Differentiability of sinc and dslope #
This file proves that the sinc function is differentiable everywhere, including at 0.
Main results #
Real.differentiableAt_sinc: sinc is differentiable at every pointReal.hasDerivAt_sinc_zero: sinc has derivative 0 at x = 0
Mathematical background #
The sinc function is defined as:
sinc x = sin x / xforx ≠ 0sinc 0 = 1
Equivalently, sinc = dslope sin 0.
The derivative of sinc at 0 can be computed using Taylor expansion:
sin x = x - x³/6 + x⁵/120 - ...sinc x = 1 - x²/6 + x⁴/120 - ...sinc'(0) = 0
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.
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:
- For x ≠ 0: ∫₀¹ cos(tx) dt = [sin(tx)/x]₀¹ = sin(x)/x = sinc(x)
- For x = 0: ∫₀¹ cos(0) dt = ∫₀¹ 1 dt = 1 = sinc(0)
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^∞.
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.