Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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.

Euler global state and log-density error on a bounded C² tube

Statement

Assume Countable Choice and the flow-density hypotheses of A continuous flow changes log density by minus its divergence. Let T>0, h=T/N, N≥1, and tn=nh. Let K⊂Rd be compact and convex, and let U⊂R1+d be a bounded convex open subset of the ODE domain containing [0,T]×K. Suppose f∈C2(U,Rd), all partial derivatives of f through order two are bounded on U, and the exact path and every Euler node lie in K. Thus [0,T]×K is a compact tube with positive margin inside U. Put q(t,x):=−div⁡xf(t,x) and define Bf:=sup⁡U∥f∥2,Lf:=sup⁡U∥Dxf∥,Mf:=sup⁡U∥∂tf∥2,Lq:=sup⁡U∥Dxq∥,Mq:=sup⁡U∣∂tq∣, where matrix and linear-functional norms are induced by Euclidean norms. These constants are finite by the hypotheses. Apply forward Euler to the state and to the continuous-flow log-density equation: xn+1=xn+hf(tn,xn),ℓ^n+1=ℓ^n+hq(tn,xn). If ∥x0−x(0)∥2≤Cx,0h and ∣ℓ^0−log⁡p0(x(0))∣≤Cℓ,0h, define Cx:=Mf+LfBf2,Dq:=Mq+LqBf2, and Kx:={Cx,0exp⁡(LfT)+Cxexp⁡(LfT)−1Lf,Lf>0,Cx,0+CxT,Lf=0. Then, for every 0≤n≤N, ∥xn−x(tn)∥2≤Kxh, and ∣ℓ^n−log⁡ptn(x(tn))∣≤(Cℓ,0+TLqKx+DqT)h. This is a first-order Euler result; it asserts no adjoint-gradient error bound.

Facts & Assumptions

Given: Countable Choice, a positive initial density and C¹ diffeomorphic flow, f∈C2 with the stated bounded derivatives on a common convex tube, and the displayed Euler state and scalar updates.

[A1]

The state and log-density initial errors are at most Cx,0h and Cℓ,0h, respectively; exact, numerical, and local flow segments stay in the stated tube.

[A2]

Countable Choice is used only to identify the continuous flow's pushforward density (The Axiom of Countable Choice (ACω)).

[F1]

A solution of x′=f(t,x) satisfies x(t)=x(t0)+∫t0tf(s,x(s)) ds (A first-order initial value problem is equivalent to its Volterra integral equation).

[F2]

The derivative of a composite is the composite of the derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[F3]

On a convex open set, a uniform derivative bound ∥Df(z)v∥2≤M∥v∥2 implies a Lipschitz bound ∥f(y)−f(x)∥2≤M∥y−x∥2 (On a convex open set, a uniform bound ∥Df(z)v∥2≤M∥v∥2 implies ∥f(y)−f(x)∥2≤M∥y−x∥2).

[F5]

A uniform one-step defect Chp+1 and stability factor 1+Lh give the stated fixed-time global error bound, including the L=0 case (Global error from uniform one-step defect and stability).

[F6]

If the divergence update has state sensitivity hKq and local quadrature defect Dhm+1, then its global error is the sum of the initial, state, and quadrature contributions (Global continuous-flow log-density error from state and quadrature bounds).

[F7]

exp⁡ is the real exponential defined by its factorial power series, with e:=exp⁡(1) (The real exponential function and the number e by a power series).

Proof

1.1A1F3algebra

For fixed tn, the open convex fiber Utn contains K. Applying [F3] there gives ∥f(tn,u)−f(tn,v)∥2≤Lf∥u−v∥2 for u,v∈K. Hence the Euler state map satisfies ∥u+hf(tn,u)−v−hf(tn,v)∥2≤(1+Lfh)∥u−v∥2.

1.2A1F1F2F4algebra

Let r(s):=f(tn+s,x(tn+s)) for 0≤s≤h. By [F2], r′(s)=∂tf(tn+s,x(tn+s))+Dxf(tn+s,x(tn+s))f(tn+s,x(tn+s)), whose norm is at most Mf+LfBf. By [F4], ∥r(s)−r(0)∥2≤(Mf+LfBf)s. Using [F1] and the integral norm bound in [F4], the exact-start Euler defect is ∥x(tn+1)−x(tn)−hf(tn,x(tn))∥2=∥∫0h(r(s)−r(0)) ds∥2≤Cxh2.

1.3A1F2F3F4algebra

Since f∈C2(U), q=−div⁡xf is C1 on U. The open convex fiber Utn contains K, so [F3] gives ∣hq(tn,u)−hq(tn,v)∣≤hLq∥u−v∥2 for u,v∈K. Along the exact trajectory, [F2] gives ddsq(tn+s,x(tn+s))=∂tq+Dxq f, with absolute value at most Mq+LqBf, hence ∣q(tn+s,x(tn+s))−q(tn,x(tn))∣≤(Mq+LqBf)s by [F4]. The left-Euler exact-start defect is ∣∫0h(q(tn+s,x(tn+s))−q(tn,x(tn))) ds∣, which [F4] bounds by (Mq+LqBf)h2/2=Dqh2.

2.1A1F5F7step 1.1step 1.2

The Euler state map from step 1.1 and defect from step 1.2 satisfy Global error from uniform one-step defect and stability with order p=1, C=Cx, L=Lf, and initial constant Cx,0. Therefore max⁡n∥xn−x(tn)∥2≤Kxh, with the stated formula when Lf>0 and its separate zero branch when Lf=0.

3.1A2F6step 2.1step 1.3∎

Apply [F6] with state order p=1, quadrature order m=1, state constant Kx from step 2.1, sensitivity Kq=Lq, defect constant D=Dq, and initial log-density error Cℓ,0h. This yields ∣ℓ^n−log⁡ptn(x(tn))∣≤(Cℓ,0+TLqKx+DqT)h for every grid index.

Depends on

Used by

Dependency tree · two levels

79 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources