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.)


#include <stdlib.h>
#include <string.h>
#include <stdio.h>
 
double* fwd_diff(double* x, unsigned int len, unsigned int order)
{
	unsigned int i, j;
	double* y;
 
	/* handle two special cases */
	if (order >= len) return 0;
 
	y = malloc(sizeof(double) * len);
	if (!order) {
		memcpy(y, x, sizeof(double) * len);
		return y;
	}
 
	/* first order diff goes from x->y, later ones go from y->y */
	for (j = 0; j < order; j++, x = y)
		for (i = 0, len--; i < len; i++)
			y[i] = x[i + 1] - x[i];
 
	y = realloc(y, sizeof(double) * len);
	return y;
}
 
int main(void)
{
	double *y, x[] = {90, 47, 58, 29, 22, 32, 55, 5, 55, 73};
	int i, len = sizeof(x) / sizeof(x[0]);
 
	y = fwd_diff(x, len, 1);
	for (i = 0; i < len - 1; i++)
		printf("%g ", y[i]);
	putchar('\n');
 
	return 0;
}