How is Ramanujan's recurrence relation for his nested radical solved?

Question

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.

Answer 1

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.

Answer 2

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

http://vixra.org/pdf/1310.0177v1.pdf

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.