“…...,sr),(j1,...,jr )q deg ℘−β d −1 ∏(q deg ℘−β d − 1) = β d >⋯>β0>0 deg ℘−(d+r−l−1)≥β l for each l 1 L deg ℘−β0 ⋯ 1 L deg ℘−β d−1 BC (s1,...,sr ),(j1,...,jr ) q deg ℘−β d −1 ∏(q deg ℘−β d − 1) = deg ℘−(r−1)≥β d >⋯>β0>0 1 L deg ℘−β0 ⋯ 1 L deg ℘−β d−1 BC (s1,...,sr),(j1,...,jr ) q deg ℘−β d −1 ∏(q deg ℘−β d − 1) by putting i l = deg ℘ − β l (d ≥ l ≥ 0), = deg ℘>i0>⋯>i d ≥r−s1,...,sr),(j1,...,jr ) q i d −...,sr),(j1,...,jr ) q i d −Substituting this into the equation(20) and by Γ 1 = 1, we have1 Γ s1 ⋯Γ sr j∈Js a j (θ) deg ℘>i0>⋯>i d ≥r−1 1 L i0 ⋯L i d BC (s1,...,sr ),(j1,...,jr ) q i d −1BC q i d −1 = ζ A k (s) ℘ mod ℘.…”