The modular lambda function λ(τ)λ(τ)=(√2η(τ2)η2(2τ)η3(τ))8 given by Mathematica as ModularLambda[tau] can solve, 2F1(12,12,1,1−λ(τ))2F1(12,12,1,λ(τ))=−τ√−1 For the special case τ=√−11 2F1(12,12,1,1−u)2F1(12,12,1,u)=√11 we can can use the tribonacci constant T, the real root of T3−T2−T−1=0. The solution in terms of T is u=λ(τ)=14(2−√2T+152T+1)=14(2−√17S+23S+2)=0.00047741… or in terms of S as the nice nested cube roots S=1T−1=3√12+3√12+3√12+3√12+… the cubic version of the golden ratio's ϕ=√1+√1+√1+√1+…
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