Topic 10.12 Notes – Lagrange Error Bound

1. The Remainder in a Taylor Polynomial

If $P_n(x)$ is the degree-$n$ Taylor polynomial for $f(x)$ centered at $x=a$, then

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

• $P_n(x)$ is your approximation.
• $R_n(x)$ is the remainder (error).
• The actual error at a specific value is $$R_n(x) = f(x) - P_n(x).$$

Taylor’s Theorem gives a formula for that remainder, called Lagrange’s form:

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

for some number $c$ between $a$ and $x$.

You never know the exact value of $c$. That’s why we don’t compute the exact error. Instead, we bound it.

Lagrange Error Bound

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

where
$M = \max |f^{(n+1)}(x)|$ on the interval between $a$ and $x$.

Two ideas matter:

• The error behaves like the next term in the Taylor series (degree $n+1$).
• We control it by finding the largest possible value of the $(n+1)$th derivative on the interval.

That’s the whole structure.

2. How to Find a Lagrange Error Bound

When you’re asked to bound the error, the process is very consistent.

Step 1. Identify what you’re given

• Center $a$
• Degree $n$
• Approximation point $x$
• The $(n+1)$th derivative

Step 2. Determine the interval

The interval is between $a$ and the x-value.

• Centered at 2, approximating at 1.7 → interval is $[1.7,2]$.
• Centered at 0, approximating at 0.4 → interval is $[0,0.4]$.

Step 3. Find $M$

Compute $f^{(n+1)}(x)$, then find the maximum of its absolute value on the interval.

Typical strategies:

• If the derivative is increasing (like $e^x$), check the right endpoint.
• If decreasing, check the left.
• If unsure, check endpoints or use basic reasoning about the function’s behavior.

Step 4. Plug into the inequality

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

That number is a guaranteed maximum error.

On FRQs, you must clearly state what $M$ is and why. Just plugging a number without justification usually loses credit.

3. Using the Error Bound to Control Accuracy

This is where the bound becomes useful.

A. Show the error is less than a number

Compute the bound and compare:

If your bound is $0.0008$ and the question asks you to show it’s less than $0.001$, you’re done once you state the inequality clearly.

B. Find the degree needed for a certain accuracy

Set up:

$$\frac{M}{(n+1)!}|x-a|^{n+1} < \text{tolerance}$$

Then test increasing values of $n$.

Factorials grow fast. That’s why Taylor polynomials become accurate quickly.

C. Find the maximum interval for a given error

Treat the bound like an inequality in $|x-a|$:

$$\frac{M}{(n+1)!}|x-a|^{n+1} < \text{tolerance}$$

Solve for $|x-a|$.
This gives you a radius around the center where the approximation stays within the required accuracy.

That’s explicitly part of what you’re expected to be able to do in this unit.

4. Alternating Series Error Bound vs. Lagrange Error Bound

Sometimes there’s an easier option.

If the Taylor series is:

• Alternating
• Terms decreasing in magnitude
• Terms approaching 0

Then you can use the Alternating Series Error Rule:

$$|R_n| \le \text{first omitted term}$$

So you just:

• Look at the next term in the series.
• Take its absolute value.

Here’s how they compare:

If both apply, alternating is usually quicker. But always verify the conditions first.

5. Common Mistakes and Exam Traps

• Forgetting absolute value around $f^{(n+1)}(x)$
• Using the derivative at one point instead of the maximum on the interval
• Choosing the wrong interval
• Dropping the factorial
• Calling the bound the “actual error”

The bound is a worst-case guarantee. The real error is usually smaller.