inf 4 1 ==== Gamma (-) \ n 4 1 1
> ----------- = --------- - ----- + --.
/ %pi n 3 4 %pi 24 ==== %e - 1 64 %pi n = 1
Still open: eta'(e^-(pi sqrt r)),
'at('diff(eta(q),q,1),q = %e^-(2*sqrt(3)*%pi)) = %e^(2*sqrt(3)*%pi)*Gamma(1/3)^(3/2)*(2^(1/3)*sqrt(3)*gamma(1/3)^6+32*%pi^3)/(512*2^(1/3)*3^(3/8)*%pi^5)
| d | -- (eta(q))| dq | - 2 sqrt(3) %pi |q = %e
2 sqrt(3) %pi 3/2 1 1/3 6 1 3 %e gamma (-) (2 sqrt(3) gamma (-) + 32 %pi ) 3 3 = -------------------------------------------------------------- 1/3 3/8 5 512 2 3 %pi
'at('diff(eta(q),q,1),q = %e^-(2*%pi/sqrt(3))) = 3*%e^(2*%pi/sqrt(3))*Gamma(1/3)^(3/2)*(32*%pi^3-2^(1/3)*sqrt(3)*Gamma(1/3)^6)/(512*2^(1/3)*3^(1/8)*%pi^5)
| d | 2 %pi -- (eta(q))| - ------- dq | sqrt(3) |q = %e
2 %pi ------- sqrt(3) 3/2 1 3 1/3 6 1 3 %e Gamma (-) (32 %pi - 2 sqrt(3) Gamma (-)) 3 3 = ---------------------------------------------------------- 1/3 1/8 5 512 2 3 %pi
and thus the corresponding sums like the above.
'sum(n/(%e^(2*%pi*n/sqrt(3))-1),n,1,inf) = 3*2^(1/3)*Gamma(1/3)^6/(256*%pi^4)-3/(8*sqrt(3)*%pi)+1/24
inf 1/3 6 1 ==== 3 2 Gamma (-) \ n 3 3 1
------------- = ---------------- - ------------- + -- / 2 %pi n 4 8 sqrt(3) %pi 24 ==== ------- 256 %pi n = 1 sqrt(3) %e - 1
'sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) = -2^(1/3)*Gamma(1/3)^6/(256*%pi^4)-1/(8*sqrt(3)*%pi)+1/24
inf 1/3 6 1 ==== 2 Gamma (-) \ n 3 1 1
--------------------- = - -------------- - ------------- + -- / 2 sqrt(3) %pi n 4 8 sqrt(3) %pi 24 ==== %e - 1 256 %pi n = 1
These were a lot of work, so they'd better be new!
Joerg Arndt>You need formula 31.1-10a on p.628 of
http://www.jjj.de/fxt/#fxtbook (note my eta(q) is defined as prod(n=1,infty, 1-q^n))
YOW! I didn't know that!
and the evaluation of K(k_4) at http://mathworld.wolfram.com/EllipticIntegralSingularValue.html
And I was exposed to these within the last year, and completely failed to make the connection.
Attached a script, remains to show algebraically that 2^(-5/12) *k^(1/12)*kp^(1/3) * (sqrt(2)+1)^(1/2) == 1
The other evaluations K(k_n) on Eric's page will give you more eta-evaluations, but likely more complicated expressions. I suggest to try k_3 first. Well, yes, I've been tediously rediscovering these a few per day.
Is there a reason that all K(k_n) with n a square only involve gamma(1/4)?
Well, it's the same reason that eta(q^a), eta(q^b), and eta(q^c) are algebraically (and homogeneously) related when a, b, and c are positive integers. E.g., using h_n to mean (conventional) eta(q^n), 16*h[1]^8*h[4]^16+h[1]^16*h[4]^8-h[2]^24 8 16 16 8 24 16 h h + h h - h 1 4 1 4 2 387420489*h[1]^12*h[2]^24*h[3]^48+19131876*h[1]^24*h[2]^24*h[3]^36+19 6830*h[1]^36*h[2]^24*h[3]^24-16777216*h[2]^72*h[3]^12-196608*h[1]^24*h[2]^48*h[3]^12-12*h[1]^48*h[2]^24*h[3]^12-h[1]^72*h[3]^12+h[1]^60*h[2]^24 12 24 48 24 24 36 36 24 24 387420489 h h h + 19131876 h h h + 196830 h h h 1 2 3 1 2 3 1 2 3 72 12 24 48 12 48 24 12 72 12 - 16777216 h h - 196608 h h h - 12 h h h - h h 2 3 1 2 3 1 2 3 1 3 60 24 + h h 1 2 etc. These would be messier with your definition lacking the q^(1/24). Another formula that would be messier is eta(q) = theta_1(pi/3,q^(1/6))/sqrt(3). And I'm afraid your simplified definition will lead to needless confusion, especially since we already have the concise notation (q;q)_oo.
Still open: eta'', eta(e^-(pi phi)), e.g..
Well, certainly not the latter, which is a singular value. Can you damn my entire week with news that the eta' valuations are also old hat? --rwg