6
$\begingroup$

I became interested in this nested radical from another question and thought I would have a go at trying to come up with a formula for it. It is $$G(0)=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+\cdots}}}}.$$ This can be written as a recurrence relation when evaluated at $x=0$: $$G(x)^2=2^x+G(x+1).$$ Due to Ramanujan, we know that for some function $F(x)$ such that $$F(x)^2=ax+(a+n)^2+xF(x+n)$$ we have $$F(x)=a+n+x.$$ We can put $G(x)$ and $F(x)$ into equatable forms by making $x=1, n=1, a=\frac{\sqrt{13}-3}{2}.$ Summing these and correcting for $G(0)$ we should have $$G(0)=\sqrt{1+\frac{\sqrt{13}-3}{2}+1+1}=\sqrt{\frac{\sqrt{13}+3}{2}}.$$ this gives $1.817354021023971.$ However the correct value for $G(0)$ is around $1.78316580926410.$ What is the error in my reasoning?

$\endgroup$
2
$\begingroup$

Your mistake was assuming that $F(x)$ and $G(x)$ are the same for any $x$ simply because they are the same for $x=1$:

$$G(1)=\sqrt{2^1+F(1+1)}=\sqrt{2+F(2)} \\ F(1)=\sqrt{\left(\frac{\sqrt{13}-3}{2}\right)(1)+\left(\left(\frac{\sqrt{13}-3}{2}\right)+1\right)^2+(1)F(1+1)}=\sqrt{2+F(2)}$$

But look at $x=2$: $$G(2)=\sqrt{2^2+F(2+1)}=\sqrt{4+F(3)} \\ F(2)=\sqrt{\left(\frac{\sqrt{13}-3}{2}\right)(2)+\left(\left(\frac{\sqrt{13}-3}{2}\right)+1\right)^2+(2)F(2+1)} \approx \sqrt{2.3028+2F(3)}$$

So if you replace $G(x+1)$ with $F(x+1)$ in your definition of $G$ (as you did to solve), then $G(0)$ actually looks like this when expanded: $$G(0)=\sqrt{\frac{\sqrt{13}+3}{2}} \approx\sqrt{1+\sqrt{2+\sqrt{2.3028+2\sqrt{2.6055+\dots}}}}$$

$\endgroup$
  • $\begingroup$ Okay. I see how it goes awry at $x\ne1$ but I really only care about it at the point where $x=1$. It seems to me that my case at that point should still hold, right? I then easily convert $G(1)$ to $G(0)$. I still don't see what's wrong. $\endgroup$ – tyobrien Jul 23 '15 at 21:22
  • $\begingroup$ I'm sorry if this confused you. I just edited my answer for a mistake in the recurrence relation. I changed the $F(x+1)$ to $G(x+1)$. $\endgroup$ – tyobrien Jul 23 '15 at 21:27
  • $\begingroup$ The recurrence relation $G(x)^2=2^x+G(x+1)$ is indeed equivalent to your original nested radical expression when evaluated at $x=0$. But this is not the case for relation $G(x)^2=2^x+F(x+1)$. These are not equivalent beyond the second layer, as I showed in my answer. The issue is that you assumed you had put $G(x)$ and $F(x)$ into equitable forms when you had not. $\endgroup$ – hexaflexagonal Jul 23 '15 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.