Unit 10 · Topic 10.12

BC only

10.12 Lagrange Error Bound

BC only. The Lagrange error bound caps how far a degree-n Taylor polynomial can be from the true function value. The error is at most M·|x − a|ⁿ⁺¹/(n + 1)!, where M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x.

Key terms

Lagrange error bound
remainder
degree n
maximum of the (n + 1)th derivative

The bound

This whole unit is BC only. Let Pₙ be the degree-n Taylor polynomial for f centered at a. If |f⁽ⁿ⁺¹⁾(t)| ≤ M for every t between a and x, then

|f(x) − Pₙ(x)| ≤ M · |x − a|ⁿ⁺¹/(n + 1)!.

Notice it looks like the next term of the Taylor polynomial, but with M in place of the (n + 1)th derivative at the center. Since you don't know exactly where between a and x the error behaves worst, you use the biggest the derivative could be on that interval.

Finding M

M must bound |f⁽ⁿ⁺¹⁾| on the whole interval between a and x, not just at a.
For sin x and cos x, every derivative is bounded by 1, so M = 1 always works.
For eˣ on [0, b] with b > 0, the max of every derivative is eᵇ. If you're approximating eᵇ itself, use a simple upper bound instead, like e^(0.5) < 2 or e < 3.
Free-response questions often just give you a bound, like “|f⁽⁴⁾(x)| ≤ 6 for all x in the interval.” Use it directly.
Any M that's an upper bound works. A smaller valid M gives a tighter (better) bound.

Using the bound

A typical conclusion: “|f(1.5) − P₃(1.5)| ≤ 0.015625, so the approximation is within 0.0157 of the true value.” If a question asks you to show the error is less than some number, compute the bound and compare.

The bound gets smaller quickly when x is close to a (because of the power) and when n is large (because of the factorial). You can also turn it into an interval: if P₃(1.5) = 2.4 and the bound is 0.0157, then f(1.5) is between about 2.384 and 2.416.

Lagrange or alternating series bound?

If the Taylor series, evaluated at your x, is an alternating series whose terms decrease to 0, you can use the alternating series error bound (10.10). It's usually simpler: the error is at most the first omitted term. For example, sin x and cos x series at small x alternate. If the series doesn't alternate, as for eˣ at a positive x, use Lagrange. Read the question: it often names which bound to use.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Bounding the error for sin(0.2)
Use the third-degree Maclaurin polynomial for sin x to approximate sin(0.2), and use the Lagrange error bound to bound the error.
Show the solutionHide the solution
1. Step 1: P₃(x) = x − x³/6. P₃(0.2) = 0.2 − 0.008/6 ≈ 0.198667.
2. Step 2: The fourth derivative of sin x is sin x, and |sin t| ≤ 1, so M = 1.
3. Step 3: Error ≤ 1 · (0.2)⁴/4! = 0.0016/24 ≈ 0.0000667.
4. Step 4: (The actual error is about 0.0000027, well within the bound.)
Answer: sin(0.2) ≈ 0.198667, with error at most about 6.67 × 10⁻⁵.
Example 2
Using a given derivative bound
f has derivatives of all orders, and |f⁽⁴⁾(x)| ≤ 6 for 1 ≤ x ≤ 1.5. P₃ is the third-degree Taylor polynomial for f about x = 1. Find a bound for |f(1.5) − P₃(1.5)|.
Show the solutionHide the solution
1. Step 1: n = 3, so use the fourth derivative: M = 6.
2. Step 2: |x − a| = 1.5 − 1 = 0.5.
3. Step 3: Error ≤ 6 · (0.5)⁴/4! = 6 · 0.0625/24 = 0.015625.
Answer: |f(1.5) − P₃(1.5)| ≤ 0.015625
Example 3
Trap: M can't use the value you're estimating
Use P₂(x) = 1 + x + x²/2 to approximate e^(0.5), and bound the error with the Lagrange error bound.
Show the solutionHide the solution
1. Step 1: P₂(0.5) = 1 + 0.5 + 0.125 = 1.625.
2. Step 2: f‴(t) = eᵗ, which on [0, 0.5] is largest at t = 0.5: e^(0.5). But that's the number you're trying to estimate, so use a simple bound: e^(0.5) = √e < √4 = 2. Take M = 2.
3. Step 3: Error ≤ 2 · (0.5)³/3! = 2 · 0.125/6 ≈ 0.0417.
4. Step 4: (Actual: e^(0.5) ≈ 1.64872, so the error is about 0.0237, within the bound.)
Answer: e^(0.5) ≈ 1.625, with error at most about 0.0417.

Common mistakes

Using the nth derivative instead of the (n + 1)th.
Evaluating the derivative only at the center instead of bounding it on the whole interval.
Forgetting the factorial, or using n! instead of (n + 1)!.
Confusing the Lagrange bound with the alternating series bound, or using the alternating bound on a series that doesn't alternate.

On the exam

Lagrange error bound questions on free response usually give you a bound on the needed derivative. Write the formula with numbers substituted: “error ≤ 6(0.5)⁴/4! = 0.015625.”
When asked to explain why an error is less than some value, show the bound and then say “which is less than …”.

Connected topics

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Check yourself

4 questions on 10.12 Lagrange Error Bound. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

The second-degree Taylor polynomial for f(x) = eˣ about x = 0 is used to approximate e^0.6. What is the Lagrange error bound for this approximation?

Question 2 of 4

The third-degree Maclaurin polynomial for f(x) = eˣ is used to approximate e¹. Using the fact that e < 3, what is the Lagrange error bound for this approximation?

Question 3 of 4Calculator allowed

The third-degree Taylor polynomial P₃(x) for f(x) = ln x about x = 1 is used to approximate ln 1.5. What is |ln 1.5 − P₃(1.5)|?

n	f⁽ⁿ⁾(2)
0	3
1	−2
2	7
3	−15

Selected derivatives of f at x = 2 (f⁽⁰⁾ means f)

Question 4 of 4Calculator allowed

For 2 ≤ x ≤ 2.35, |f⁽⁴⁾(x)| ≤ 60. Using M = 60, what is the Lagrange error bound for |f(2.35) − P₃(2.35)|?

0 of 4 answered