Problem Statement

Provide code that produces a list of numbers which is the   nth  order forward difference, given a non-negative integer (specifying the order) and a list of numbers.

The first-order forward difference of a list of numbers   A   is a new list   B,   where   Bn = An+1 - An. List   B   should have one fewer element as a result. The second-order forward difference of   A   will be: The same as the first-order forward difference of   B. That new list will have two fewer elements than   A   and one less than   B. The goal of this task is to repeat this process up to the desired order. For a more formal description, see the related   Mathworld article.

Let's start with the solution:

: forwardDiff(l)     l right(l size 1 -) l zipWith(#-) ;
: forwardDiffN(n, l)  l #[ forwardDiff dup println ] times(n) ;