The Wikipedia article, here, describes in some detail the derivation of Ramanujan's famous nested radical, $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}.$$ In the Wikipedia article it provides a generalized expression: $$F(x)=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\cdots}}}$$ where the first expression is defined for $a=0$, $n=1$, $x=2$. Taking the square root of both sides allows for the recurrence relation, $$F(x)^2=ax+(n+a)^2+xF(x+n).$$ The article then solves the relation, simply saying, "it can then be shown that" $$F(x)=x+n+a.$$ How is this kind of recurrence relation solved? I don't want to see Ramanujan's method. I want to know exactly what steps were taken to get from recurrence to no recurrence.

More importantly, can a recurrence relation be solved such as $$F(x)^2=ax+(n+a)^2+g(x)F(x+n)?$$ By all means, let $a=0$ and $n=1$ for simplification.

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    $\begingroup$ Knowing Ramanujan, that could well have been the proof he gave. $\endgroup$ – preferred_anon Jul 13 '15 at 20:19
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    $\begingroup$ @DanielLittlewood: Bragging again about personally knowing Ramanujan, Littlewood ? ;-$)$ $\endgroup$ – Lucian Jul 13 '15 at 21:30

I'm not very familiar with functional equations, but I was able to get some seemingly meaningful results. Someone feel free to correct if wrong. So, clearly $$F(0)=a+n$$ Then, we can rewrite the equation as $$F(x)=g(x)+a+n$$ With $g(0)=0$. Now, after a substitution with $F(0)$, we find $$F(-n)=a$$ And then plugging in again, $$F(-n)=g(-n)+a+n=a$$ So, $g(-n)=-n$. A fixed point! Then, $$F(x)=x+n+a$$ Of course plugging in will confirm the results.

As for the other part of the question, I seriously doubt that this functional equation can be solved in general. Notice how solving it depended greatly on the form of the equation.

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  • $\begingroup$ Didn't you just go in a circle without providing any significant proof? You started with $F(x)=x+n+a$ and then went round-about, plugging in $x=0$ and so on and ended up again saying $F(x)=x+n+a$. $\endgroup$ – tyobrien Jul 14 '15 at 18:42
  • $\begingroup$ I'm sorry, I'm still not clear. The functional equation is dependent on $F(x+n)$ which is furthermore dependent on $n$. Just setting $x$ to zero does not get to $F(0)=a+n$. Even if you knew $n$ (like I said by all means, let $n=1$) you still wouldn't be able to get a LHS $F(x)$ in closed form this easily. $\endgroup$ – tyobrien Jul 14 '15 at 22:21
  • $\begingroup$ Oh wait. I see what you did. Let me think about this for a sec. $\endgroup$ – tyobrien Jul 14 '15 at 22:23
  • $\begingroup$ I see what I was thinking wrong. I +1 your answer. I'll award it when I've decided. Thanks! $\endgroup$ – tyobrien Jul 14 '15 at 22:25
  • $\begingroup$ I think there may be a logical flaw. Consider your function $g(x)$. You are saying that this function must be defined as $g(x)=x$. But couldn't it be defined differently. For eg. it could be that $g(x)=2^x-1$. This still lets $F(0)=a+n$. I see where you were going but I think some assumptions were made. $\endgroup$ – tyobrien Jul 14 '15 at 22:53

This may be of interest to you, given the pattern to your questions, I had invented a generalization of this:


See bottom of page 1 to page 2.

The idea is that if you encounter a continuous increase between the coefficients in front of the radicals.

Then a simple test can be done to the floating additive constants (in this case the 1's) that allows you to verify if it has a good closed form.

Ramanujan's radical falls out of here as a special case, but this method doesn't directly line up with his, so it's better to say that I have a different mechanism of finding the same results as his.

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