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:

defmodule Diff do
  def forward(list,i\\1) do
    forward(list,[],i)
  end
  
  def forward([_],diffs,1), do: IO.inspect diffs
  def forward([_],diffs,i), do: forward(diffs,[],i-1)
  def forward([val1,val2|vals],diffs,i) do
    forward([val2|vals],diffs++[val2-val1],i)
  end
end

Enum.each(1..9, fn i ->
  Diff.forward([90, 47, 58, 29, 22, 32, 55, 5, 55, 73],i)
end)