divided differences of powers

In this entry, we will prove the claims about divided differencesMathworldPlanetmath of polynomials. Because the divided difference is a linear operator, we can focus our attention on powers.

Theorem 1.

If f(x)=xn and mn, then


If m>n, then Δmf[x0,xm]=0.


We proceed by induction. The formula is trivially true when m=0. Assume that the formula is true for a certain value of m. Then we have

Δm+1f[x0,xm+1] =Δmf[x1,x2xm]-Δmf[x0,x2xm]x1-x0

Using the identity for the sum of a geometric series,


this becomes

Δm+1f[x0,xm+1] =k+k2++km=n-mk0+k1=k-1x0k0x1k1x2k2xm+1km+1

Note that when k=0, we have x1k-x0k=0, which is consistent with the formula given above because, in that case, there are no solutions to k1+k2=k, so the sum is empty and, by convention, equals zero. Likewise, when n=m, then the only solution to k0++km=0 is k0==km=0, so the sum only consists of one term, x00xm0=1 so Δnf[x0,xn]=1, hence taking further differences produces zero. ∎

Title divided differences of powers
Canonical name DividedDifferencesOfPowers
Date of creation 2013-03-22 16:47:59
Last modified on 2013-03-22 16:47:59
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 11
Author rspuzio (6075)
Entry type Theorem
Classification msc 39A70