# Closed form of an infinitely nested radical.

1. Dec 9, 2012

### Pzi

Hello.

Does anybody happen to know a closed form of this infinitely nested radical?
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 [Broken]

Also here is a convergence plot:
http://img100.imageshack.us/img100/64/convergence.jpg [Broken]
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?

Best regards,
Pranas.

Last edited by a moderator: May 6, 2017
2. Dec 9, 2012

### 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.......

3. Dec 9, 2012

### Pzi

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

4. Dec 11, 2012

### Pzi

Still looking for any insightful ideas!

5. Dec 12, 2012

### haruspex

I get a good fit to 1.19 x1/4

6. Dec 12, 2012

### Dick

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