proof of Ruffa’s formula for continuous functions

Define sn to be the following sum:


Making the substitution m=2m and using the fact that 1+(-1)m=0 when m is odd to express the sum over even values of m as a sum over all values of m, this becomes


Subtracting this sum from sn+1 and simplifying gives


Using the telescoping sum trick, we may write


To completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof, we must investigate the limit as k. Since f is assumed continuousMathworldPlanetmathPlanetmath and the interval [a,b] is compactPlanetmathPlanetmath, f is uniformly continuousPlanetmathPlanetmath. This means that, for every ϵ>0, there exists a δ>0 such that |x-y|<δ implies |f(x)-f(y)|<ϵ. By the Archimedean property, there exists an integer k>0 such that 2kδ>|a-b|. Hence, |f(x)-f(a+m(b-a)/2n)|ϵ when x lies in the interval [a+(m-1)(b-a)/2n,a+(m+1)(b-a)/2n]. Thus, (a-b)sk+|a-b|ϵ is a Darboux upper sum for the integral


and (b-a)sk-|a-b|ϵ is a Darboux lower sum. (Darboux’s definition of the integral may be thought of as a modern incarnation of the ancient method of exhaustion.) Hence


Taking the limit ϵ0, we see that

Title proof of Ruffa’s formulaMathworldPlanetmathPlanetmath for continuous functions
Canonical name ProofOfRuffasFormulaForContinuousFunctions
Date of creation 2013-03-22 14:56:41
Last modified on 2013-03-22 14:56:41
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 8
Author rspuzio (6075)
Entry type Proof
Classification msc 30B99
Classification msc 26B15
Classification msc 78A45