From: vmhjr@frii.com (Virgil Hancher)
Newsgroups: sci.math
Subject: Re: finite-series formulas
Date: Thu, 24 Dec 1998 16:45:08 -0600

In article <3681DCBE.D3B77C4B@jps.net>, Steve Leman <ktkat@jps.net> wrote:

>  I am not familiar with calculus of
> finite differences (could you please explain). 
 
It is too large a field to explain in any detail, But I will illustrate.

Let f(n) be a function of the integer variable n.
Let Df(n) = f(n+1)-f(n) be the first differences function
let DDf(n) = D( Df(n+1) - Df(n) ) = f(n+2) - 2*f(n+1) + f(n) 
be the second differences function, and so on.

It can be shown that a function is a polynomial if and only if 
there is a kth differences function which is identically zero.
Furthermore, the original function can then be reconstructed from 
the various difference functions all evaluated at the same point.

This is ultimately equivalent to saying that a function is 
a polynomial if and only if its first difference function 
is a polynomial ( as it happens, of one smaller degree).

Note that if f(n) is the sum of kth powers of i from i = 1 to i = n
then Df(n) = n^k is a polynomial.

There is a lot more to the calculus of finite differences 
than I have indicated here, but this outlines the part 
relevant to your question.

Posted and Mailed