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.

✓ 7 results · all verified · 5 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Gamma Function — Examples

1 · Prerequisites

2 · Summary

The companion page keeps the complex formulas concrete. It computes special half-integer values, a sample pole residue, the classical value $B (1 / 2, 1 / 2)$ , and the simplest numerical Stirling check.

Its negative examples isolate two common misunderstandings. The recurrence and factorial values do not characterize Gamma by themselves, and Gamma is meromorphic rather than entire because its continuation has poles at the nonpositive integers.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

Half-integer and negative-half-integer values of Gamma

Example

$Γ (3 / 2) = \frac{\sqrt{π}}{2}, Γ (5 / 2) = \frac{3 \sqrt{π}}{4}, Γ (- 1 / 2) = - 2 \sqrt{π} .$

Facts & Assumptions

Given: $Γ (1 / 2) = \sqrt{π}$ and the functional equation.

[L1]

$Γ (1 / 2) = \sqrt{π}$ (The value of Gamma at one half).

[L2]

$Γ (z + 1) = z Γ (z)$ (The Gamma functional equation).

Verification

technique · direct

1.1

Applying [L2] at $z = 1 / 2$ and $z = 3 / 2$ gives $Γ (3 / 2) = \frac{1}{2} Γ (1 / 2)$ and $Γ (5 / 2) = \frac{3}{2} Γ (3 / 2)$ .

L2given

2.1

Applying [L2] at $z = - 1 / 2$ gives $Γ (1 / 2) = (- 1 / 2) Γ (- 1 / 2)$ . Substituting [L1] into step 1.1 and this last identity yields the three displayed values.

step 1.1L1L2algebra∎

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

The residue of Gamma at z=-2

Example

$Res (Γ, - 2) = \frac{1}{2} .$

Facts & Assumptions

Given: The pole-residue formula for Gamma.

[L1]

$Res (Γ, - n) = (- 1)^{n} / n!$ for every integer $n \geq 0$ (Meromorphic continuation of Gamma).

Verification

technique · direct

1.1

Substitute $n = 2$ into [L1]. Then $Res (Γ, - 2) = (- 1)^{2} / 2!$ .

L1given

2.1

Since $(- 1)^{2} = 1$ and $2! = 2$ , step 1.1 simplifies to $Res (Γ, - 2) = 1 / 2$ .

step 1.1algebra∎

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

$B (1 / 2, 1 / 2) = π$

Example

$B (1 / 2, 1 / 2) = π .$

Facts & Assumptions

Given: The Beta-Gamma identity and the value $Γ (1 / 2) = \sqrt{π}$ .

[L1]

$B (p, q) = Γ (p) Γ (q) / Γ (p + q)$ on the right half-planes (The Beta-Gamma identity).

[L2]

$Γ (1 / 2) = \sqrt{π}$ (The value of Gamma at one half).

[L3]

$Γ (1) = 1$ (Gamma at the positive integers).

Verification

technique · direct

1.1

Substitute $p = q = 1 / 2$ into [L1] to obtain $B (1 / 2, 1 / 2) = Γ (1 / 2)^{2} / Γ (1)$ .

L1given

2.1

By [L2], $Γ (1 / 2)^{2} = π$ , and [L3] gives $Γ (1) = 1$ . Therefore step 1.1 simplifies to $B (1 / 2, 1 / 2) = π$ .

step 1.1L2L3algebra∎

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Checking the reflection formula at z=1/2

Example

At $z = 1 / 2$ , the reflection formula reads $Γ (1 / 2)^{2} = π$ .

Facts & Assumptions

Given: The reflection formula and the value of Gamma at one half.

[L1]

$Γ (z) Γ (1 - z) = π / \sin (π z)$ (Euler's reflection formula).

[L2]

$Γ (1 / 2) = \sqrt{π}$ (The value of Gamma at one half).

Verification

technique · direct

1.1

Substituting $z = 1 / 2$ into [L1] gives $Γ (1 / 2)^{2} = π / \sin (π / 2) = π$ .

L1given

2.1

This agrees with [L2], since $(\sqrt{π})^{2} = π$ .

step 1.1L2algebra∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Stirling's approximation for 10!

Example

Using Stirling's leading term at $z = 11$ gives

$10! = Γ (11) \approx \sqrt{2 π} 11^{21 / 2} e^{- 11} \approx 3.60 \times 10^{6},$

while the exact value is $3,628,800$ .

Facts & Assumptions

Given: Stirling's asymptotic and the factorial values of Gamma.

[L1]

$Γ (n + 1) = n!$ for integers $n \geq 0$ (Gamma at the positive integers).

[L2]

Stirling's formula for Gamma is $Γ (z) \sim \sqrt{2 π} z^{z - 1 / 2} e^{- z}$ on the positive real axis as part of the sectorial asymptotic (Stirling's formula for Gamma).

Verification

technique · direct

1.1

By [L1], $10! = Γ (11)$ .

L1given

2.1

Applying [L2] at $z = 11$ gives the displayed leading-term approximation. The comparison with the exact factorial from step 1.1 shows the scale and the size of the first neglected correction.

step 1.1L2algebra∎

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

A periodic perturbation preserves the Gamma recurrence and factorial values

Statement refuted

Every meromorphic function with the Gamma recurrence and the Gamma factorial values must equal Gamma.

Facts & Assumptions

Given: The perturbed function $F (z) = Γ (z) e^{\sin (2 π z)}$ .

[L1]

The false-statement refutation already proves that $F$ satisfies the Gamma recurrence and the Gamma factorial values, but is not equal to Gamma (FALSE: the Gamma recurrence and factorial values characterize Gamma).

Counterexample

technique · direct

1.1

The function $F (z) = Γ (z) e^{\sin (2 π z)}$ is meromorphic, because Gamma is meromorphic and the exponential factor is entire.

given

2.1

By [L1], this same $F$ has the Gamma recurrence and the Gamma factorial values while still differing from Gamma. Hence it is a counterexample to the statement refuted.

step 1.1L1∎

False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

FALSE: the Gamma function is entire

Statement

False claim: The Gamma function is entire.

Facts & Assumptions

Given: The meromorphic continuation theorem for Gamma.

[L1]

Gamma extends meromorphically to $C$ with simple poles at $0, - 1, - 2, \dots$ (Meromorphic continuation of Gamma).

Refutation

technique · direct

1.1

By [L1], Gamma has a pole at $0$ and therefore is not holomorphic there.

L1given

2.1

An entire function is holomorphic on all of $C$ , so step 1.1 contradicts the claim. Hence Gamma is not entire.

step 1.1algebra∎

The Gamma Function — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Half-integer and negative-half-integer values of Gamma

Example

Facts & Assumptions

Verification

The residue of Gamma at z=-2

Example

Facts & Assumptions

Verification

B(1/2,1/2)=πB(1/2,1/2)=π

Example

Facts & Assumptions

Verification

Checking the reflection formula at z=1/2

Example

Facts & Assumptions

Verification

Stirling's approximation for 10!

Example

Facts & Assumptions

Verification

A periodic perturbation preserves the Gamma recurrence and factorial values

Statement refuted

Facts & Assumptions

Counterexample

FALSE: the Gamma function is entire

Statement

Facts & Assumptions

Refutation

Sources

$B (1 / 2, 1 / 2) = π$