If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.