Lagrange Error Bound Calculator (Taylor polynomial remainder estimate)

Understanding the Lagrange error bound calculator — concept and principle

This calculator bounds the error of a Taylor polynomial approximation with the Lagrange remainder. For sin x, cos x, eˣ or your own bound M, it gives the error bound at a point, the Taylor polynomial value, the true value, the actual error and the smallest degree that meets a tolerance. Use it for calculus homework on Taylor series and numerical approximation.

The bound uses the largest possible next derivative, so the true error is usually smaller.

How it is calculated

f(x) = Pₙ(x) + Rₙ(x), with Rₙ(x) = f⁽ⁿ⁺¹⁾(c) ÷ (n + 1)! × (x − a)ⁿ⁺¹ for some c between a and x

Lagrange error bound: |Rₙ(x)| ≤ M × |x − a|ⁿ⁺¹ ÷ (n + 1)!

M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x. For sin x and cos x, M = 1.
The smallest degree n with bound ≤ tolerance is found by trying n = 0, 1, 2, …

The principle

A Taylor polynomial matches a function's value and first n derivatives at the centre a. The error that remains behaves like the next term of the series, but with the derivative evaluated at some unknown point c. Since c is unknown, using the largest possible size of that derivative, M, gives a guaranteed upper limit. The factorial in the denominator grows quickly, which is why a few more terms usually shrink the error dramatically near the centre.

Worked example

Default inputs: f(x) = sin x, a = 0, x = 0.5, n = 3, M = 1, tolerance 0.0001.

(n + 1)! = 4! = 24; |x − a|⁴ = 0.5⁴ = 0.0625; M used 1.
Error bound = 1 × 0.0625 ÷ 24 = 0.00260417.
Taylor polynomial P₃(0.5) = 0.5 − 0.5³ ÷ 6 = 0.4791666667.
True value sin 0.5 = 0.4794255386; actual error = 0.0002589, well inside the bound.
Smallest degree with bound ≤ 0.0001: 5 (bound 0.5⁶ ÷ 720 ≈ 0.0000217).

Each point is the calculator re-run with only “Point x where you approximate” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Point x where you approximate	Error bound \|Rₙ(x)\| ≤
0.125	0.0000101725
0.25	0.00016276
0.375	0.000823975
0.5	0.00260417
0.625	0.00635783
0.75	0.0131836
1	0.0416667

How to read the result

The bound is a worst case. Here the actual error is about ten times smaller, because for sin x the degree-3 and degree-4 polynomials are the same (the x⁴ coefficient is 0). Use the bound to decide how many terms you need for a guaranteed accuracy, not to estimate the exact error.

Common mistakes

Using the n-th derivative instead of the (n + 1)-th.
Choosing M at the centre only, instead of the largest value on the whole interval.
Forgetting the factorial.
Applying the bound outside the interval where M was found.

Keep in mind: For eˣ and other growing derivatives, M must be taken at the end of the interval farther from zero; a too-small M gives a false guarantee.

Frequently asked questions

What is the Lagrange error bound?

|Rₙ(x)| ≤ M|x − a|ⁿ⁺¹ ÷ (n + 1)!, a guaranteed upper limit on a Taylor polynomial's error.

How do I choose M?

Take the largest absolute value of the (n + 1)-th derivative between a and x.

Why is the actual error smaller than the bound?

The bound assumes the worst possible derivative value; the true one is usually smaller.

Sources: OpenStax, Calculus Volume 2 (2016), §6.3 Taylor and Maclaurin Series

Checked 9 Nov 2026. OpenStax, Calculus Volume 2 (2016), §6.3 Taylor and Maclaurin Series — “Taylor’s Theorem with Remainder” (Rₙ(x) = f⁽ⁿ⁺¹⁾(c)/(n + 1)! · (x − a)ⁿ⁺¹ for some c between a and x; if |f⁽ⁿ⁺¹⁾| ≤ M then |Rₙ(x)| ≤ M/(n + 1)! · |x − a|ⁿ⁺¹), as reproduced in Lumen Learning Calculus II.

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