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A Function Based on Nested Radicals

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Post Thu Sep 12, 2013 12:15 am

Posts: 8
Location: Athens, Georgia
I've been studying the function \(\displaystyle \rho (x) = \sqrt{x+\sqrt{x^2+\sqrt{x^3+...}}}\), defined on \(\displaystyle x=[0, \infty]\), and it has proved itself very uncooperative. So far I've gotten some pretty basic bounds; \(\displaystyle \sqrt{2x}\leq \rho(x)\leq \sqrt{3x}\) which works for all values of x bigger than one I think.

Additionally, I've managed to get a couple explicit values for \(\displaystyle \rho(x)\). \(\displaystyle \rho(0)\) is trivially equal to 0, and \(\displaystyle \rho(1)=\phi\) is a relatively famous result as well (in the context of the golden ratio itself). I've also managed to discover a value of \(\displaystyle \rho'(x)\), the derivative of the function in question. If we use the chain rule recursively, we get the expansion
\(\displaystyle \rho'(x)=\frac{1}{2\sqrt{x+\sqrt{x^2+...}}}\left ( 1+\frac{1}{2\sqrt{x^2+\sqrt{x^3+...}}}\left ( 2x+\frac{1}{2\sqrt{x^3+\sqrt{x^4+...}}}\left ( 3x^2+... \right ) \right ) \right )\).

Plugging in x=1 and multiplying out the terms, we can see that \(\displaystyle \rho'(1)=\sum_{k=0}^{\infty}k(2\phi)^{-k} = \frac{1+\sqrt5}{5}\).

That's all I've got though. Anybody else here think they can make any headway? :0

EDIT: sorry, it looks like the brackets get a bit messy when this transfers over to the board. I hope that part with the derivative isn't too hard to read!

Post Thu Sep 12, 2013 9:18 am

Posts: 138
Location: North Londinium, UK
This is way beyond my meagre skills... :oops:

Have you looked into the link between hypergeometric functions and continued fractions? That might be of some use...

Post Tue Sep 24, 2013 4:28 pm

Posts: 8
Location: Athens, Georgia
New info! It turns out \(\displaystyle \rho (4)\) has a closed form value as well!
Check this out guys:

We start with this:
\(\displaystyle (2^{n}+x)^2=2^{2n}+x(2^{n+1}+x) \Rightarrow x(2^n+x)=x\sqrt{2^{2n}+x(2^{n+1}+x)}\)

set n=1, so \(\displaystyle x(2+x)=\sqrt{2^2+x(2^2+x)}=\sqrt{2^2+x\sqrt{2^4+x(2^3+x)}}\)

It's pretty evident at this point that the formula can be applied recursively forever. Expanding it out into an infinite radical, we get

\(\displaystyle x(2+x)=\sqrt{2^2+x\sqrt{2^4+x\sqrt{2^6+x\sqrt{2^8+\cdots}}}}\)

and thus \(\displaystyle 3=\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\sqrt{2^8+\cdots}}}}\)
which is \(\displaystyle \rho (4)\) by definition.

I feel like this method could be generalized to find a few more values, but I can't figure out any others.

Post Sat Sep 28, 2013 4:28 pm

Posts: 138
Location: North Londinium, UK
Well played!! :D

Please DO keep us updated if you find any more variants...

Post Wed Aug 27, 2014 3:52 pm

Posts: 24
Nimee wrote:
\(\displaystyle x(2+x)=\sqrt{2^2+x\sqrt{2^4+x\sqrt{2^6+x\sqrt{2^8+\cdots}}}}\)

I think you made a typo - it should be:$$\displaystyle 2+x=\sqrt{2^2+x\sqrt{2^4+x\sqrt{2^6+x\sqrt{2^8+\cdots}}}}$$You accidentally introduced an extra $x$ earlier. (Otherwise, plugging in $x=0$ gives 0, while it should give 2.)

By the way - Hi, I'm new.

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