When we approximate a function using a Taylor polynomial, there is always some error. The Lagrange Error Bound (also called Taylor's Theorem with Remainder) gives us an upper bound on this error. This is essential for determining how many terms are needed to achieve a desired accuracy.
🎯 In this section you will learn
- The Lagrange Error Bound formula
- How to find $M$, the maximum of $|f^{(n+1)}(x)|$ on the interval
- How to use the error bound to guarantee accuracy
- How to find the number of terms needed for a given error tolerance
📌 The Lagrange Error Bound Formula
$$ |R_n(x)| = |f(x) - P_n(x)| \le \frac{M}{(n+1)!} |x-a|^{n+1} $$
where $M$ is the maximum of $|f^{(n+1)}(t)|$ for $t$ between $a$ and $x$
💡 Understanding the Formula
- $R_n(x)$ is the remainder (error) after using the $n$th degree Taylor polynomial.
- $M$ is the maximum value of the $(n+1)$st derivative on the interval between $a$ and $x$.
- The error bound decreases as $n$ increases (factorial in denominator grows fast).
Find the maximum possible error when using $P_3(x)$ for $e^x$ to approximate $e^{0.5}$.
①
Identify $n$, $a$, $x$
$n=3$, $a=0$, $x=0.5$
②
Find $f^{(4)}(t)$
$f(t) = e^t$, $f^{(4)}(t) = e^t$
③
Find $M$ on $[0, 0.5]$
$e^t$ is increasing, so maximum at $t=0.5$: $M = e^{0.5} \approx 1.6487$
④
Apply the bound
$|R_3(0.5)| \le \frac{1.6487}{4!} (0.5)^4 = \frac{1.6487}{24} \cdot 0.0625 = \frac{0.10304}{24} \approx 0.00429$
Find the error bound when using $P_3(x) = x - \frac{x^3}{6}$ to approximate $\sin(0.3)$.
①
Identify $n$, $a$, $x$
$n=3$, $a=0$, $x=0.3$
②
Find $f^{(4)}(t)$
$f(t) = \sin t$, $f^{(4)}(t) = \sin t$ (since derivatives cycle every 4)
③
Find $M$ on $[0, 0.3]$
$|\sin t| \le 1$, so $M = 1$
④
Apply the bound
$|R_3(0.3)| \le \frac{1}{4!} (0.3)^4 = \frac{1}{24} \cdot 0.0081 = 0.0003375$
How many terms of the Maclaurin series for $e^x$ are needed to approximate $e^{0.5}$ with error less than $0.0001$?
①
Set up the inequality
We need $n$ such that $\frac{M}{(n+1)!} (0.5)^{n+1} < 0.0001$ with $M = e^{0.5} \approx 1.6487$
②
Try $n=4$
$\frac{1.6487}{5!}(0.5)^5 = \frac{1.6487}{120} \cdot 0.03125 = 0.01374 \cdot 0.03125 \approx 0.000429$ (too high)
③
Try $n=5$
$\frac{1.6487}{6!}(0.5)^6 = \frac{1.6487}{720} \cdot 0.015625 = 0.00229 \cdot 0.015625 \approx 0.0000358$ ($< 0.0001$)
④
Conclusion
$n=5$ (degree 5 polynomial) is enough. That means terms up to $x^5/5!$.
For $\sin x$, the alternating series error bound gives $|R_n| \le \frac{|x|^{n+1}}{(n+1)!}$ (since the first omitted term's absolute value). Compare with Lagrange bound (which gives the same here).
⚠️ Important Notes
- $M$ is the maximum of $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$. If you don't know the exact maximum, use an overestimate.
- For alternating series with decreasing terms, the alternating series error bound is often easier and gives the same result.
- The Lagrange error bound is usually an overestimate — the actual error is smaller.
🔍 Key Takeaways
- Lagrange Error Bound: $|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$.
- $M$ is the maximum of $|f^{(n+1)}(t)|$ between $a$ and $x$.
- Use the error bound to determine how many terms are needed for a given accuracy.
- For alternating series, the alternating series error bound is often simpler.
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