Processing math: 100%

Wednesday, September 28, 2016

Entry 25

Let η(τ) be the Dedekind eta function. In his Lost Notebook, Ramanujan came up with the nice evaluationsη(i)=Γ(14)2π3/4η(2i)=η(i)23/8η(3i)=η(i)33/8(2+3)1/12η(4i)=η(i)213/16(1+2)1/4
Extending this list, we find η(5i)=η(i)5(1+52)1/2η(6i)=η(i)63/8(53233/42)1/6η(7i)=η(i)7(72+7+127+47)1/4η(8i)=η(i)241/32(1+42)1/2(1+2)1/8η(16i)=η(i)2113/64(1+42)1/4(1+2)1/16(25/8+1+2)1/2
It seems that for prime p=4m+1, then x=(pη(pi)η(i))2 is an algebraic number of degree p12, while if p=4m+3 for p>3, then y=(pη(pi)η(i))4  has degree p+12. For the special case p=7, note that y=(72+7+127+47)0.092192
is one of the four quartic roots such that the polynomial on the RHS vanishes (y2+5y+1)3(y2+13y+49)y=123+(y4+14y3+63y2+70y7)2y
and the LHS assumes the value 123=1728.

No comments:

Post a Comment