# 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!

• Ramanujan's nested radical is very similar to this, and has a simple, linear solution. Maybe you could try $f(x)=ax+b$ first, and see whether it works. – Arthur Mar 31 '18 at 5:19
• Since $f(-1)^2=-1$, the function is not real-valued, or at least is not defined everywhere. Did you mean positive reals only? – András Salamon Mar 31 '18 at 5:27
• Ramanujan's nested radical is actually where I got the idea of the problem from! I tried his technique, but it didn't work. The solution also seems to converge to 3.598544540688569 which the Inverse Symbolic Calculator is not able to find an expression for, so there might not be a closed form expression for $f(x)$ or $f(1)$. – Johannes Mar 31 '18 at 5:29
• Yes, positive reals! My bad! – Johannes Mar 31 '18 at 5:30

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.

• Then the simplest solution is $f(x)=0$ on $[0,\infty)$. If the nested radical indeed converges to the vicinity of $3.598544540688569$ then the formulation of $f(x)$ evidently does not capture the essence of the nested radical. Curious. – John Wayland Bales Mar 31 '18 at 7:05
• That's not correct : the derivatives are also subject to bounds: see my answer. – G Cab Apr 2 '18 at 13:49
• @JohnWaylandBales: the "curiosity" is actually due to that also the derivatives of $f(x)$ are related : see my answer. – G Cab Apr 2 '18 at 13:51
• @GCab Yes, agreed. – John Wayland Bales Apr 2 '18 at 14:25

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.

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.

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.

• ".. does not converge": you mean for $x \to \infty$ ? – G Cab Apr 2 '18 at 13:53
• Yes, thanks for pointing out the ambiguity. – András Salamon Apr 3 '18 at 0:31