Improving computations by using low complexity matrix algebras

(the proof of the fact that Bn is uniquely defined is left to the reader, it could be a further exercise). First observe that any Bn is monic, in fact, if Bn(x) = anx n + an−1xn−1 + . . . then Bn(x+ 1)−Bn(x) = (an(x + 1) + an−1(x+ 1)n−1 + . . .)− (anx + an−1x + . . .) = [an(x n + nxn−1 + . . .) + an−1(x n−1 + (n− 1)xn−2 + . . .) + . . .]− (anx + an−1x + . . .) = annx n−1 + (·)xn−2 + . . . and thus the condition Bn(x+ 1)−Bn(x) = nxn−1 implies ann = n, i.e. an = 1. Let us compute B1(x) and B2(x). B1(x) is of the form B1(x) = x + β, thus B1(x + 1) − B1(x) = (x + 1 + β) − (x + β) = 1, i.e. the first condition is satisfied ∀β. Moreover, ∫ 1 0 (x + β) dx = [x /2 + βx]0 = 1 2 + β must be zero, so B1(x) = x− 12 . B2(x) is of the form B2(x) = x +αx+β, thus B2(x+1)−B2(x) = ((x+1)2+α(x+1)+β)−(x2+αx+β) = 2x+1+α, and the first condition is satisfied if α = −1. Moreover, ∫ 1 0 (x2−x+β) dx = [x/3−x/2+βx]0 = 1/3− 1/2 + β must be zero, so B2(x) = x − x+ 1/6.