* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 12. 2008 16:01]:
[...]
but still nothing about plain old
inf ==== n \ q > ------, / n ==== q - 1 n = 1
with which we could sum the reciprocal Fibonacci numbers.
Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy.
Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.) --rwg
I wonder if the Borweins knew this generalization:
inf %pi - x/3 ==== theta (---, %e ) \ cosh(n y) 1 3 x > ----------- = log(--------------------) + -- - log(sqrt(3)). / n sinh(n x) %i y - x 12 ==== theta (----, %e ) n = 1 4 2
Note d/dy gives a nice sum(sinh(ny)/sinh(nx)) in terms of theta'. Specializing x=2*y, and using the amazing Abel-Plana formula from a paper by Almkvist via Steve Finch: inf inf inf ==== / / \ [ h(%i y) - h(- %i y) [ h(0) > h(n) = %i I ------------------- dy + I h(x) dx + ----, / ] 2 %pi y ] 2 ==== / %e - 1 / n = 0 0 0
we get a peculiar theta' valued integral
inf / [ sin(t y) I ---------------------------------------- dt = ] 2 %pi t / (%e - 1) (cos(2 t y) + cosh(2 y)) 0
%i y - 2 y y theta'(----, %e ) acot(%e ) 4 2 csch(2 y) - ----------- + ------------------------------- - ---------. 2 y sinh(y) %i y - 2 y 4 4 sinh(y) theta (----, %e ) 1 2
Note that by last week's halfangle formulae,
4 %i y - 2 y - 2 y 3 - 2 y theta (----, %e ) = (- %i theta (0, %e ) theta (0, %e ) 1 2 1 4
- 2 y 3 - 2 y + theta (0, %e ) theta (0, %e ) 2 3
3 - 2 y - 2 y y/2 - theta (0, %e ) theta (0, %e )) %e /2, 2 3
but I don't know any special values of theta' except theta'_1(0). --rwg
theta'(0, q) = theta (0, q) theta (0, q) theta (0, q) 1 2 3 4 and the conjugate points, e.g., 9/4 3 %i log(q) %pi = - %i q theta'(----------- - ---, q) 3 2 2 1/4 %i log(q) = %i q theta'(---------, q) = - q theta' (%i log(q), q) 4 2 1 %pi = - theta'(---, q) = . . . 2 2 But, duh, Macsyma already knew that theta_s'(0) = 0, s>1, (because those thetas are even). So now we can differentiate the halfangle formulae and get all the theta_s'(pi/2^n,q) as algebraic functions of the three theta_s(0,q). E.g., %pi %pi theta'(---, q) = - theta'(---, q) 1 4 2 4 4 1/4 1/4 theta (0, q) theta (0, q) theta (0, q) 2 3 4 = ------------------------------------------. 5/4 2 2 3/4 2 (theta (0, q) - theta (0, q)) 3 4 Tayloring these gives 1/4 9/4 25/4 49/4 sqrt(2) q + 3 sqrt(2) q - 5 sqrt(2) q - 7 sqrt(2) q 81/4 + 9 sqrt(2) q + . . ., so they are all just theta'(0, %i q) 1 = ---------------. 1/4 sqrt(2) %i I still need to look at the fractional pseudoperiods. --rwg