A curious function in quantum gravity

[marcus]

quantum gravity research seems to be undergoing some changes
so just because this function comes up today doesn't mean it
will be around for eternity, but that said, look at this function


L(x) = \sum_{p = semi-integer > 0}\frac{1}{x^\sqrt{p(p+1)}}

It comes up in connection with Black Hole entropy, and LQG.
The Immirzi parameter (a useful number in LQG) can be found
by solving the equation

L(x) = 1/2

what does this function look like? Would it be convenient for anyone at PF to plot it? has anyone seen this function before. I have not.

It is vaguely reminding me of the Riemann zeta, but it really is not anything like the Riemann zeta

you sum over all half-integer spins-----1/2, 1, 3/2, 2, 5/2,...

and you take the geometric mean of successive pairs of spins

and use that as an exponent for x.

--------------------------

suppose you solved numerically and discovered the X such that
L(X) = 1/2
(do we know that a solution exists? is it unique?)

then if you took the natural logarithm of X and divided by 2 pi
that would be what Christoph Meissner at Warsaw (and Jerzy L, and friends) are saying is the right Immirzi number.

maybe even without the quantum gravity connection this function
L(x) is a nice function. or who knows maybe it is just some throwaway
function with a boring shape.

like 1/x^3/2

Oh, if you could find an analytic expression for this function, or rather its inverse, then you would have a research paper
following on the coattails of Meissner gr-qc/0407052

but no one will find such an analytic expression because Poles are known to be very good at that sort of thing and Meissner would have already made great efforts to do it and couldnt. so it is a kind of novelty, a nice-looking
easy to define function which Poles cannot solve.

 

[marcus]

I had better re-write this more clearly


L(x) = \sum_{p= half integer > 0}x^{-\sqrt{p(p+1)}

that way there is less chance of mistake.

Being unable to find a formula for this, Meissner resorted to the computer and by brute calculation found the number X such that

L(X) = 1/2

It turns out that X must be 4.448...
but he determined it to more than ten decimal places.

Once having found 4.448 one should take the natural logarithm
of it, which turns out to be around 1.492...(reminding us of Columbus)
and then divide by 2 pi (because he proved the Earth is round)
and that, says Meissner, is the highly-esteemed Barbero-Immirzi parameter

 

[Stingray]

It looks quite boring: monotonically decreasing away from x=1 (where its infinite of course). It converges very quickly past x=3 or so, so its not hard to make a plot of it even with some free software.

 

[marcus]

It looks quite boring...

thanks much for checking Stingray
I was afraid it would be