regarded as a branch of Integral Calculus, that the doubly periodic functions arose. As in the familiar case
z = ∫ ζ0 (1 − ζ²)−1/2 dζ,
where ζ = sin z, it has proved finally to be simpler to regard ζ as a function of z. We shall come to the other point of view below, under § 20, Elliptic Integrals.
To prove that any doubly periodic function F(z) with periods ω, ω′, having poles at the points z = a1, ... z = am of a parallelogram, these being, for simplicity of explanation, supposed to be all of the first order, is rationally expressible in terms of φ(z) and ƒ(z), and we proceed as follows:—
Consider the expression
| Φ(z) = | (ζ, 1)m + η(ζ, 1)m−2 |
| (ζ − A1) (ζ − A2)...(ζ − Am) |
where As = ƒ(as), ζ is an abbreviation for ƒ(z) and η for φ(z), and (ζ, 1)m, (ζ, 1)m−2, denote integral polynomials in ζ, of respective orders m and m − 2, so that there are 2m unspecified, homogeneously entering, constants in the numerator. It is supposed that no one of the points a1, ... am is one of the points mω + m′ω′ where f(z) = ∞. The function Φ(z) is a monogenic function of z with the periods ω, ω′, becoming infinite (and having singularities) only when (1) ζ = ∞ or (2) one of the factors ζ-As is zero. In a period parallelogram including z = 0 the first arises only for z = 0; since for ζ = ∞, η is in a finite ratio to ζ3/2; the function Φ(z) for ζ = ∞ is not infinite provided the coefficient of ζm in (ζ, 1)m is not zero; thus Φ(z) is regular about z = 0. When ζ − As = 0, that is ƒ(z) = f(as), we have z = ±as + mω + m′ω′, and no other values of z, m and m′ being integers; suppose the unspecified coefficients in the numerator so taken that the numerator vanished to the first order in each of the m points −a1, −a2, ... −am; that is, if φ(as) = Bs, and therefore φ(−as) = −Bs, so that we have the m relations
(As, 1)m − Bs(As, 1)m−2 = 0;
then the function Φ(z) will only have the m poles a1, ... am. Denoting further the m zeros of F(z) by a1′, ... am′, putting ƒ(as′) = As′, φ(as′) = Bs′, suppose the coefficients of the numerator of Φ(z) to satisfy the further m − 1 conditions
(As′, 1)m + Bs′ (As′, 1)m−2 = 0