Closed form of an infinitely nested radical.

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?

Best regards,

Pranas.

Re: Closed form of an infinitely nested radical.

Still looking for any insightful ideas!

Re: Closed form of an infinitely nested radical.

I'm not sure your equations are correct, but I think you mean that , and you would like to find .

If you can prove that the limit exists, then it's easy to determine what that limit is. Take the limit of the recurrence relation , giving . Then you can solve for a(x).

How do you prove that the limit exists? You can probably prove that , so it's bounded below. Then you would need to prove that it is decreasing: . And there's a theorem that a decreasing sequence that is bounded below always converges.

You might have to vary your approach for different ranges of x and make some assumptions about .

- Hollywood

Re: Closed form of an infinitely nested radical.

Re: Closed form of an infinitely nested radical.

Quote:

Originally Posted by

**Pranas** Still looking for any insightful ideas!

Please do not bump your thread. It is rude.

-Dan

Re: Closed form of an infinitely nested radical.

So for example,

I think we can still prove the sequence is bounded, and it's obviously increasing, but you're right - there seems to be no obvious recurrence relation.

- Hollywood