Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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Gaussian product theorem and finite polynomial shifts

Statement

Let a,b>0, A,B∈R3, p=a+b, μ=ab/p, and P=(aA+bB)/p. Then for every x∈R3, exp⁡(−a∥x−A∥22)exp⁡(−b∥x−B∥22)=exp⁡(−μ∥A−B∥22)exp⁡(−p∥x−P∥22). For any multi-indices α,β∈N03, the polynomial factor (x−A)α(x−B)β is a finite polynomial in x−P of total degree at most ∣α∣+∣β∣. Moreover, P=(a/p)A+(b/p)B lies on the closed segment from A to B.

Facts & Assumptions

Given: a,b>0, centers A,B∈R3, and multi-indices α,β∈N03 as in Cartesian Gaussian primitives, contractions, and integral conventions.

[F1]

Cartesian Gaussian primitives, contractions, and integral conventions: a primitive has exponential factor exp⁡(−a∥x−A∥22) and polynomial factor (x−A)α.

Proof

1.1givenF1algebra

Since p=a+b>0, expand the left exponent using ∥x−A∥2=∥x∥2−2A⋅x+∥A∥2. The quadratic and linear terms agree with p∥x−P∥2 because pP=aA+bB. The remaining constant satisfies a∥A∥2+b∥B∥2−p∥P∥2=abp∥A−B∥2=μ∥A−B∥2, by substituting P=(aA+bB)/p and expanding the inner products. Thus a∥x−A∥2+b∥x−B∥2=p∥x−P∥2+μ∥A−B∥2, which gives the exponential identity.

1.2givenF1algebra

Each coordinate obeys xj−Aj=(xj−Pj)+(Pj−Aj) and xj−Bj=(xj−Pj)+(Pj−Bj). The finite binomial theorem expands each integer power as a polynomial in xj−Pj of no larger degree; multiplying the six coordinate polynomials gives a finite polynomial of total degree at most ∣α∣+∣β∣.

2.1givenalgebra∎

The coefficients a/p and b/p are positive and sum to one. Hence P is a convex combination of A and B, so it lies on their closed line segment, including the case A=B.

Depends on

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Sources