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Multivariable Taylor formula with $o(\|h\|^k)$ remainder

Statement

Let $k\in\mathbb N$ with $k\ge1$ , let $V\subseteq\mathbb R^m$ be open and convex, let $a\in V$ , and let $f\in C^k(V)$ . Then, as $h\to0$ with $a+h\in V$ ,

$f(a+h)=T_kf(a;h)+o(\|h\|^k).$

Facts & Assumptions

Given: The hypotheses of the statement and small $h$ with $a+h\in V$ .

[L1]

By Multivariable Taylor formula with a Lagrange remainder along a line segment, applying the multivariable Lagrange formula with degree $k-1$ gives some $\theta_h\in(0,1)$ such that

$f(a+h)=T_{k-1}f(a;h)+\sum_{|\alpha|=k}\frac{D^\alpha f(a+\theta_hh)}{\iota(\alpha!)}h^\alpha$

[L2]

For every $|\alpha|=k$ , $D^\alpha f$ is continuous at $a$ ( $C^k$ maps and multi-index derivative notation in Euclidean space).

Proof

technique · direct

1.1

Subtract the degree- $k$ Taylor polynomial from the equality of [L1]. The remainder is the following.

L1algebra

$\sum_{|\alpha|=k}\frac{D^\alpha f(a+\theta_hh)-D^\alpha f(a)}{\iota(\alpha!)}h^\alpha.$

2.1

For $h\ne0$ , divide the absolute value in step 1.1 by $\|h\|^k$ . Since $|h^\alpha|\le\|h\|^k$ , it is bounded by the finite sum of the coefficient differences divided by $\iota(\alpha!)$ . As $h\to0$ , also $a+\theta_hh\to a$ , so every term tends to zero by [L2].

step 1.1L2algebra

3.1

This proves that the remainder in step 1.1 is $o(\|h\|^k)$ , and the subtracted polynomial is $T_kf(a;h)$ by The multivariable Taylor polynomial in multi-index notation.

step 2.1∎

Depends on

Used by

Second-order Taylor expansion f(a+h)=f(a)+∇ f(a)· h+tfrac12h^TH_f(a)h+o(‖h‖²) Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 54 results over 17 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

MAT237 notes: Taylor's theorem in several variables (standard reference, not scraped)

Multivariable Taylor formula with o(∥h∥k)o(\|h\|^k)o(∥h∥k) remainder