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A uniform derivative bound gives a uniform Taylor remainder bound

Statement

Let $n\in\mathbb N$ , and suppose $f$ has derivatives through order $n+1$ on the closed interval between $a$ and $x$ , with the usual endpoint continuity. If $|f^{(n+1)}(t)|\le M$ throughout that interval, then $|R_{n,a}f(x)|\le \frac{M}{\iota((n+1)!)}|x-a|^{n+1}.$

Facts & Assumptions

Given: The stated regularity and derivative bound.

[L1]

The Lagrange remainder formula is The Lagrange and Cauchy forms of Taylor's remainder.

[L2]

Absolute value respects products and powers (Basic properties of the absolute value), and factorials are positive (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ , Canonical naturals are positive and strictly increasing, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Proof

technique · direct

1.1

If $x=a$ , then $R_{n,a}f(a)=0$ , so the estimate is immediate. If $x\ne a$ , [L1] gives $R_{n,a}f(x)=\frac{f^{(n+1)}(\xi)}{\iota((n+1)!)}(x-a)^{n+1}$ for some point $\xi$ strictly between $a$ and $x$ .

2.1

In the case $x\ne a$ , take absolute values in step 1.1, use $|f^{(n+1)}(\xi)|\le M$ , and divide by the positive factorial. Together with the case $x=a$ , this proves the estimate.

step 1.1L2algebra∎

Depends on

Used by

The Taylor polynomial of (1-x)⁻¹ at 0 has the exact geometric remainder xⁿ⁺¹/(1-x) Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)
Taylor's theorem (Wikipedia): Lagrange remainder and error estimate (standard reference, not scraped)