Hermite upper envelope

Abstract

A positive third derivative makes the Hermite quadratic an upper bound, and matching two moments evaluates its sum.

Theorem 1.1 (Remainder on an open domain).

$$\exists z\in (L,H):f\left(x\right)-p\left(x\right)=\frac{\mathrm{iteratedDeriv}\left(3,f,z\right)}{6}(x-L{)}^2(x-H)\wedge f\left(x\right)-p\left(x\right)<0$$

Proof. Machine-checked in Lean as D5/S3/Analytic/Interpolation/HermiteUpperEnvelope.hermite_two_point_remainder_on (✓ std3). ∎

Source. Repository-derived.

Commentary.

Let s be an open subset of the real line containing the closed interval from L to H, and let x lie strictly between L and H. Assume f and p are three times continuously differentiable on s, the third derivative of p is zero on s, p and f agree in value and first derivative at L, and they agree in value at H. If the third derivative of f is positive between the nodes, there is a point z strictly between them with the following remainder and strict sign. Replacing f-p by a globally smooth function agreeing near the closed interval permits the usual Hermite remainder formula to apply.

Theorem 1.2 (An upper bound determined by the mean and variance).

$$\forall k:\mathrm{Nat}\left(\right),\forall x:\mathrm{Fin}\left(k\right)\to \mathrm{Real}\left(\right),(2\le k\wedge (\forall i:\mathrm{Fin}\left(k\right),0<x\left(i\right)))⟹\sum _{i\in \mathrm{Fin}\left(k\right)}f\left(x\left(i\right)\right)\le f\left(H\right)+(k-1)f\left(L\right)$$

Proof. Machine-checked in Lean as D5/S3/Analytic/Interpolation/HermiteUpperEnvelope.hermite_upper_envelope (✓ std3). ∎

Source. Repository-derived.

Commentary.

Let k be a natural number at least two and let every coordinate x indexed by Fin k be a positive real number. Let m be the arithmetic mean and V the total squared deviation, without division by k. Define the radius r and nodes L and H as follows.

$m=\frac{{\sum }_{i\in \mathrm{Fin}\left(k\right)}x\left(i\right)}{k},V={\sum }_{i\in \mathrm{Fin}\left(k\right)}(x\left(i\right)-m{)}^2,r=\sqrt{\frac{V}{k(k-1)}},L=m-r,H=m+(k-1)r$

$f(t)=\log (1-\exp (-t))$

The moment bounds give a positive L and place every coordinate at or below H. If V is zero, all coordinates equal m and both sides coincide. Otherwise L is strictly below H. Take the quadratic p agreeing with f in value and first derivative at L and in value at H.

For a coordinate between the nodes the remainder formula gives f less than p. For a positive coordinate to the left of L, suppose f minus p were nonnegative. Two applications of the mean value theorem then give a nonnegative second derivative to the left of L, while two applications of Rolle’s theorem give a zero second derivative to its right. This contradicts strict increase of the second derivative. At either node the values agree.

Writing the quadratic in powers of t-L reduces its sum to the first and second displacement moments. These equal kr and k squared times r squared, respectively, since V equals k(k-1) times r squared. Thus the quadratic sum equals p(H)+(k-1)p(L), which gives the stated bound after substitution.

References

• Truth anchor: D5/S3/Analytic/Interpolation/HermiteUpperEnvelope.hermite_two_point_remainder_on
• Truth anchor: D5/S3/Analytic/Interpolation/HermiteUpperEnvelope.hermite_upper_envelope
• Dependency: D5/S3/Analytic/Interpolation/HermiteMomentBounds
• Dependency: D5/S3/Analytic/Interpolation/HermiteTwoPointRemainder
• Dependency: D5/S3/Analytic/Interpolation/LogOneSubExpDerivatives