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Mathlib.Analysis.SpecialFunctions.Stirling

Stirling's formula #

This file proves Stirling's formula for the factorial. It states that $n!$ grows asymptotically like $\sqrt{2\pi n}(\frac{n}{e})^n$.

Proof outline #

The proof follows: https://proofwiki.org/wiki/Stirling%27s_Formula.

We proceed in two parts.

Part 1: We consider the sequence $a_n$ of fractions $\frac{n!}{\sqrt{2n}(\frac{n}{e})^n}$ and prove that this sequence converges to a real, positive number $a$. For this the two main ingredients are

Part 2: We use the fact that the series defined in part 1 converges against a real number $a$ and prove that $a = \sqrt{\pi}$. Here the main ingredient is the convergence of Wallis' product formula for π.

Part 1 #

https://proofwiki.org/wiki/Stirling%27s_Formula#Part_1

source
noncomputable def Stirling.stirlingSeq (n : ℕ) :
ℝ

Define stirlingSeq n as $\frac{n!}{\sqrt{2n}(\frac{n}{e})^n}$. Stirling's formula states that this sequence has limit $\sqrt(π)$.

Equations
Instances For
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    @[simp]
    theorem Stirling.stirlingSeq_zero :
    stirlingSeq 0 = 0
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    @[simp]
    theorem Stirling.stirlingSeq_one :
    stirlingSeq 1 = Real.exp 1 / √2
    source
    theorem Stirling.log_stirlingSeq_formula (n : ℕ) :
    Real.log (stirlingSeq n) = Real.log ↑n.factorial - 1 / 2 * Real.log (2 * ↑n) - ↑n * Real.log (↑n / Real.exp 1)
    source
    theorem Stirling.log_stirlingSeq_diff_hasSum (m : ℕ) :
    HasSum (fun (k : ℕ) => 1 / (2 * ↑(k + 1) + 1) * ((1 / (2 * ↑(m + 1) + 1)) ^ 2) ^ (k + 1)) (Real.log (stirlingSeq (m + 1)) - Real.log (stirlingSeq (m + 2)))

    The sequence log (stirlingSeq (m + 1)) - log (stirlingSeq (m + 2)) has the series expansion ∑ 1 / (2 * (k + 1) + 1) * (1 / 2 * (m + 1) + 1)^(2 * (k + 1))

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    theorem Stirling.log_stirlingSeq'_antitone :
    Antitone (Real.log ∘ stirlingSeq ∘ Nat.succ)

    The sequence log ∘ stirlingSeq ∘ succ is monotone decreasing

    source
    theorem Stirling.log_stirlingSeq_diff_le_geo_sum (n : ℕ) :
    Real.log (stirlingSeq (n + 1)) - Real.log (stirlingSeq (n + 2)) ≤ (1 / (2 * ↑(n + 1) + 1)) ^ 2 / (1 - (1 / (2 * ↑(n + 1) + 1)) ^ 2)

    We have a bound for successive elements in the sequence log (stirlingSeq k).

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    theorem Stirling.log_stirlingSeq_sub_log_stirlingSeq_succ (n : ℕ) :
    Real.log (stirlingSeq (n + 1)) - Real.log (stirlingSeq (n + 2)) ≤ 1 / (4 * ↑(n + 1) ^ 2)

    We have the bound log (stirlingSeq n) - log (stirlingSeq (n+1)) ≤ 1/(4 n^2)

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    theorem Stirling.log_stirlingSeq_bounded_aux :
    ∃ (c : ℝ), ∀ (n : ℕ), Real.log (stirlingSeq 1) - Real.log (stirlingSeq (n + 1)) ≤ c

    For any n, we have log_stirlingSeq 1 - log_stirlingSeq n ≤ 1/4 * ∑' 1/k^2

    source
    theorem Stirling.log_stirlingSeq_bounded_by_constant :
    ∃ (c : ℝ), ∀ (n : ℕ), c ≤ Real.log (stirlingSeq (n + 1))

    The sequence log_stirlingSeq is bounded below for n ≥ 1.

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    theorem Stirling.stirlingSeq'_pos (n : ℕ) :
    0 < stirlingSeq (n + 1)

    The sequence stirlingSeq is positive for n > 0

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    theorem Stirling.stirlingSeq'_bounded_by_pos_constant :
    ∃ (a : ℝ), 0 < a ∧ ∀ (n : ℕ), a ≤ stirlingSeq (n + 1)

    The sequence stirlingSeq has a positive lower bound.

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    theorem Stirling.stirlingSeq'_antitone :
    Antitone (stirlingSeq ∘ Nat.succ)

    The sequence stirlingSeq ∘ succ is monotone decreasing

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    theorem Stirling.stirlingSeq_has_pos_limit_a :
    ∃ (a : ℝ), 0 < a ∧ Filter.Tendsto stirlingSeq Filter.atTop (nhds a)

    The limit a of the sequence stirlingSeq satisfies 0 < a

    Part 2 #

    https://proofwiki.org/wiki/Stirling%27s_Formula#Part_2

    source
    theorem Stirling.tendsto_self_div_two_mul_self_add_one :
    Filter.Tendsto (fun (n : ℕ) => ↑n / (2 * ↑n + 1)) Filter.atTop (nhds (1 / 2))

    The sequence n / (2 * n + 1) tends to 1/2

    source
    theorem Stirling.stirlingSeq_pow_four_div_stirlingSeq_pow_two_eq (n : ℕ) (hn : n ≠ 0) :
    stirlingSeq n ^ 4 / stirlingSeq (2 * n) ^ 2 * (↑n / (2 * ↑n + 1)) = Real.Wallis.W n

    For any n ≠ 0, we have the identity (stirlingSeq n)^4 / (stirlingSeq (2*n))^2 * (n / (2 * n + 1)) = W n, where W n is the n-th partial product of Wallis' formula for π / 2.

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    theorem Stirling.second_wallis_limit (a : ℝ) (hane : a ≠ 0) (ha : Filter.Tendsto stirlingSeq Filter.atTop (nhds a)) :
    Filter.Tendsto Real.Wallis.W Filter.atTop (nhds (a ^ 2 / 2))

    Suppose the sequence stirlingSeq (defined above) has the limit a ≠ 0. Then the Wallis sequence W n has limit a^2 / 2.

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    theorem Stirling.tendsto_stirlingSeq_sqrt_pi :
    Filter.Tendsto stirlingSeq Filter.atTop (nhds √Real.pi)

    Stirling's Formula

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    theorem Stirling.factorial_isEquivalent_stirling :
    Asymptotics.IsEquivalent Filter.atTop (fun (n : ℕ) => ↑n.factorial) fun (n : ℕ) => √(2 * ↑n * Real.pi) * (↑n / Real.exp 1) ^ n

    Stirling's Formula, formulated in terms of Asymptotics.IsEquivalent.