Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Gradient of a finite sum is the finite sum of gradients

Statement

Let U be open in R^d, let x in U, let n>=1, and let f_1,...,f_n:U->R be totally differentiable at x. Put F=n^{-1}sum_{i=1}^n f_i. Then F is totally differentiable at x and

∇F(x)=1n∑i=1n∇fi(x).

Facts & Assumptions

Given: The open domain U, x, n, and the functions in the Statement.

[F1]

Finite sums and finite products, by recursion: finite sums are recursively defined, with each successor sum equal to the preceding sum plus its next term.

[F2]

Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives: the sum and scalar multiple of totally differentiable maps are totally differentiable, with derivatives equal to the corresponding sum and scalar multiple.

[F3]

A total derivative computes every directional derivative, and its matrix is the Jacobian: for each standard basis vector e_j, the partial derivative is DF(x)e_j, and the matrix of DF(x) is the Jacobian.

[F4]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: the gradient of a scalar function is the vector of its partial derivatives.

[F5]

The principle of mathematical induction: a property true at zero and inherited by each successor holds for every natural index.

[F6]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: vectors in R^d are equal when their d coordinates agree.

[F7]

The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder: total differentiability means a linear first-order approximation with a remainder whose norm divided by ||h||_2 tends to zero.

[F8]

Canonical naturals are positive and strictly increasing: the canonical real image of every natural n>=1 is positive.

Proof

Given: The domain and functions in the Statement.

1.1F1F2F5F7givenalgebrainduction

For 0<=m<=n define S_m:U->R by S_m(y)=sum_{i<m}f_{i+1}(y), using [F1]. The base S_0 is the zero function, whose total derivative is the zero linear map directly from [F7]. If S_m is totally differentiable at x with derivative sum_{i<m}Df_{i+1}(x), then S_{m+1}=S_m+f_{m+1}; [F2] makes it differentiable with derivative sum_{i<m}Df_{i+1}(x)+Df_{m+1}(x)=sum_{i<m+1}Df_{i+1}(x). Induction [F5] proves this formula through m=n.

2.1F1F2F8step 1.1givenalgebra

Since the canonical real image of n is positive by [F8], it is nonzero; the field structure in [F1] therefore supplies its inverse n^{-1}. The scalar-multiple rule [F2] gives DF(x)=n^{-1}sum_{i<n}Df_{i+1}(x).

3.1F3F4F6step 2.1givenalgebra∎

For each coordinate j, [F3] gives partial_j F(x)=DF(x)e_j=n^{-1}sum_{i<n}Df_{i+1}(x)e_j=n^{-1}sum_{i<n}partial_j f_{i+1}(x). By [F4] these are exactly the coordinates of n^{-1}sum_{i=1}^n grad f_i(x); [F6] then identifies the vectors. Hence the displayed gradient formula holds.

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Sources