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楕圓モヅラル函數

$Q_{1} = \prod_{n = 1}^{\infty} (1 - q^{2 n})$
$Q_{2} = \prod_{n = 1}^{\infty} (1 + q^{2 n})^{2}$
$Q_{3} = \prod_{n = 1}^{\infty} (1 + q^{2 n - 1})^{2}$
$Q_{4} = \prod_{n = 1}^{\infty} (1 - q^{2 n - 1})^{2}$
ただし $q = e^{π i τ}$ $, τ = \frac{ω_{1}}{ω_{0}}$

$θ_{2} = 2 q^{\frac{1}{4}} Q_{1} Q_{2}$
$θ_{3} = Q_{1} Q_{3}$
$θ_{4} = Q_{1} Q_{4}$

$Q_{2} Q_{3} Q_{4} = 1$
$(θ_{2} θ_{3} θ_{4})^{4} = 16 q Q_{1}^{12}$
$(θ_{2})^{4} + (θ_{4})^{4} = (θ_{3})^{4}$

$a := \frac{1}{3} {(\frac{π}{ω_{0}})}^{2} (θ_{4})^{4}, b := \frac{- 1}{3} {(\frac{π}{ω_{0}})}^{2} (θ_{3})^{4}, c := \frac{1}{3} {(\frac{π}{ω_{0}})}^{2} (θ_{2})^{4}$
$e_{1} := a - b$ $, e_{2} := b - c$ $, e_{3} := c - a$

性質
$e_{1} - e_{2} = - 3 b$ $, e_{2} - e_{3} = - 3 c$ $, e_{3} - e_{1} = - 3 a$
$a + b + c = 0$
$e_{1} + e_{2} + e_{3} = 0$

$u = e_{1}, e_{2}, e_{3}$ は $4 u^{3} - g_{2} u - g_{3} = 0$ の根とする.

$a + b + c = 0$ とする.
1. $a^{2} + b^{2} + c^{2} = - 2 (a b + b c + c a)$
2. $(a - b)^{2} (b - c)^{2} (c - a)^{2} = - 4 (a b + b c + c a)^{3} - 27 (a b c)^{2}$

$Δ = g_{2}^{3} - 27 g_{3}^{2} = 16 {(\frac{π}{ω_{0}})}^{12} (θ_{2} θ_{3} θ_{4})^{8} = {(\frac{2 π}{ω_{0}})}^{12} q^{2} (Q_{1})^{24}$

$u = e_{1}, e_{2}, e_{3}$ は $4 u^{3} - g_{2} u - g_{3} = 0$ の根なので,
$3 (a b + b c + c a) = a b + b c + c a - (a^{2} + b^{2} + c^{2}) = \frac{(a - b)^{2} + (b - c)^{2} + (c - a)^{2}}{- 2} = \frac{e_{1}^{2} + e_{2}^{2} + e_{3}^{2}}{- 2} = e_{1} e_{2} + e_{2} e_{3} + e_{3} e_{1} = \frac{- g_{2}}{4}$
$(a - b) (b - c) (c - a) = e_{1} e_{2} e_{3} = \frac{g_{3}}{4}$
$(- 27 a b c)^{2} = (e_{1} - e_{2})^{2} (e_{2} - e_{3})^{2} (e_{3} - e_{1})^{2} = - 4 (e_{1} e_{2} + e_{2} e_{3} + e_{3} e_{1})^{3} - 27 (e_{1} e_{2} e_{3})^{2} = - 4 (\frac{- g_{2}}{4})^{3} - 27 (\frac{g_{3}}{4})^{2} = \frac{Δ}{16}$

また $- 27 a b c = {(\frac{π}{ω_{0}})}^{6} (θ_{2} θ_{3} θ_{4})^{4}$
より.

$Δ (τ) = (2 π)^{12} q^{2} (Q_{1})^{24}$
$η (τ) = q^{\frac{1}{12}} Q_{1}$
$j (τ) = \frac{(12 g_{2})^{3}}{Δ}$
$γ (τ) = j (τ)^{\frac{1}{3}}$
$f (τ) = \sqrt{\frac{θ_{3}}{η (τ)}}, f_{1} (τ) = \sqrt{\frac{θ_{4}}{η (τ)}}, f_{2} (τ) = \sqrt{\frac{θ_{2}}{η (τ)}}$

$f_{1} (τ)^{8} + f_{2} (τ)^{8} = f (τ)^{8}$
$j (τ) = \frac{32 (θ_{2}^{8} + θ_{3}^{8} + θ_{4}^{8})^{3}}{(θ_{2} θ_{3} θ_{4})^{8}}$

$\frac{a^{2} + b^{2} + c^{2}}{- 2} = a b + b c + c a = \frac{- g_{2}}{12}$
より
$\frac{g_{2}}{6} = \frac{1}{9} {(\frac{π}{ω_{0}})}^{4} [(θ_{2})^{8} + (θ_{3})^{8} + (θ_{4})^{8}]$
$(g_{2})^{3} = \frac{8}{27} {(\frac{π}{ω_{0}})}^{12} {[(θ_{2})^{8} + (θ_{3})^{8} + (θ_{4})^{8}]}^{3}$
ゆえに
$j (τ) = 12^{3} \frac{(g_{2})^{3}}{Δ} = 12^{3} \frac{8}{27} \frac{1}{16} \frac{(θ_{2}^{8} + θ_{3}^{8} + θ_{4}^{8})^{3}}{(θ_{2} θ_{3} θ_{4})^{8}}$

$γ (τ) = \frac{f (τ)^{24} - 16}{f (τ)^{8}} = \frac{f_{1} (τ)^{24} + 16}{f_{1} (τ)^{8}} = \frac{f_{2} (τ)^{24} + 16}{f_{2} (τ)^{8}}$

$x = - f (τ)^{8}$ $, y = f_{1} (τ)^{8}$ $, z = f_{2} (τ)^{8}$ とおくと, $x + y + z = 0$

$(θ_{2} θ_{3} θ_{4})^{4} = 16 q Q_{1}^{12}$ より $x y z = - 16$

$γ (τ)^{3} = \frac{32 (θ_{2}^{8} + θ_{3}^{8} + θ_{4}^{8})^{3}}{(16 η (τ)^{12})^{2}} = \frac{(θ_{2}^{8} + θ_{3}^{8} + θ_{4}^{8})^{3}}{8 η (τ)^{24}} = {(\frac{x^{2} + y^{2} + z^{2}}{2})}^{3} = - (x y + y z + z x)^{3}$
解と係数の関係により $x, y, z$ は $T^{3} - γ (τ) T + 16 = 0$ の根である.
$T = x, y, z$ は $γ (τ) = \frac{T^{3} + 16}{T}$ を満たす.
$T = - x$ は
$γ (τ) = \frac{T^{3} - 16}{T}$ を満たす.