Error Bounds

You’ll be able to

Bound the error of an alternating series by the first omitted term
Apply the Lagrange error bound for a Taylor polynomial
Interpret an error bound as a guarantee on approximation accuracy

How good is the approximation?

When you truncate a series to finitely many terms, you introduce an error — the difference between the true value and the partial sum. Two tools bound this error. For an alternating series that meets the alternating series test, the error is remarkably simple to control. For a general Taylor polynomial, the Lagrange error bound provides the guarantee. Both give an upper bound on |error|, certifying the approximation is at least that accurate.

Alternating series error bound

| S − Sₙ | ≤ | a_{n+1} |

For a convergent alternating series, the error after n terms is no larger than the absolute value of the first omitted term.

The alternating series remainder

For a series that alternates in sign with terms decreasing in magnitude to 0, the truncation error is at most the first term you left out: |error| ≤ |a_{n+1}|. This is astonishingly clean — no derivatives, no factorials to bound, just the size of the next term. It is the go-to method whenever the series in question alternates, such as the Maclaurin series for sin x, cos x, or arctan.

Lagrange error bound

| Rₙ(x) | ≤ ( M / (n+1)! ) · | x − a |ⁿ⁺¹

M is a bound on |f^(n+1)| on the interval between a and x. Controls the error of the nth-degree Taylor polynomial.

The Lagrange bound for any Taylor polynomial

When the series is not alternating (or you want a general guarantee), the Lagrange error bound applies: the remainder Rₙ(x) after an nth-degree Taylor polynomial is at most (M/(n+1)!)·|x − a|ⁿ⁺¹, where M is the maximum of the (n+1)th derivative’s magnitude on the interval between a and x. The factorial in the denominator makes this bound shrink fast as you add terms, which is why Taylor approximations are so accurate near the center.

Worked example

The Maclaurin series for cos x is used to approximate cos(0.5) with terms up to x². Bound the error using the alternating series estimate.

1.The series cos x = 1 − x²/2! + x⁴/4! − … alternates, so the error is at most the first omitted term.
2.Keeping terms through x² means the first omitted term is x⁴/4!.
3.Evaluate that term at x = 0.5: (0.5)⁴/4! = 0.0625/24.
4.Compute: 0.0625/24 ≈ 0.0026.

Answer: The error is at most about 0.0026 — the size of the first omitted term x⁴/4! at x = 0.5. The true approximation is that accurate or better.

Watch out

The alternating series bound is the first omitted term, a_{n+1}, not the last term you kept. Off-by-one here is the most common error-bound mistake: count carefully which term is the first one left out.

Checkpoint

An alternating series is approximated by its first 4 terms. The 5th term is 0.02 and the 6th is 0.008. The error is at most:

Checkpoint

In the Lagrange error bound |Rₙ(x)| ≤ (M/(n+1)!)·|x − a|ⁿ⁺¹, what does M represent?

On the exam

Choose the bound that fits: if the series alternates with decreasing terms, use the simple "first omitted term" bound. Otherwise reach for Lagrange, and take M as the maximum of the (n+1)th derivative on the relevant interval — often just its largest possible value there.

Answer the 2 checkpoints as you read.