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Saturday, May 24, 2025

Entry 89

The modular lambda function λ(τ)λ(τ)=(2η(τ2)η2(2τ)η3(τ))8 given by Mathematica as ModularLambda[tau] can solve, 2F1(12,12,1,1λ(τ))2F1(12,12,1,λ(τ))=τ1 For the special case τ=11 2F1(12,12,1,1u)2F1(12,12,1,u)=11 we can can use the tribonacci constant T, the real root of T3T2T1=0. The solution in terms of T is u=λ(τ)=14(22T+152T+1)=14(217S+23S+2)=0.00047741 or in terms of S as the nice nested cube roots S=1T1=312+312+312+312+ the cubic version of the golden ratio's ϕ=1+1+1+1+

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