PropositionStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Differentiation of symmetric periodic pair potentials

Statement

Let $X$ be a nonempty finite set, let $w_{x y} = w_{y x}$ be arbitrary real numbers for $x, y \in X$ , and let $v : R \to R$ be even, twice continuously differentiable and $2 π$ -periodic. Even means $v (- t) = v (t)$ , and $2 π$ -periodic means $v (t + 2 π) = v (t)$ for every real $t$ . For $θ \in R^{X}$ define $U (θ) = \frac{1}{2} \sum_{x, y \in X} w_{x y} v (θ_{x} - θ_{y})$ . Write $(\nabla U)_{x} = \partial U / \partial θ_{x}$ and let $\nabla^{2} U$ be the matrix of second partial derivatives. Then, for every $f \in R^{X}$ ,

$(\nabla U)_{x} = \sum_{y \in X} w_{x y} v^{'} (θ_{x} - θ_{y}),$

$f^{T} \nabla^{2} U f = \frac{1}{2} \sum_{x, y \in X} w_{x y} v^{''} (θ_{x} - θ_{y}) (f_{x} - f_{y})^{2} .$

The terms with $x = y$ contribute zero to both displayed expressions. The identities hold for signed weights; nonnegativity is an additional hypothesis when needed for a later inequality.

Facts & Assumptions

[F1]

Finite sums of differentiable real-valued functions may be differentiated termwise (Phasors, trigonometric identities and periodic derivatives). For a real constant $c$ and a differentiable real function $g$ , the product rule gives $(c g)^{'} = c g^{'}$ : the constant function has derivative zero because all its difference quotients are zero (Sum, product, quotient, and chain rules from difference quotients).

[F2]

The derivative of a composite is the composite of its derivatives when the constituent maps are differentiable (The chain rule for total derivatives: D(g∘ f)(a)=Dg(f(a))∘ Df(a)).

[F3]

A coordinate partial derivative $\partial_{z} F (θ)$ is the derivative at $t = 0$ of the real line restriction $t \mapsto F (θ + t e_{z})$ , where $(e_{z})_{x} = δ_{x z}$ . Iterated partial derivatives apply this definition again to the first partial maps (Directional derivatives and partial derivatives of a map U⊆ℝᵐ→ℝⁿ).

[A1]

The set, weights, function and vectors satisfy the hypotheses in the Statement.

Proof

Given: $X$ , $w$ , $v$ , $θ$ and $f$ as in the Statement.

Goal: Prove the gradient and Hessian quadratic-form identities, including the diagonal and signed-weight assertions.

1.1A1F1F2F3algebra

Differentiate $v (- t) = v (t)$ using [F2]. This gives $- v^{'} (- t) = v^{'} (t)$ , so $v^{'}$ is odd and in particular $v^{'} (0) = 0$ . By [F3], compute coordinate derivatives on $θ + t e_{z}$ . On this line $θ_{x} - θ_{y}$ becomes $θ_{x} - θ_{y} + t (δ_{x z} - δ_{y z})$ , whose derivative is $δ_{x z} - δ_{y z}$ directly from its difference quotient; here $δ_{x z}$ is $1$ if $x = z$ and $0$ otherwise. By [F1] and [F2], $\partial_{z} U = \frac{1}{2} \sum_{x, y} w_{x y} v^{'} (θ_{x} - θ_{y}) (δ_{x z} - δ_{y z})$ .

2.1A1step 1.1algebra

The first Kronecker-delta sum is $\frac{1}{2} \sum_{y} w_{z y} v^{'} (θ_{z} - θ_{y})$ . In the second, exchange the two indices and use symmetry of $w$ and oddness of $v^{'}$ from step 1.1; it becomes the same sum. Their sum is the asserted gradient component. The term $y = z$ is zero by $v^{'} (0) = 0$ .

3.1A1F1F2F3step 1.1step 2.1algebra∎

Since $v$ is twice continuously differentiable, differentiating the first partial maps along coordinate lines as in [F3] and applying [F1] and [F2] gives $\partial_{a} \partial_{b} U = \frac{1}{2} \sum_{x, y} w_{x y} v^{''} (θ_{x} - θ_{y}) (δ_{x a} - δ_{y a}) (δ_{x b} - δ_{y b})$ . Multiplying by $f_{a} f_{b}$ and summing over $a, b$ gives the stated quadratic form, because $\sum_{a} f_{a} (δ_{x a} - δ_{y a}) = f_{x} - f_{y}$ . For $x = y$ this factor is zero. No step uses a sign restriction on $w$ , completing the proof.

Depends on

Used by · 8

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.