Like Calculus, these “finite difference” methods also apply in the reverse direction, i.e., there is also a “pseudo-anti-derivative” which is the reverse of the “pseudo-derivative” process.

Given an integer polynomial p(x), let the pseudo-anti-derivative of p be marked as ‘p(x). The value of ‘p(x) is unknown, since we must know the value of ‘p(0) to determine it. This fits nicely with how our other methods operate, so use ‘p(0) to denote this unknown value. Also, let (x/k) be my representation of the binomial term “x choose k”:

‘p(x) = ‘p(0) + p(0)*(x/1) + p`(0)*(x/2) + …

To show that this is a correct “pseudo-anti-derivative,” we calculate ‘p(x+1) – ‘p(x):

‘p(x+1)-‘p(x) = p(0)(((x+1)/1) – (x/1)) + p`(0)*(((x+1)/2) – (x/2)) + …

= p(0)*1 + p`(0)*(x/1) + p“(0)*(x/2) + …

Note that ((x+1)/k) – (x/k) = (x/(k-1)). Since this is what the above proof is based on, here is why it’s true:

((x+1)/k) – (x/k) = (x+1)x(x-1)(x-2)…(x-k+2)/k! – x(x-1)(x-2)(x-3)…(x-k+1)/k! = (x+1 – (x-k+1)) * x(x-1)(x-2)…(x-k+2)/k! = kx(x-1)(x-2)…(x-k+2)/k! = (x/(k-1)).

This form of “anti-derivative” creates polynomials with all-integer results, but does not have the restriction that all coefficients are integers. For example, if p(x) = x + 1, p(0) = p`(0) = 1, then:

‘p(x) = ‘p(0) + x + x(x-1)/2 = ‘p(0) + x/2 + x^2/2.