Lagrange Error Bound

The Lagrange error bound tells you how far a Taylor polynomial can be from its function. If |f^{(n+1)}(z)| \le M for every z between a and x, then the error of the nth-degree Taylor polynomial is at most |R_n(x)| \le \dfrac{M}{(n+1)!}|x - a|^{n+1}.

Before this lesson: Taylor Polynomials, Alternating Series Error Bound and Higher-Order Derivatives

Where the error comes from

The remainder R_n(x) = f(x) - P_n(x) is the error of a Taylor polynomial. Taylor’s theorem says the remainder looks just like the next term of the series, except that the derivative is taken at some unknown point z between a and x:

R_n(x) = \dfrac{f^{(n+1)}(z)}{(n+1)!}(x - a)^{n+1}

You can’t find z, but you can find the largest value M that |f^{(n+1)}| reaches between a and x. Using M in place of the unknown derivative gives an upper bound on the error.

The Lagrange error bound formula

|R_n(x)| = |f(x) - P_n(x)| \le \dfrac{M}{(n+1)!}|x - a|^{n+1}

• n is the degree of the Taylor polynomial, and a is its center.
• M is any number with |f^{(n+1)}(z)| \le M for every z between a and x. It is usually the largest value of the (n + 1)th derivative on that interval.
• The bound uses the next derivative and the next factorial: (n + 1), not n.

How to use the Lagrange error bound

1. Find the (n + 1)th derivative of f.
2. Find M, the largest value of its absolute value between a and x. A simple number that is a bit larger also works.
3. Put M, n + 1 and |x − a| into the formula.
4. To find the degree you need, try n = 1, 2, 3, … until the bound is small enough.

Lagrange error bound examples

Example 1: Estimating e to the 0.5

Use the third-degree Maclaurin polynomial P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} to estimate e^{0.5}, and bound the error.

P_3(0.5) = 1 + 0.5 + 0.125 + 0.0208333 \approx 1.6458333

f^{(4)}(z) = e^z

e^z \le e^{0.5} \lt 2 \text{ for } 0 \le z \le 0.5

|R_3(0.5)| \le \dfrac{2}{4!}(0.5)^4

\dfrac{2}{24}(0.0625) \approx 0.0052

Answer: e^{0.5} \approx 1.64583 with error at most 0.0052. (The true value is 1.64872, so the actual error is about 0.0029.)

Example 2: A sine estimate

Use P_3(x) = x - \frac{x^3}{6} to estimate sin 0.3, and bound the error with the Lagrange error bound.

P_3(0.3) = 0.3 - \dfrac{0.027}{6} = 0.3 - 0.0045 = 0.2955

f^{(4)}(z) = \sin z, \; |\sin z| \le 1

|R_3(0.3)| \le \dfrac{1}{4!}(0.3)^4 = \dfrac{0.0081}{24}

\dfrac{0.0081}{24} = 0.0003375

Answer: sin 0.3 ≈ 0.2955 with error at most 0.0003375. (The true value is 0.2955202.)

Example 3: Centered at a = 1

Use P_2(x) = (x - 1) - \frac{(x - 1)^2}{2}, the second-degree Taylor polynomial of ln x at a = 1, to estimate ln 1.2. Bound the error.

P_2(1.2) = 0.2 - \dfrac{0.04}{2} = 0.18

f'''(z) = \dfrac{2}{z^3}

\dfrac{2}{z^3} \le \dfrac{2}{1^3} = 2 \text{ for } 1 \le z \le 1.2

|R_2(1.2)| \le \dfrac{2}{3!}(0.2)^3 = \dfrac{2(0.008)}{6}

\dfrac{0.016}{6} \approx 0.00267

Answer: ln 1.2 ≈ 0.18 with error at most 0.00267. (The true value is 0.18232.)

Example 4: Choosing the degree

What degree of Maclaurin polynomial for e^x is needed to estimate e (that is, x = 1) with error less than 0.001?

M = 3

\dfrac{3}{(n+1)!}(1)^{n+1} \lt 0.001

(n + 1)! \gt 3000

6! = 720, \; 7! = 5040

n + 1 = 7, \text{ so } n = 6

Answer: a sixth-degree polynomial. P_6(1) = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720} \approx 2.71806, and e ≈ 2.71828.

Example 5: A pendulum simulation (word problem)

A pendulum simulation replaces cos θ with 1 - \frac{\theta^2}{2}, the second-degree Maclaurin polynomial, for angles up to 0.2 radians. How large can the error be?

f'''(z) = \sin z

|\sin z| \le 1

|R_2(\theta)| \le \dfrac{1}{3!}|\theta|^3 \le \dfrac{(0.2)^3}{6}

\dfrac{(0.2)^3}{6} = \dfrac{0.008}{6} \approx 0.00133

Answer: the error is at most about 0.00133 for every angle up to 0.2 radians. The true largest error is only about 0.00007, so the bound is safe but cautious.

Common mistakes

Related methods

Lagrange or alternating series error bound?

If the Taylor series alternates at your x and its terms shrink, such as the series for sin x or cos x, the alternating series error bound is often easier: the error is less than the next nonzero term. For series that don’t alternate, such as the one for e^x at a positive x, use the Lagrange error bound.

A sharper bound for sin and cos

The Maclaurin polynomials of sin x skip every even power, so P_3 = P_4. You may treat x - \frac{x^3}{6} as P₄ and use the fifth derivative. For sin 0.3 this gives \frac{0.3^5}{5!} \approx 0.00002, a much smaller bound.

Practice problems

1. Use P_2(x) = 1 + x + \frac{x^2}{2} to estimate e^{0.2}, and bound the error. Use e^0.2 < 2.
2. Use P_4(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} to estimate cos 0.5, and bound the error.
3. Use P_1(x) = 2 + \frac{x - 4}{4} for \sqrt{x} at a = 4 to estimate \sqrt{4.1}, and bound the error.
4. What degree of Maclaurin polynomial for sin x guarantees an error less than 0.0001 at x = 1? Use M = 1.
5. |f^{(4)}(x)| \le 6 on 1 ≤ x ≤ 3. P₃ is the third-degree Taylor polynomial of f at a = 2. Bound the error of P₃(2.5).
6. Use P_3(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} to estimate ln 1.1, and bound the error.
7. Use P_2(x) = 1 - x + \frac{x^2}{2} (for e^{-x} at 0) to estimate e^{-0.1}, and bound the error.
8. Show that sin x ≈ x has error at most 0.001 when |x| ≤ 0.04.
9. Bound the largest error of P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} for e^x on −1 ≤ x ≤ 1. Use e < 3.
10. A rocket’s height is approximated by its second-degree Taylor polynomial at t = 0 for 0 ≤ t ≤ 2 seconds. If |h'''(t)| \le 1.5 m/s³ on that interval, what is the largest possible error?

P_2(0.2) = 1 + 0.2 + 0.02 = 1.22

|R_2(0.2)| \le \dfrac{2}{3!}(0.2)^3 = \dfrac{2(0.008)}{6}

\dfrac{0.016}{6} \approx 0.00267

P_4(0.5) = 1 - 0.125 + \dfrac{0.0625}{24} \approx 0.8776042

|R_4(0.5)| \le \dfrac{1}{5!}(0.5)^5 = \dfrac{0.03125}{120}

\dfrac{0.03125}{120} \approx 0.00026

P_1(4.1) = 2 + \dfrac{0.1}{4} = 2.025

f''(z) = -\dfrac{1}{4z^{3/2}}

|f''(z)| \le \dfrac{1}{4 \cdot 4^{3/2}} = \dfrac{1}{32}

|R_1(4.1)| \le \dfrac{1/32}{2!}(0.1)^2 = \dfrac{0.01}{64}

\dfrac{0.01}{64} \approx 0.000156

\dfrac{1}{(n+1)!} \lt 0.0001

(n + 1)! \gt 10{,}000

7! = 5040, \; 8! = 40{,}320

n + 1 = 8

|R_3(2.5)| \le \dfrac{6}{4!}(0.5)^4

\dfrac{6}{24}(0.0625) = 0.015625

P_3(1.1) = 0.1 - 0.005 + 0.0003333 \approx 0.0953333

f^{(4)}(z) = -\dfrac{6}{z^4}

|f^{(4)}(z)| \le 6

|R_3(1.1)| \le \dfrac{6}{4!}(0.1)^4 = 0.25(0.0001) = 0.000025

P_2(0.1) = 1 - 0.1 + 0.005 = 0.905

|f'''(z)| = e^{-z} \le 1 \text{ for } 0 \le z \le 0.1

|R_2(0.1)| \le \dfrac{1}{3!}(0.1)^3 = \dfrac{0.001}{6} \approx 0.000167

|R_1(x)| \le \dfrac{1}{2!}|x|^2

\dfrac{(0.04)^2}{2} = \dfrac{0.0016}{2} = 0.0008

M = 3

|R_3(x)| \le \dfrac{3}{4!}|x|^4 \le \dfrac{3}{24}

|R_2(t)| \le \dfrac{1.5}{3!}t^3

\dfrac{1.5}{6}(2)^3 = 0.25(8) = 2

Lagrange error bound FAQ

What is the Lagrange error bound?

It is an upper limit on the error of a Taylor polynomial: |R_n(x)| \le \frac{M}{(n+1)!}|x - a|^{n+1}, where M is the largest value of |f^{(n+1)}| between a and x.

How do you find M?

Find the (n + 1)th derivative, then find its largest absolute value between a and x. For sin and cos you can use M = 1. For an increasing function like e^x, use the value at the right end.

What is the difference between the Lagrange and alternating series error bounds?

The alternating series bound works only for alternating series and uses the next term. The Lagrange bound works for any Taylor polynomial but needs a bound M on a derivative.

Why does the formula use (n + 1)! and not n!?

The error acts like the first term the polynomial leaves out. That term has degree n + 1, so it has (n + 1)! in the denominator.

Summary

• Error of Pₙ: |R_n(x)| \le \frac{M}{(n+1)!}|x - a|^{n+1}.
• M bounds the (n + 1)th derivative between a and x.
• Use the next derivative and the next factorial.
• The bound is a ceiling; the real error is usually smaller.
• Raise n until the bound is small enough to reach a target accuracy.

Previous lesson: Alternating Series Error Bound

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