In the past couple posts, polynomials have taken on some new presentations. This time, I’ll construct some values of a polynomial using only its values at x=0 of p(x) and all of its pseudo-derivatives.
In particular, I’ve established that p`(x)=p(x+1)-p(x). This means that p(x+1)=p(x)+p`(x). This has an expansion matching Pascal’s Triangle, because p(x+2) = p(x)+2p`(x)+p“(x), and p(x+3) = p(x)+3p`(x)+3p“(x)+p“`(x).
This pattern continues across all possible values of p(x). In fact, it is possible to express every value of p(x) as p(0) + ap`(0) + bp“(0) + … where a, b, … coefficients are the entries of the xth row of Pascal’s Triangle.