TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

Uniform entropy approximation for type-class sizes

Statement

For a fixed finite alphabet $A$ of size $k \geq 1$ , uniformly over every $n$ -type $q \in P_{n} (A)$ , $n H (q) - (k - 1) \ln (n + 1) \leq \ln ∣ T_{q} ∣ \leq n H (q) .$ Consequently, $0 \leq H (q) - \frac{1}{n} \ln ∣ T_{q} ∣ \leq \frac{(k - 1) \ln (n + 1)}{n} ⟶ 0$ uniformly over $q \in P_{n} (A)$ as $n \to \infty$ .

Facts & Assumptions

Given: A finite alphabet $A$ , $k = ∣ A ∣ \geq 1$ , and an attainable type $q$ for each $n \geq 1$ .

[F1]

Polynomially sharp bounds for type-class cardinality: $(n + 1)^{- (k - 1)} e^{n H (q)} \leq ∣ T_{q} ∣ \leq e^{n H (q)}$ for every $n$ -type $q$ .

[F2]

Finite distributions, entropy, divergence, and stochastic matrices: $H (q)$ is finite and uses the natural logarithm.

[F3]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: for $x > 0$ , $\ln x = \int_{1}^{x} d s / s$ .

Proof

1.1F1F2algebra

Each type class is nonempty because $q$ is attainable, so $∣ T_{q} ∣ > 0$ . Taking natural logarithms in [F1] gives $n H (q) - (k - 1) \ln (n + 1) \leq \ln ∣ T_{q} ∣ \leq n H (q)$ .

2.1F3step 1.1algebra∎

Divide by $n$ and subtract from $H (q)$ to get the uniform error interval in the statement. For $x \geq 1$ , [F3] and $1 / s \leq 1 / \sqrt{s}$ give $\ln x \leq \int_{1}^{x} d s / \sqrt{s} = 2 (\sqrt{x} - 1)$ , hence $\ln (n + 1) / n \to 0$ . The bound is independent of $q$ , so convergence is uniform.

Depends on

Used by

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Dependency tree · two levels

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, author-hosted book draft (standard reference, not scraped)
Thomas M. Cover and Joy A. Thomas, Elements of Information Theory, second edition, Chapter 11 (standard reference, not scraped)