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Second-order centered and unequal-step tangent estimates

Statement

Let R:J→Rd be a unit-speed C3 curve on an open interval J containing [0,1]. Let J0⊆J be an interval on which max⁡jsup⁡u∈J0∣Rj′′′(u)∣≤M3/d. For the centered estimate, take s and h>0 with [s−h,s+h]⊆J0 and define

dh:=R(s+h)−R(s−h)2h,t^h:=dh∥dh∥2.

If δh:=M3h2/6≤1/2, then dh≠0 and

∥dh−R′(s)∥2≤δh,∥t^h−R′(s)∥2≤4δh.

For the unequal-step estimate, take positive h−,h+ with [s−h−,s+h+]⊆J0 and define

dh−,h+:=h+h−(h−+h+)(R(s)−R(s−h−))+h−h+(h−+h+)(R(s+h+)−R(s)).

If δh−,h+:=M3h−h+/6≤1/2, then it is nonzero and its unit normalization differs from R′(s) by at most 4δh−,h+. In particular, equal steps give an O(h2) unit-tangent error; quasiuniform unequal steps give an O(h−h+) error. No second-order claim is made for a one-sided tangent branch.

Facts & Assumptions

Given: The unit-speed C3 curve, stencil bounds and steps in the Statement.

[F1]

Regular minimum-energy paths on a declared collision-free tube defines the unit tangent of a regular path as R′(s)/∥R′(s)∥2. For the present curve, the separate unit-speed hypothesis means ∥R′(s)∥2=1 on all of J, so its unit tangent there is R′(s).

[F2]

Multivariable Taylor formula with a Lagrange remainder along a line segment: applied coordinatewise to a C3 curve, the degree-two Taylor remainder at displacement u is bounded in Euclidean norm by M3∣u∣3/6 under the stated coordinate bound.

[A1]

The interval J0 lies in the domain of the C3 curve and every coordinate third derivative satisfies ∣Rj′′′(u)∣≤M3/d for u∈J0; the C3 convention is Ck maps and multi-index derivative notation in Euclidean space.

Given: The hypotheses of the Statement.

Proof

1.1A1F2givenalgebra

Taylor's formula [F2] at s+h and s−h gives R(s±h)=R(s)±hR′(s)+h22R′′(s)+r± with ∥r±∥2≤M3h3/6 by [A1]. Subtracting cancels the second-derivative terms, so dh=R′(s)+(r+−r−)/(2h) and ∥dh−R′(s)∥2≤M3h2/6=δh.

1.2A1F2givenalgebra

For unequal steps, write R(s)−R(s−h−)=h−R′−h−22R′′+r− and R(s+h+)−R(s)=h+R′+h+22R′′+r+ with ∥r±∥2≤M3h±3/6 by [A1, F2]. Substitution into dh−,h+ gives coefficient 1 on R′(s) and 0 on R′′(s); the two remainder coefficients yield ∥dh−,h+−R′(s)∥2≤M3h−h+/6=δh−,h+.

2.1F1step 1.1algebra

Since ∥R′(s)∥2=1, step 1.1 gives ∥dh∥2≥1−δh≥1/2, so dh≠0. Moreover, ∣∥dh∥2−1∣≤δh, and therefore ∥dh/∥dh∥2−R′(s)∥2≤∥dh−R′(s)∥2/∥dh∥2+∣1−∥dh∥2∣/∥dh∥2≤4δh.

3.1F1step 2.1step 1.2algebra∎

The same normalization argument as in step 2.1, using δh−,h+≤1/2, gives nonzero dh−,h+ and normalized error at most 4δh−,h+. Setting h−=h+=h recovers the centered formula, and if both steps are bounded by a fixed multiple of their maximum mesh size, their product is O(h2).

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