LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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.

Quantitative shallow and deep bounds for the hyperbolic tangent

Statement

For every real $s \geq 0$ ,

$0 \leq s - \tanh s \leq \frac{s^{3}}{3}, 0 \leq 1 - \tanh s \leq 2 e^{- 2 s} .$

Statement references

The linear gravity-wave shallow and deep limits use $\tanh (k h)$ : Feddersen, Linear Surface Gravity Waves 1, §2.4, PDF pp. 5–6.

The finite-depth surface-gravity dispersion relation and its limiting regimes: Firing, OCN 620, §19.4.

The six hyperbolic functions and their natural domains

Addition formulas, identities, parity, and derivatives of the hyperbolic functions

Facts & Assumptions

Given: A real number $s \geq 0$ .

[F1]

For real $x$ , $\sinh x = (e^{x} - e^{- x}) / 2$ , $\cosh x = (e^{x} + e^{- x}) / 2$ , $\tanh x = \sinh x / \cosh x$ , and $sech x = 1 / \cosh x$ ; $\cosh x > 0$ (The six hyperbolic functions and their natural domains).

[F2]

$\cosh^{2} x - \sinh^{2} x = 1$ , $\tanh^{'} (x) = {sech}^{2} x$ , $\tanh$ is strictly increasing, and $\tanh (0) = 0$ (Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

[F3]

If $a < b$ and $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , some $c \in (a, b)$ satisfies $f (b) - f (a) = f^{'} (c) (b - a)$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F4]

$e^{x + y} = e^{x} e^{y}$ for real $x, y$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[F5]

$e^{x} > 0$ and $e^{- x} = 1 / e^{x}$ for real $x$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Proof

1.1givenF1F2F3algebra

Divide $\cosh^{2} s - \sinh^{2} s = 1$ by $\cosh^{2} s > 0$ to get $1 - {sech}^{2} s = \tanh^{2} s$ . Let $f (s) = s - \tanh s$ . Then $f (0) = 0$ and $f^{'} (s) = \tanh^{2} s \geq 0$ . If $s > 0$ , apply [F3] on $[0, s]$ to obtain $f (s) - f (0) = f^{'} (c) s \geq 0$ for some $c \in (0, s)$ ; at $s = 0$ equality holds. Thus $s - \tanh s \geq 0$ .

2.1givenF2F3step 1.1algebra

Let $g (s) = s^{3} / 3 - f (s)$ . By step 1.1 and monotonicity in [F2], $s - \tanh s \geq 0$ and $s + \tanh s \geq 0$ for $s \geq 0$ . Hence $g (0) = 0$ and $g^{'} (s) = s^{2} - \tanh^{2} s = (s - \tanh s) (s + \tanh s) \geq 0$ . For $s > 0$ , [F3] on $[0, s]$ gives $g (s) - g (0) = g^{'} (c) s \geq 0$ for some $c \in (0, s)$ ; at $s = 0$ equality holds. Therefore $s - \tanh s \leq s^{3} / 3$ .

3.1givenF1F4F5step 1.1algebra∎

The definitions [F1] and reciprocal identity [F5] give $1 - \tanh s = 2 e^{- s} / (e^{s} + e^{- s})$ . Multiplying numerator and denominator by $e^{s} > 0$ and using [F4] gives $1 - \tanh s = 2 / (e^{2 s} + 1)$ . This is nonnegative, and $e^{2 s} + 1 \geq e^{2 s} > 0$ , so $2 / (e^{2 s} + 1) \leq 2 e^{- 2 s}$ . Together with step 1.1 this proves both claimed bounds.

Source notes

Hyperbolic functions and derivative identities: The six hyperbolic functions and their natural domains and Addition formulas, identities, parity, and derivatives of the hyperbolic functions.

Quantitative shallow and deep bounds for the hyperbolic tangent

Statement

Statement references

Facts & Assumptions

Proof

Source notes

Depends on

Dependency tree · 2 levels

Sources