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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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Multivariable Taylor formula with o(∥h∥k) remainder

Statement

Let k∈N with k≥1, let V⊆Rm be open and convex, let a∈V, and let f∈Ck(V). Then, as h→0 with a+h∈V,

f(a+h)=Tkf(a;h)+o(∥h∥k).

Facts & Assumptions

Given: The hypotheses of the statement and small h with a+h∈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 θh∈(0,1) such that

f(a+h)=Tk−1f(a;h)+∑∣α∣=kDαf(a+θhh)ι(α!)hα

[L2]

For every ∣α∣=k, Dαf is continuous at a (Ck 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

∑∣α∣=kDαf(a+θhh)−Dαf(a)ι(α!)hα.

2.1

For h≠0, divide the absolute value in step 1.1 by ∥h∥k. Since ∣hα∣≤∥h∥k, it is bounded by the finite sum of the coefficient differences divided by ι(α!). As h→0, also a+θhh→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 Tkf(a;h) by The multivariable Taylor polynomial in multi-index notation.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources