1π=12K(k15)√∞∑n=0(2n)!3n!61(212ϕ8)n
1π=12K(k25)√∞∑n=0(2n)!3n!61(26ϕ24)n
The denominators having different powers of ϕ are the exact values of (η(τ)η(2τ))24 for τ=1+√−52,1+√−152,1+√−252, respectively. (Using d=−35 would already need a sextic.) The last implies eπ√25≈26ϕ24−24.00004…
And K(kn) can be expressed as a product of gamma functions as given in the link above. For example K(k5)=ϕ3/4√Γ(120)Γ(320)Γ(720)Γ(920)4√10π=1.576390…
which can also be numerically evaluated by Wolfram Alpha.
No comments:
Post a Comment