Task

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:

tdefmodule Diff do
	def forward(arr,i\\1) do
		forward(arr,[],i)
	end

	def forward([_|[]],diffs,i) do
		if i == 1 do
			IO.inspect diffs
		else 
			forward(diffs,[],i-1)
		end
	end

	def forward([val1|[val2|vals]],diffs,i) do
		forward([val2|vals],diffs++[val2-val1],i) 
	end
end 

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.



Algorithmic options

 Iterate through all previous forward differences and re-calculate a new array each time.
 Use this formula (from Wikipedia):


 




Δ

n


[
f
]
(
x
)
=

∑

k
=
0


n





(


n
k


)



(
−
1

)

n
−
k


f
(
x
+
k
)


{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}




 (Pascal's Triangle   may be useful for this option.)


 
# Project : Forward difference
 
s = [90, 47, 58, 29, 22, 32, 55, 5, 55, 73]
for p = 1 to 9 
      s = fwddiff(s)
      showarray(s) 
next
 
func fwddiff(s)
        for j=1 to len(s)-1
             s[j] = s[j+1]-s[j]
        next
        n = len(s)
        del(s, n)
        return s
 
func showarray(vect)
        see "{"
        svect = ""
        for n = 1 to len(vect)
              svect = svect + vect[n] + ", "
        next
        svect = left(svect, len(svect) - 2)
        see svect
        see "}" + nl
 