(p\pm q)_i = (p)_i \pm (q)_i

(pq)_i = \frac{1}{i!}\sum_{r=0}^{i}\frac{i!}{r! (i-r)!}p^{(r)}q^{(i-r)}=\sum_{r=0}^{i} (p)_r (q)_{i-r}