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LeanCert.Contrib.Sinc

Differentiability of sinc and dslope #

This file proves that the sinc function is differentiable everywhere, including at 0.

Main results #

Mathematical background #

The sinc function is defined as:

Equivalently, sinc = dslope sin 0.

The derivative of sinc at 0 can be computed using Taylor expansion:

theorem Real.abs_sin_sub_self_le (x : ) (hx : |x| 1) :
|sin x - x| |x| ^ 3 / 4

Bound on |sin x - x| in terms of |x|³. For x > 0: sin x < x and x - x³/4 < sin x, so 0 < x - sin x < x³/4. By symmetry (sin(-x) = -sin(x)), |sin x - x| ≤ |x|³/4 for all x with |x| ≤ 1.

The derivative of sinc at 0 is 0.

The proof uses the squeeze theorem with the bound |sinc x - 1| ≤ |x|² / 4, which follows from sin bounds in Mathlib.

sinc is differentiable everywhere.

sinc is a differentiable function.

theorem Real.sinc_eq_integral_cos (x : ) :
sinc x = (t : ) in 0..1, cos (x * t)

Integral representation of sinc: sinc x = ∫ t in 0..1, cos (x * t). This is the standard average-cosine form used for derivative bounds.

theorem Real.hasDerivAt_sinc_of_ne_zero {x : } (hx : x 0) :
HasDerivAt sinc ((x * cos x - sin x) / x ^ 2) x

The derivative of sinc at a nonzero point.

theorem Real.deriv_sinc {x : } :
deriv sinc x = if x = 0 then 0 else (x * cos x - sin x) / x ^ 2

The derivative of sinc. For x = 0, deriv sinc 0 = 0. For x ≠ 0, deriv sinc x = (x cos x - sin x) / x².

@[simp]

Bound: |x cos x - sin x| ≤ x² for all x. Proof uses monotonicity of auxiliary functions on [0, ∞).

deriv sinc x is in [-1, 1] for all x

Integral representation and smoothness of sinc #

The sinc function is smooth (C^∞). The key insight is the integral representation: sinc(x) = ∫ t in 0..1, cos(t * x) dt

This works because:

Smoothness follows from differentiation under the integral sign (Leibniz rule): since cos(t*x) is C^∞ in x and the domain [0,1] is compact, the integral is C^∞.

theorem Real.sinc_eq_integral (x : ) :
sinc x = (t : ) in 0..1, cos (t * x)

The integral representation of sinc: sinc(x) = ∫ t in 0..1, cos(t * x)

The derivative of sinc equals dslope (cos - sinc) at 0.

For x ≠ 0: dslope (cos - sinc) 0 x = (cos x - sinc x) / x = (x cos x - sin x) / x² = deriv sinc x

For x = 0: dslope (cos - sinc) 0 0 = deriv (cos - sinc) 0 = -sin 0 - deriv sinc 0 = 0 = deriv sinc 0

sinc is smooth at every nonzero point.

sinc is analytic at 0.

The proof uses the order theory of analytic functions. Since sin is analytic at 0 with order ≥ 1 (because sin(0) = 0), there exists an analytic function g such that sin(z) = z • g(z) near 0. This g must equal sinc away from 0, and by continuity of both functions at 0, g = sinc everywhere near 0. Therefore sinc is analytic at 0.

sinc is analytic at every point.

sinc is smooth (infinitely differentiable).

The proof uses that sinc is analytic everywhere (analyticAt_sinc), and analytic functions are smooth.

sinc is an analytic function.