An Arithmetic and Geometric Mean Invariant

A positive real interval, [a, b] can be partitioned into sub-intervals such that sub-interval widths divided by sub-interval ”‘average”’ values remains constant. That both Arithmetic Mean and Geometric Mean ”‘average”’ values produce constant ratios for the same log scale is the stated invariance proved in this short note. The continuous analog is briefly considered and shown to have similar properties. 1 Problem and Summary Define a relative weight for an interval, [a, b], with 0 < a < b, as the interval length divided by the average value in the interval. We explore interval partitions of [a, b] that have equal relative weights for consecutive sub-intervals and show that there is a surprising invariance to whether we use arithmetic or geometric averages, or even an interval endpoint, for the definition of ”‘average.”’ The ratios shown in figure 1 for a partition of [1, 16] are seen to be constant when using either the arithmetic or geometric means of sub-interval endpoints. 2 Main Result Theorem 1: Given an interval [a, b], there is a partition a = x0 < x1 < ... < xn = b (1) 1 ar X iv :1 20 3. 46 17 v1 [ cs .N A ] 2 0 M ar 2 01 2 such that the following equations are satisfied: xi−1 − xi (xi+1 + xi)/2 = xj+1 − xj (xj+1 + xj)/2 ;∀i < j ∈ {0, 1...n} (2) xi+1 − xi √ xi+1xi = xj+1 − xj √ xj+1xj ;∀i < j ∈ {0, 1...n} (3) Moreover, the partition satisfying the conditions is given by: xi = b i/na(n−i)/n (4) 3 Proof The invariance follows by considering the ratio between consecutive terms: zi+1 = xi+1 xi (5) 3.1 The AM case The arithmetic mean (AM) case follows from: zi+1 − 1 zi+1 + 1 = zi − 1 zi + 1 (6) Reduction yields: (zi+1 − 1)(zi + 1) = (zi+1 + 1)(zi − 1)⇒ (7) zi+1 − zi − 1 = −zi+1 + zi − 1⇒ zi+1 = zi (8)