Thursday, October 31, 2019

Entry 40

Let q=e2πiτ. Given τ=n or τ=1+n2, then the solutions α,β,γ to the following equations,
2F1(12,12;1;1α)2F1(12,12;1;α)i=4τ
2F1(13,23;1;1β)2F1(13,23;1;β)i=3τ
2F1(14,34;1;1γ)2F1(14,34;1;γ)i=2τ
are given by, α=16(η(τ)η(4τ))8+16=(2η(τ)η2(4τ)η3(2τ))8=(ϑ2(q)ϑ3(q))4
β=27(η(τ)η(3τ))12+27=(3(η(τ/3)η(3τ))3+3)3=(c(q)a(q))3
γ=64(η(τ)η(2τ))24+64=(8(η(τ/2)η(2τ))8+8)2=(f(q)d(q))2
where the functions of q=e2πiτ are the Jacobi and Borwein theta functions discussed in Entry 38. Also, α=λ(2τ) and λ(τ) is the modular lambda function.

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