series everywhere: for any $x$, $|f^{(n+1)}(t)| = e^t \leq e^{|x|}$ on $[0, x]$. So $|R_n(x)| \leq e^{|x|}|x|^{n+1}/(n+1)! \to 0$ as $n \to \infty$ (since $|x|^{n+1}/(n+1)! \to 0$ for any fixed $x$). Thus $e^x = \sum x^n/n!$ for all $x$.


Key Concepts

1. Using Taylor’s Inequality to Bound Error

Example 1. How accurately does $T_3(x) = 1 + x + x^2/2! + x^3/3!$ approximate $e^x$ on $[-1, 1]$? (Stewart 11.10, Example 5.)

For $x \in [-1, 1]$, $|f^{(4)}(t)| = e^t \leq e^1 = e$. Taylor’s Inequality with $n = 3$, $a = 0$, $M = e$: \[ |R_3(x)| \leq \frac{e}{4!}|x|^4 \leq \frac{e}{24} \approx \frac{2.718}{24} \approx 0.113. \]

So $T_3$ approximates $e^x$ on $[-1,1]$ with error at most about $0.113$.

Boxed answer: $|R_3(x)| \leq e/24 \approx 0.113$ on $[-1, 1]$.


2. Finding $n$ to Achieve a Given Accuracy

Example 2. How many terms of the $e^x$ Maclaurin series are needed to compute $e$ to within $0.0001$? (Stewart 11.10, Example 6.)

We want $|R_n(1)| < 0.0001$. By Taylor’s Inequality with $M = e \leq 3$: \[ |R_n(1)| \leq \frac{3}{(n+1)!}. \] Need $3/(n+1)! < 0.0001$, i.e., $(n+1)! > 30000$. Since $9! = 362880 > 30000$ and $8! = 40320 > 30000$, and $7! = 5040 < 30000$, we need $n + 1 \geq 8$, so $n \geq 7$.

Using $T_7$ (terms through $x^7/7!$) suffices.


3. Proving $R_n(x) \to 0$ for $\sin$ and $\cos$

For both $\sin x$ and $\cos x$, all derivatives satisfy $|f^{(n+1)}(t)| \leq 1$ for every $t$. So: \[ |R_n(x)| \leq \frac{1}{(n+1)!}|x|^{n+1} \to 0 \quad \text{as } n \to \infty. \] (The sequence $r^n/n! \to 0$ for any fixed $r$.) Therefore $\sin x$ and $\cos x$ equal their Maclaurin series for all $x$.


Common misconception

using $M$ = the maximum on the wrong interval. Taylor’s Inequality requires $|f^{(n+1)}(t)| \leq M$ for $t$ between $a$ and $x$. If you are approximating near $x = 2$ with center $a = 0$, you need $M$ to bound the $(n+1)$-th derivative on $[0, 2]$, not just at $x = 2$ or at $x = 0$. Taking $M$ too small produces an incorrect (too optimistic) error bound.


Common Errors Summary

Error Correction
Using $n!$ instead of $(n+1)!$ in the denominator The bound is $M|x-a|^{n+1}/(n+1)!$; the power and factorial match the next term
Choosing $M$ as $f^{(n+1)}(a)$ instead of the max on the interval $M$ must bound $|f^{(n+1)}(t)|$ for all $t$ between $a$ and $x$
Concluding $f = T_n$ from a small error bound $|R_n| \leq \varepsilon$ means the approximation is within $\varepsilon$, not that $f = T_n$

Leveled Practice

Problem 1. Use Taylor’s Inequality to bound the error in approximating $\cos(0.3)$ by $T_4(0.3) = 1 - (0.3)^2/2 + (0.3)^4/24$.

Show answer

The next term is degree $6$, so use $|R_4| \leq M|x|^5/5!$. For $\cos x$, all derivatives satisfy $|f^{(5)}(t)| = |\pm\sin t| \leq 1$, so $M = 1$. $|R_4(0.3)| \leq (0.3)^5/120 = 0.00243/120 \approx 0.0000202$.


Problem 2. Find $n$ such that $T_n$ approximates $\sin x$ on $[-\pi/4, \pi/4]$ with error at most $10^{-6}$.

Show answer

All derivatives of $\sin$ have absolute value $\leq 1$, so $|R_n(x)| \leq |x|^{n+1}/(n+1)! \leq (\pi/4)^{n+1}/(n+1)!$. Need $(\pi/4)^{n+1}/(n+1)! < 10^{-6}$. Compute: $(\pi/4)^9/9! \approx 0.785^9/362880 \approx 0.12/362880 < 10^{-6}$. So $n = 8$ suffices.


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