Closed form of an infinitely nested radical.

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December 9th 2012, 01:51 AM
Pranas

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.
December 11th 2012, 05:00 AM
Pranas

Re: Closed form of an infinitely nested radical.

Still looking for any insightful ideas!
December 11th 2012, 06:43 AM
hollywood

Re: Closed form of an infinitely nested radical.

I'm not sure your equations are correct, but I think you mean that $a_{n+1}(x)=\sqrt{x+a_n(x)}$ , and you would like to find $a(x)=\lim_{n\rightarrow\infty}a_n(x)$ .

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

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

- Hollywood
December 11th 2012, 06:58 AM
Pranas

Re: Closed form of an infinitely nested radical.

Quote:

Originally Posted by hollywood

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

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 $a_{n+1}(x)$ in terms of $a_{n}(x)$ .

Example:
${a_3}(x) = \sqrt {{x^0} + \sqrt {{x^1} + \sqrt {{x^2} + \sqrt {{x^3}} } } }$
${a_4}(x) = \sqrt {{x^0} + \sqrt {{x^1} + \sqrt {{x^2} + \sqrt {{x^3} + \sqrt {{x^4}} } } } }$

So $a_{4}(x)$ in terms of $a_{3}(x)$ is... ???
December 11th 2012, 07:12 AM
topsquark

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
December 12th 2012, 01:45 AM
hollywood

Re: Closed form of an infinitely nested radical.

So for example,

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

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

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

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

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