About Division by 1

The Euclidean division of two formal series in one variable produces a sequence of series that we obtain explicitly, remarking that the case where one of the two initial series is 1 is sufficiently generic. As an application, we define a Wronskian of symmetric functions. The Euclidean division of two polynomials P (z), Q(z), in one variable z, of consecutive degrees, produces a sequence of linear factors (the successive quotients), and a sequence of successive remainders, both families being symmetric functions in the roots of P and Q separately. Euclidean division can also be applied to formal series in z, but it never stops in the generic case, leaving time enough to observe the law of the coefficients appearing in the process. Moreover, since the quotient of two formal series is also a formal series, it does not make much difference if we suppose that one of the two initial series is 1. This renders the division of series simpler than that of polynomials; in fact the latter could be obtained from the former. By formal series we mean a unitary series f(z) = 1 + c1z + c2z 2 + · · · . We shall moreover formally factorize it f(z) = σz(A ) := ∏ a∈A (1 − za)−1 = ∞ ∑ i=0 z Si(A ) , ∗Written during the conference Applications of the Macdonald Polynomials, at the Newton Institute in April 2001. the electronic journal of combinatorics 8 (2001), #N8 1 the alphabet A being supposed to be an infinite set of indeterminates, or of complex numbers, the coefficients Si(A ) being called the complete functions in A . Given two series, dividing f−1(z) = σz(A ) by f0(z) = σz(B ) means finding the unique coefficients α, β such that ( σz(A ) − (1 + αz) σz(B ) ) 1 β z−2 (1) is a unitary series f1(z) = σz(C ). Dividing in turn f0(z) by f1(z), one obtains f2(z), and iterating one gets from the initial pair (f−1, f0) an infinite sequence of series f−1, f0, f1, f2, f3, . . ., calling fk the k-th remainder. However, all the equations fk−1(z) = (1 + αk z) fk(z) + βk z fk+1(z) (2) can be divided by f0(z) = σz(B ). If the k-th remainder for the pair (σz(A ), σz(B )) is σz(C ), then the k-th remainder for the pair ( σz(A )/σz (B ) , 1 ) is