LemmaStatement: AI-adaptedProof: 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.
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A finite product of quotients is the quotient of finite products

Statement

Let $n \in N$ , and let $u_{0}, \dots, u_{n - 1} \in R$ and nonzero $d_{0}, \dots, d_{n - 1} \in R$ . For $d \neq 0$ , use $u / d$ to denote $u d^{- 1}$ . Then $\prod_{k < n} \frac{u_{k}}{d_{k}} = \frac{\prod_{k < n} u_{k}}{\prod_{k < n} d_{k}} .$ The denominator on the right is nonzero by A finite product of nonzero reals is nonzero. The identity includes $n = 0$ .

Facts & Assumptions

Given: A finite list of real numerators and a same-length list of nonzero real denominators.

[F1]

Finite sums and finite products, by recursion: finite products have empty value $1$ and extend by multiplying the next term.

[F2]

The principle of mathematical induction: induction holds on the natural numbers.

[F3]

The reals form a field and Field: real multiplication is associative and commutative; nonzero elements have inverses.

[F4]

Identities and inverses in a field are unique: a multiplicative inverse, when it exists, is unique.

[F5]

A finite product of nonzero reals is nonzero: a finite product of nonzero real factors is nonzero.

Proof

technique · induction

1.1basegivenF1F3F4

For $n = 0$ , both products are empty and equal $1$ by [F1]. Also $1^{- 1} = 1$ , since $1 \cdot 1 = 1$ and inverses are unique by [F3, F4]. Thus both sides equal $1$ .

1.2givenF1F3F4F5ihalgebra

Assume the identity for the first $n$ terms, and append $u_{n} \in R$ and $d_{n} \neq 0$ . Put $U = \prod_{k < n} u_{k}$ and $D = \prod_{k < n} d_{k}$ . By [F5], both $D$ and the extended denominator product $D d_{n}$ are nonzero, so $D^{- 1}$ and $(D d_{n})^{- 1}$ exist; also $d_{n}^{- 1}$ exists by [F3]. Associativity and commutativity give $(D d_{n}) (D^{- 1} d_{n}^{- 1}) = (D D^{- 1}) (d_{n} d_{n}^{- 1}) = 1$ , so uniqueness [F4] gives $(D d_{n})^{- 1} = D^{- 1} d_{n}^{- 1}$ . Using the induction hypothesis, product recursion [F1], and the quotient convention, $\prod_{k < n + 1} \frac{u_{k}}{d_{k}} = (\prod_{k < n} \frac{u_{k}}{d_{k}}) \frac{u_{n}}{d_{n}} = \frac{U}{D} \frac{u_{n}}{d_{n}} = \frac{U u_{n}}{D d_{n}} = \frac{\prod_{k < n + 1} u_{k}}{\prod_{k < n + 1} d_{k}} .$ The equality $(U / D) (u_{n} / d_{n}) = (U u_{n}) / (D d_{n})$ follows from associativity and commutativity together with $(D d_{n})^{- 1} = D^{- 1} d_{n}^{- 1}$ .

2.1step 1.1step 1.2F2discharge-induction∎

Induction [F2] applied to steps 1.1 and 1.2 proves the identity for every finite list length.

Depends on

Used by

Cardinal interpolation functions are polynomials of the stated degrees Lemma

Dependency tree · two levels

25 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

Mark Embree, Lecture Notes on Numerical Analysis, Math 5466 (standard reference, not scraped)
Jean-Paul Berrut and Lloyd N. Trefethen, Barycentric Lagrange Interpolation (standard reference, not scraped)