Closed form of an infinitely nested radical.

[Pzi]

Hello.

Does anybody happen to know a closed form of this infinitely nested radical?
http://imageshack.us/a/img268/6544/radicals.jpg
By any chance, maybe you even saw it somewhere?

I haven't had too much success so far. At the moment I am so desperate that I'm even willing to try and guess the solution, then prove that it is equal to my nested radical. For any real positive x the limit indeed exists (various criteria can be found for that very reason). Numerical limits can be seen in the plot:
http://imageshack.us/a/img842/876/plote.jpg

Also here is a convergence plot:
http://img100.imageshack.us/img100/64/convergence.jpg
It is made in a sense that using double precision variables computer sees no difference between a_{k}(x) and a_{k+1}(x) which in turn means that ~16 decimal digits have already been found. In fact it's so nasty that a{6}(50000) - a{5}(50000) < 10^(-24).
Notably the bigger my argument, the faster it converges (although I'm not sure what useful conclusions I can draw from that).

Pretty much the only known elegant cases: a(1) is equal to golden ratio, a(4)=2.

What would you suggest?



Pranas.

 

[I like Serena]

Hi Pzi!

You can write:
$$\sqrt{x^1 + \sqrt{x^2 + ...}} = \sqrt x \cdot \sqrt{x^0 + \sqrt{x^1 + ...}} = \sqrt x \cdot a(x)$$

If you substitute that in the expression for a(x), you get an equation that you can solve...

 

[Pzi]

Hi Pzi!

You can write:
$$\sqrt{x^1 + \sqrt{x^2 + ...}} = \sqrt x \cdot \sqrt{x^0 + \sqrt{x^1 + ...}} = \sqrt x \cdot a(x)$$

If you substitute that in the expression for a(x), you get an equation that you can solve...

Not really.
Your idea requires powers like 1, 2, 4, 8, 16, 32... whereas we actually have 1, 2, 3, 4, 5, 6...

 

[Pzi]

Still looking for any insightful ideas!

 

[haruspex]

I get a good fit to 1.19 x^1/4

 

[Dick]

Still looking for any insightful ideas!

You say you are desperate for a solution. Do you have any reason to believe there is an analytically exact solution?