ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A positive mode multiplier too small to certify

Statement

Use the dyadic interval format with precision $p = 5$ and exponent limits $e_{\min} = - 2$ , $e_{\max} = 8$ , and select the matching operationwise model with a valid unit-roundoff bound $u = 1 / 32$ . Set the common range truncation index to $J = 64$ , twiddle bound to $ϵ_{t w} = 1 / 32$ , phase tolerance to $δ_{ϕ} = 1 / 32$ , phase precisions to $p_{ϕ, 0} = 5$ and $p_{ϕ, \max} = 8$ , and the FFT recursion cap to $d_{F F T} = 4$ . Give the algorithm one type-1 source $x_{0} = 0, c_{0} = 1$ , mode bound $K = 7$ , one candidate $τ = 1 / 40$ , $M = 16$ , $w = L = 1$ , and requested positive absolute tolerance $ε_{r e q} = 1$ . Its least positive subnormal is $η = 2^{- 6} = 1 / 64$ . The exact smallest deconvolution multiplier $d_{\min} = \sqrt{1 / 40} e^{- 49 π / 40}$ is positive but less than $η$ . Therefore a finite outward dyadic lower endpoint for this multiplier cannot be positive, and the algorithm must reject the candidate as inconclusive at its positive-lower-bound check (or at an earlier range check); it cannot return a certified result using this candidate.

Facts & Assumptions

Given: The displayed one-dimensional candidate and dyadic format, with $p = 5$ , $e_{\min} = - 2$ , $e_{\max} = 8$ , and $u = 1 / 32$ .

[F1]

Dyadic floating-point format and directed rounding: the least positive subnormal is $2^{e_{\min} - p + 1}$ , and every positive finite representable number is at least that value.

[F2]

Real elementary functions and their domains: $e^{x} = \sum_{n = 0}^{\infty} x^{n} / n!$ .

[F3]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ and The exponential function is strictly increasing: $e^{x} > 0$ , $e^{- x} = 1 / e^{x}$ , and the real exponential is strictly increasing.

[F4]

Certified Gaussian gridding for type 1 and type 2: the candidate requires a positive certified lower endpoint for $d_{\min}$ and rejects it if the lower endpoint is nonpositive; a non-finite elementary-range result is also inconclusive.

[F5]

Nonnegative real square roots exist uniquely: the nonnegative square root of a positive number is positive and order comparisons follow from squaring nonnegative quantities.

[F6]

Real elementary functions and their domains: $π = 4 \int_{0}^{1} (1 + t^{2})^{- 1} d t \geq 2 > 0$ , since the integrand is at least $1 / 2$ on $[0, 1]$ .

Proof

1.1givenF1F2F3F5F6algebra

The format's subnormal spacing is $η = 2^{- 2 - 5 + 1} = 2^{- 6} = 1 / 64$ by [F1], and the exact multiplier is positive by [F3], [F5] and $τ = 1 / 40 > 0$ . Since $1 / 40 < 1 / 36 = (1 / 6)^{2}$ , [F5] gives $\sqrt{1 / 40} < 1 / 6$ ; also $π \geq 2$ by [F6]. Thus $d_{\min} = \sqrt{1 / 40} e^{- 49 π / 40} < \frac{1}{6} e^{- 49 / 20} .$ For $x = 49 / 20 > 0$ , the positive exponential series gives $e^{x} > \sum_{n = 0}^{5} \frac{x^{n}}{n!} = \frac{4,277,427,349}{384,000,000} > \frac{32}{3} .$ Taking positive reciprocals therefore yields $d_{\min} < \frac{1}{6} \cdot \frac{3}{32} = 1 / 64 = η$ .

2.1F1F4step 1.1algebra∎

By [F1], any positive finite dyadic endpoint is at least $η$ . A sound lower endpoint for the exact multiplier must be at most $d_{\min} < η$ by step 1.1, so it cannot be positive. The algorithm's explicit check [F4] consequently rejects this candidate as inconclusive if its finite multiplier enclosure is reached; if an earlier range operation is non-finite, [F4] gives the same inconclusive outcome. Thus this candidate cannot yield a certified transform result.

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