If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8...}}}}$. This can be written more generally as $f(x)=\sqrt{x+(x+1)f(x+2)}$ where $x=1$. This is where the problem arises from. If anybody can find an expression for the nested radical or find $f(x)$ I would be very happy!
 A: Rewrite the equation to get $f(x+2)=(f(x)^2-x)/(x+1)$. Thus you may define $f$ arbitrarily in the interval $[0,2)$ and extend it to $[0,4)$ by using the equation. And so on. Choosing $f$ continuous on $[0,2)$ such that $\lim_{x\to 2}f(x)=f(0)^2$ gives (all) continuous solution. Thus the value of $f(1)$ plays no particular role.
A: The functional equation for $f(x)$ actually implies functional
relations also for all its derivatives
$$
\left\{ \matrix{
  f(x)^{\,2}  = x + \left( {x + 1} \right)f(x + 2) \hfill \cr 
  2f(x)f'(x) = 1 + f(x + 2) + \left( {x + 1} \right)f'(x + 2) \hfill \cr 
  2f'(x)^{\,2}  + 2f(x)f''(x) = 2f'(x + 2) + \left( {x + 1} \right)f''(x + 2) \hfill \cr 
  \quad \quad  \vdots  \hfill \cr}  \right.
$$
so that
$$
\left\{ \begin{gathered}
  f(2) = f(0)^{\,2}  \hfill \\
  f'(2) = 2f(0)f'(0) - f(0)^{\,2}  - 1 \hfill \\
  f''(2) = 2f'(0)^{\,2}  + 2f(0)f''(0) - 4f(0)f'(0) + 2f(0)^{\,2}  + 2 \hfill \\
  \quad \quad  \vdots  \hfill \\ 
\end{gathered}  \right.
$$
Therefore, being $f(x)$ continuous, we are not free to fix $f(x)\quad |\;0\le x < 2$
equal to whatever continuous function respecting only $f(2)=f(0)^2$.
Instead it shall be such as to respect the functional relation, at $x$ and $x+2$, for all the derivatives. 
A: Here is a suggested alternate approach.
Consider the recursive sequence $a_k\in\mathbb{R}$
\begin{eqnarray}
a_0&=&\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+...}}}}}\\
a_{k+1}&=&\frac{a_k^2+1}{2(k+1)}-1 \tag{1}
\end{eqnarray}
This gives the increasing unbounded sequence
\begin{eqnarray}
a_1&=&\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+10\sqrt{11+...}}}}}\\
a_2&=&\sqrt{5+6\sqrt{7+8\sqrt{9+10\sqrt{11+12\sqrt{13+...}}}}}\\
a_3&=&\sqrt{7+8\sqrt{9+10\sqrt{11+12\sqrt{13+14\sqrt{15+...}}}}}\\
&\vdots&
\end{eqnarray}
This converts the problem of finding $a_0$ into the problem of finding a generating function
\begin{equation}
G(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots
\end{equation} 
and a formula for the general term $a_n$ of the coefficient sequence.
A: Suppose the function exists and converges as its argument tends to infinity.
In particular, there is some real $x_0 > 0$ and $d > 0$ such that $0 \le f(x) \le d$ for all $x \ge x_0$.
There must then also exist some $x_1 \ge x_0$ such that $x_1 > d^2$.
It follows that for any $x \ge x_1$, we have
$f(x+2) = (f(x)^2-x)/(x+1) \le (d^2 - x)/(x+1) < 0$, a contradiction.
So the function does not converge.
