and hence, if Σ′Ω−2n = σn, since Σ′Ω−(2n−1) = 0, we have, for sufficiently small z greater than zero,
ƒ(z) = z−2 + 3σ2·z2 + 5σ3·z4 + ...
and
φ(z) = −2z−3 + 6σ2·z + 20σ3·z3 + ...;
using these series we find that the function
F(z) = [φ(z)]² − 4[ƒ(z)]³ + 60σ2ƒ(z) + 140σ3
contains no negative powers of z, being equal to a power series in z² beginning with a term in z². The function F(z) is, however, doubly periodic, with periods ω, ω′, and can only be infinite when either ƒ(z) or φ(z) is infinite; this follows from its form in ƒ(z) and φ(z); thus in one parallelogram of periods it can be infinite only when z = 0; we have proved, however, that it is not infinite, but, on the contrary, vanishes, when z = 0. Being, therefore, never infinite for finite values of z it is a constant, and therefore necessarily always zero. Putting therefore ƒ(z) = ζ and φ(z) = dζ/dz we see that
| dz | = (4ζ³ − 60σ2ζ − 140σ3)−1/2. |
| dζ |
Historically it was in the discussion of integrals such as
∫ dζ (4ζ³ − 60σ2·ζ − 140σ3)−1/2,