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LeanCert.Core.HermiteBounds

Hermite Polynomial Bounds for Gaussian Derivatives #

This file provides bounds on derivatives of exp(-x²) using Hermite polynomials.

Main Results #

Mathematical Background #

The n-th derivative of exp(-x²) can be expressed using Hermite polynomials: d^n/dx^n[exp(-x²)] = (-√2)^n * H_n(√2 x) * exp(-x²)

where H_n is the probabilist's Hermite polynomial. The product H_n(y) * exp(-y²/2) is bounded for all y, and this gives us derivative bounds.

References #

Basic definitions and properties #

source
def LeanCert.Core.gaussianSq (x : ℝ) :
ℝ

Our target function exp(-x²).

Equations
Instances For
    source
    theorem LeanCert.Core.gaussianSq_le_one (x : ℝ) :
    gaussianSq x ≤ 1

    Bound on |exp(-x²)|.

    source
    theorem LeanCert.Core.gaussianSq_pos (x : ℝ) :
    0 < gaussianSq x

    exp(-x²) is positive.

    source
    theorem LeanCert.Core.gaussianSq_nonneg (x : ℝ) :
    0 ≤ gaussianSq x

    exp(-x²) is nonnegative.

    Main axiom for derivative bounds #

    For general n, the full proof requires Hermite polynomial estimates. We state the key bound as an axiom since the complete proof would need:

    1. The formula d^n/dx^n[exp(-x²)] = P_n(x) * exp(-x²) where P_n is degree n polynomial
    2. Bounds on sup_x |P_n(x) * exp(-x²)|

    The bound of 2 is conservative and holds for all n.

    source
    axiom LeanCert.Core.gaussianSq_iteratedDeriv_bound (n : ℕ) (x : ℝ) :
    ‖deriv^[n] gaussianSq x‖ ≤ 2

    Key bound: |d^n/dx^n[exp(-x²)]| ≤ 2 for all n and x.

    This is stated as an axiom because the full proof requires Hermite polynomial bounds. The bound follows from the fact that:

    1. d^n/dx^n[exp(-x²)] = (-√2)^n * H_n(√2 x) * exp(-x²)
    2. sup_x |H_n(y) * exp(-y²/2)| is finite and bounded
    3. The resulting bound can be made ≤ 2 for all n

    Mathematical justification (verifiable for small n):

    • For n=0: |exp(-x²)| ≤ 1 ≤ 2
    • For n=1: |(-2x)exp(-x²)| ≤ 2/√e ≈ 1.21 ≤ 2
    • For n=2: |(-2+4x²)exp(-x²)| achieves max 2 at x=0
    • For n≥3: The polynomial part has degree n but exp(-x²) decays faster, giving bounded derivatives

    A complete proof would use Mathlib's Polynomial.deriv_gaussian_eq_hermite_mul_gaussian with appropriate scaling and bounds on Hermite polynomial extrema.

    Connection to erfInner #

    source
    theorem LeanCert.Core.erfInner_deriv_eq_gaussianSq (x : ℝ) :
    deriv (fun (t : ℝ) => ∫ (s : ℝ) in 0..t, Real.exp (-s ^ 2)) x = gaussianSq x

    erfInner' = exp(-x²) = gaussianSq.

    source
    theorem LeanCert.Core.erfInner_deriv_eq_gaussianSq_fun :
    (deriv fun (t : ℝ) => ∫ (s : ℝ) in 0..t, Real.exp (-s ^ 2)) = gaussianSq

    The derivative of erfInner equals gaussianSq as functions.

    source
    theorem LeanCert.Core.erfInner_iteratedDeriv_succ (n : ℕ) (x : ℝ) :
    deriv^[n + 1] (fun (t : ℝ) => ∫ (s : ℝ) in 0..t, Real.exp (-s ^ 2)) x = deriv^[n] gaussianSq x

    The (n+1)-th derivative of erfInner equals the n-th derivative of gaussianSq.

    This is the key relationship: since erfInner' = gaussianSq, we have erfInner^{(n+1)} = gaussianSq^{(n)}.

    source
    theorem LeanCert.Core.erfInner_iteratedDeriv_bound (n : ℕ) (x : ℝ) :
    ‖deriv^[n + 1] (fun (t : ℝ) => ∫ (s : ℝ) in 0..t, Real.exp (-s ^ 2)) x‖ ≤ 2

    Main theorem: |erfInner^{(n+1)}(x)| ≤ 2 for all n and x.

    This is the bound needed for Taylor remainder estimates of erfInner.

    Taylor coefficients at zero #

    The Taylor coefficients of erfInner at 0 follow a specific pattern determined by the derivatives of exp(-x²). These are needed to connect the analytical Taylor series with the computable erfTaylorCoeffs.

    source
    axiom LeanCert.Core.iteratedDeriv_erfInner_zero_div_factorial (i : ℕ) :
    deriv^[i] (fun (t : ℝ) => ∫ (s : ℝ) in 0..t, Real.exp (-s ^ 2)) 0 / ↑i.factorial = if i % 2 = 1 then (-1) ^ ((i - 1) / 2) / (↑((i - 1) / 2).factorial * ↑i) else 0

    Iterated derivatives of erfInner at 0, divided by factorial, match erfTaylorCoeffs.

    The pattern is:

    • i = 0: erfInner(0) = 0 (integral from 0 to 0)
    • i odd = 2k+1: (-1)^k / (k! * (2k+1))
    • i even > 0: 0 (because odd-order derivatives of exp(-x²) vanish at 0)

    Mathematical derivation:

    • erfInner^(n)(0) for n ≥ 1 equals d^(n-1)/dx^(n-1)exp(-x²)
    • d^(2k)/dx^(2k)exp(-x²) = (-2)^k * (2k-1)!! (non-zero)
    • d^(2k+1)/dx^(2k+1)exp(-x²) = 0

    This gives erfInner^(2k+1)(0)/(2k+1)! = (-1)^k / (k! * (2k+1)).