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 $\displaystyle a_{n+1}(x)=\sqrt{x+a_n(x)}$, and you would like to find $\displaystyle a(x)=\lim_{n\rightarrow\infty}a_n(x)$.

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 $\displaystyle a_{n+1}(x)=\sqrt{x+a_n(x)}$, giving $\displaystyle a(x)=\sqrt{x_0+a(x)}$. Then you can solve for a(x).

How do you prove that the limit exists? You can probably prove that $\displaystyle a_n(x)\ge{0}$, so it's bounded below. Then you would need to prove that it is decreasing: $\displaystyle a_{n+1}(x)<{a_n(x)}$. 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 $\displaystyle x_0$.

- Hollywood

Re: Closed form of an infinitely nested radical.

Quote:

Originally Posted by

**hollywood** ... $\displaystyle a_{n+1}(x)=\sqrt{x+a_n(x)}$ ...

Thanks for your reply, but unfortunately this is not the same thing at all.

Essentially that's the main problem - there is no obvious convenient way to express $\displaystyle a_{n+1}(x)$ in terms of $\displaystyle a_{n}(x)$.

Example:

$\displaystyle {a_3}(x) = \sqrt {{x^0} + \sqrt {{x^1} + \sqrt {{x^2} + \sqrt {{x^3}} } } } $

$\displaystyle {a_4}(x) = \sqrt {{x^0} + \sqrt {{x^1} + \sqrt {{x^2} + \sqrt {{x^3} + \sqrt {{x^4}} } } } }$

So $\displaystyle a_{4}(x)$ in terms of $\displaystyle a_{3}(x)$ is... ???

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,

$\displaystyle a_0(2) = \sqrt{2^0} = \sqrt{1} = 1$

$\displaystyle a_1(2) = \sqrt{2^0+\sqrt{2^1}} = \sqrt{1+\sqrt{2}} \approx 1.554$

$\displaystyle a_2(2) = \sqrt{2^0+\sqrt{2^1+\sqrt{2^2}}} = \sqrt{1+\sqrt{2+\sqrt{4}}} = \sqrt{3} \approx 1.732$

$\displaystyle a_3(2) = \sqrt{2^0+\sqrt{2^1+\sqrt{2^2+\sqrt{2^3}}}}=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8}}}} \approx 1.774$

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