Lagrange Error Bound: Study Guide (Calculus BC 10.12)

BC-only material. The Lagrange error bound is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

Prerequisite	Topic	What you use it for here
Taylor polynomials	10.11	The approximation Pₙ(x) whose error we bound
Alternating series error bound	10.10	A second way to bound some Taylor errors
Greatest value of a function on an interval	5.2, 5.5	Finding M
Higher-order derivatives	3.6	Finding f⁽ⁿ⁺¹⁾

Review the previous topic, finding Taylor polynomial approximations, first if needed. Other prerequisites are on the Calculus BC roadmap.

The question: how wrong could the estimate be?

In Topic 10.11 you used Pₙ(x) to estimate f(x). On the exam you usually cannot check the estimate against the true value. You need a guaranteed limit on the error, worked out from information you do have.

The error (or remainder) is

Rₙ(x) = f(x) − Pₙ(x)

The Lagrange error bound gives a number E with |Rₙ(x)| ≤ E. Then the true value must lie in the interval [Pₙ(x) − E, Pₙ(x) + E].

The Lagrange error bound

If |f⁽ⁿ⁺¹⁾(z)| ≤ M for every z between a and x, then |f(x) − Pₙ(x)| ≤ M · |x − a|ⁿ⁺¹ / (n + 1)!

Read it piece by piece:

• n is the degree of the polynomial you used.
• f⁽ⁿ⁺¹⁾ is the next derivative, the first one the polynomial does not match.
• M is any upper bound for the size of that derivative on the interval between the centre a and the point x.
• |x − a|ⁿ⁺¹ / (n + 1)! looks exactly like the next Taylor term, without the derivative.

So the bound is “the next term, with the unknown derivative replaced by its worst case”. That is why it shrinks fast close to the centre (small |x − a|) and for high degree (large factorial).

Where it comes from (background). Taylor’s theorem says the error equals f⁽ⁿ⁺¹⁾(c)/(n + 1)! · (x − a)ⁿ⁺¹ for some number c between a and x. For n = 0 this is the Mean Value Theorem: f(x) − f(a) = f′(c)(x − a). We do not know c, so we replace |f⁽ⁿ⁺¹⁾(c)| by M, which is at least as big. You do not need to prove Taylor’s theorem; you need to use the bound.

Choosing M

M must be at least the largest value of |f⁽ⁿ⁺¹⁾(z)| for z between a and x. Common ways to find one:

Situation	How to get M	Example
sin or cos	The size of sin z or cos z is never more than 1	f⁽ⁿ⁺¹⁾ = ±sin z or ±cos z gives M = 1
The size of f⁽ⁿ⁺¹⁾ only increases or only decreases on the interval	Use the larger endpoint value	For f(x) = 1/(1 + x), the size of f‴(z) is 6/(1 + z)⁴, largest at the left end
eᶻ on [0, x]	eᶻ ≤ eˣ, then round up to a simple number	eᶻ ≤ e^0.6 < e < 3, so M = 3
The question gives a bound	Use it as given	“The size of f⁽⁴⁾(x) is at most 6 for 1 ≤ x ≤ 1.5” gives M = 6

A bigger M is always safe (the bound is still true, just weaker). A smaller M than the true maximum is wrong, even if the arithmetic is neat.

Worked example 1: √4.4 with a quadratic

Question. Let f(x) = √x. (a) Find P₂(x), the second-degree Taylor polynomial for f about x = 4, and use it to estimate √4.4. (b) Use the Lagrange error bound to show that the estimate differs from √4.4 by less than 0.0002. (c) Give an interval that must contain √4.4.

1. Derivatives. f(x) = x^(1/2), f′(x) = ½x^(−1/2), f″(x) = −¼x^(−3/2), f‴(x) = (3/8)x^(−5/2).
2. Values at 4. f(4) = 2, f′(4) = 1/4, f″(4) = −¼ · 1/8 = −1/32.
3. Polynomial. P₂(x) = 2 + ¼(x − 4) − (1/64)(x − 4)².
4. Estimate. x − 4 = 0.4: P₂(4.4) = 2 + 0.1 − (1/64)(0.16) = 2 + 0.1 − 0.0025 = 2.0975.
5. Find M (n = 2, so use f‴). On 4 ≤ z ≤ 4.4, f‴(z) = 3/(8z^(5/2)) is positive and decreasing, because z^(5/2) grows. So its largest value is at z = 4: M = 3/(8 · 32) = 3/256.
6. Bound. |f(4.4) − P₂(4.4)| ≤ (3/256) · 0.4³ / 3! = (3/256)(0.064)/6 = 1/8000 = 0.000125.
7. Compare and conclude. 0.000125 < 0.0002, so the estimate 2.0975 differs from √4.4 by less than 0.0002.
8. (c) Interval. √4.4 lies in [2.0975 − 0.000125, 2.0975 + 0.000125] = [2.097375, 2.097625].

Checks.

• Calculator comparison (where allowed): √4.4 ≈ 2.097618, so the actual error is about 0.000118, just under the bound 0.000125. The bound did its job.
• Over or under? By Taylor’s theorem the error is f‴(c)/3! · (0.4)³ for some c between 4 and 4.4. Both factors are positive, so √4.4 > P₂(4.4): the estimate is an underestimate. That narrows the interval to [2.0975, 2.097625]. This sign argument works only when f⁽ⁿ⁺¹⁾ keeps one sign on the interval.

Worked example 2: choosing the degree

Question. You want to estimate e^0.6 using a Maclaurin polynomial for eˣ, with error less than 0.001. Use the Lagrange error bound to find the smallest degree n that guarantees this.

1. Derivative. Every derivative of eˣ is eˣ, so f⁽ⁿ⁺¹⁾(z) = eᶻ.
2. Choose M. For 0 ≤ z ≤ 0.6, eᶻ ≤ e^0.6 < e¹ < 3. Take M = 3.
3. Bound for degree n: 3 · 0.6ⁿ⁺¹ / (n + 1)!.
4. Test values of n.

n	Bound 3 · 0.6ⁿ⁺¹ / (n + 1)!	Less than 0.001?
3	3(0.1296)/24 = 0.0162	No
4	3(0.07776)/120 ≈ 0.00194	No
5	3(0.046656)/720 ≈ 0.000194	Yes

5. Conclusion. The Lagrange bound guarantees the accuracy for n = 5, so use P₅(0.6) ≈ 1.822048.

A subtle point. The actual error of P₄(0.6) is about 0.000719, which is already less than 0.001. But the bound cannot guarantee it: even with the tightest M = e^0.6 ≈ 1.822, the n = 4 bound is about 0.00118. The bound is a guarantee, so it is always at least as big as the true error, often bigger. Answer the question that is asked: “guaranteed by the Lagrange error bound” means n = 5.

When M is given: communicating the comparison

Exam questions often give a derivative bound directly. Example. P₃ is the third-degree Taylor polynomial for f about x = 1, and |f⁽⁴⁾(x)| ≤ 6 for 1 ≤ x ≤ 1.5. Show that |f(1.5) − P₃(1.5)| < 0.02.

• Lagrange: |f(1.5) − P₃(1.5)| ≤ 6 · (0.5)⁴ / 4! = 6(0.0625)/24 = 0.015625.
• Comparison sentence: 0.015625 < 0.02, so |f(1.5) − P₃(1.5)| < 0.02.

Both lines matter. Showing the bound and then stating that it is less than the target is what “show that” requires.

The alternating series error bound as an alternative

Sometimes the Taylor polynomial is a partial sum of a series whose terms alternate in sign, decrease in size and approach 0. Then the error is at most the size of the first omitted term (Topic 10.10), and the sign of that term tells you whether the estimate is too high or too low.

Example: sin 0.4 with P₃(x) = x − x³/6.

• The Maclaurin series for sin x is x − x³/3! + x⁵/5! − … (Topic 10.14 shows it equals sin x). At x = 0.4 the terms alternate, shrink and tend to 0.
• Alternating series bound: the first omitted term is 0.4⁵/120 ≈ 0.0000853. It is positive, so P₃(0.4) ≈ 0.389333 is an underestimate.
• Lagrange with n = 3: |f⁽⁴⁾(z)| = |sin z| ≤ 1, giving 0.4⁴/24 ≈ 0.00107. That is valid but weaker.
• Lagrange with n = 4: the x⁴ coefficient of sin x is 0, so P₃ = P₄. Then |f⁽⁵⁾(z)| = |cos z| ≤ 1 gives 0.4⁵/120 ≈ 0.0000853, the same as the alternating bound.
• Calculator check (where allowed): sin 0.4 ≈ 0.389418, actual error ≈ 0.0000850. Both bounds hold.

When you cannot use the alternating bound. For eˣ at x = 0.6, every term 0.6ᵏ/k! is positive: nothing alternates, so use Lagrange. Always state the conditions (alternating, decreasing in size, limit 0) if you use the alternating bound.

Common misconceptions

• Using the wrong derivative. For Pₙ you need f⁽ⁿ⁺¹⁾, not f⁽ⁿ⁾.
• Using n! instead of (n + 1)!, or |x − a|ⁿ instead of |x − a|ⁿ⁺¹.
• Taking M at the centre without checking. If |f⁽ⁿ⁺¹⁾| is largest at the other end of the interval, the value at a is too small and the “bound” may be false.
• Taking M at x only. M must cover every z between a and x.
• Thinking the bound is the actual error. It is an upper limit. The true error is usually smaller.
• Leaving out the comparison. “Show the error is less than 0.02” needs the number and the sentence “which is less than 0.02”.
• Using the alternating series bound on non-alternating terms, or without checking that the terms decrease.
• Forgetting the absolute values. The bound limits the size of the error in either direction.

Where this leads

So far each estimate uses a polynomial. Topic 10.13, radius and interval of convergence of power series, asks what happens when the polynomial goes on forever: for which x does the series converge? Topic 10.14 then writes full Taylor and Maclaurin series. See the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.