Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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Repeated derivatives along a line expand by the multinomial formula

Statement

Let k∈N, let U⊆Rm be open, f∈Ck(U), and let I⊆R be an open interval such that a+th∈U for every t∈I. Write ι:N→R for the canonical-natural map of The canonical natural ι(n)=n⋅1F of a field. For g(t)=f(a+th) and every 0≤r≤k,

g(r)(t)=∑∣α∣=rι(r!)ι(α!)Dαf(a+th)hα(t∈I).

Facts & Assumptions

Given: The stated open-domain, open-interval, Ck, and direction hypotheses.

[L1]

A function with continuous first partial derivatives near a point is totally differentiable there, and the total chain rule then applies to the affine line map t↦a+th (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[L2]

Ordered mixed derivatives through order k commute under permutation (Continuous mixed partials of order k are invariant under permutations).

[L3]

The multi-index conventions ∣α∣, α!, hα, and the canonical derivative Dαf are those of Ck maps and multi-index derivative notation in Euclidean space.

[L4]

The canonical-natural map carries finite natural sums and products to the corresponding real sums and products (The canonical natural ι(n)=n⋅1F of a field, Laws of finite sums and products in N, and ι(∑k<nak)=∑k<nι(ak)).

Proof

technique · induction
1.1

For r=0 the displayed sum consists of the zero multi-index and equals f(a+th)=g(t).

baseL3
1.2

Fix r<k and assume the formula at order r.

ih
1.3

Each Dαf with ∣α∣=r has continuous first partials, so [L1] differentiates its composition with the affine line. By [L3], we use the canonical multi-index notation for the resulting derivatives. When r=0 the resulting first derivatives are already canonical; when r≥1, [L2] permits the resulting derivatives to be written as Dα+eif. By [L4], collecting the coefficient of a fixed β with ∣β∣=r+1 gives

∑i: βi>0ι(r!)ι((β−ei)!)=ι(r!)ι(β!)∑i<mι(βi)=ι((r+1)!)ι(β!).

Thus the formula at order r+1 follows. [step 1.2, L1, L2, L3, L4, algebra]

2.1

Steps 1.1--2.1 prove the formula successively for every r≤k.

step 1.1step 1.2step 1.3discharge-induction∎

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