# Nonlinear inhomogeneous recurrence $f(x)^2=f(x+1)+S(x)$ to find nested radical

I want to solve the recurrence relation

$$f(x)^2=f(x+1)+S(x)$$

where $S(x)$ is a given polynomial. The background is to find nested radicals expressions of the form $$\sqrt{S(x)+\sqrt{S(x+1)+\sqrt{...}}}$$

I am especially - but not only - interested in the case $S(x)=\tfrac{1}{2}x+\tfrac{1}{2}x^2$.

I have no standard techniques, as the problem is nonlinear.

What I tried so far:

• I found in the homogeneous case $S(x)=0$ possible solutions are $$f(x)=1,\qquad f(x)=e^{c 2^x}.$$ However, as the problem is nonlinear in $f$, I doubt that this can be of use.

• I defined $f(x,t)$ as the solution of $$f(x,t)^2=f(x+1,t)+t S(x)$$ and tried to derive an equation of motion in "time" $t$ which i could integrate with the "intial condition" for $t=0$ given by the above homogeneous solution.

• Another approach was to allow continous values $x\in\mathbb{R}$, interpret $f$ as a function and write it as a power series in $x$, i.e. $$f(x)=\sum_{n=0}^\infty a_nx^n.$$ Then the equation leads to an infinite set of nonlinear equations for the sequence $(a_n)$. These equations contain an infinite number of unknowns and are also nonlinear, doesn't seem promising...

• I defined the operator $$Df(x):=f(x)^2-f(x+1)$$ and found some properties, like $$D(f+g)=Df+Dg+2fg$$ $$Df=0\Rightarrow D(fg)=f^2Dg$$ These properties allow to relate solutions of different inhomogenities $S(x)$ to each other, however I din't find a way to exploit this.

• Also found this thread about the case $S(x)=x+1$

• Any other ideas??

Thanks for any help.

• Very interesting question! This looks like a difference analog of the Riccati differential equation. However in contrast to the continuous case it is not clear how to transform it to a linear form... – Start wearing purple Jul 14 '14 at 14:19
• Just realized how old this question was, but I felt it need closure. – Zach466920 Apr 9 '15 at 14:46

You know an recurrence relation is going to be hard to solve when you get the Mandelbrot set,(its more or less a death kiss). $$f^2(x)=f(x+1)+S(x)$$ $$f(x+1)=f^2(x)-S(x)$$ however S(x) can be any polynomial so alternatively... $$f(x+1)=f^2(x)+Q(x)$$ as mentioned in your question, even the case where the "polynomial" is a constant is hard. Here's why. Replace Q(x) with c... $$f(x+1)=f^2(x)+C$$ here's the solution color coded to show which solutions go to infinity the fastest, red and which don't, black. • Thanks for this perspective! I intended to use this approach to determine nested radicals $f(0)=\sqrt{S(0)+\sqrt{S(1)+\sqrt{...}}}$. For this question, the only relevant initial condition is $x=0$ and the dynamics is on the real axis. Although the full image suggests complexity, I have the feeling that it might be possible. For example, in the trivial, constant case, which in your opinion is hard, the radical is solvable straight-forwardly by self-similarity, which gives $f(0)=\sqrt{K+\sqrt{K+\sqrt{...}}}=(1+\sqrt{1+4K})/2$ (where $C=-K<0$ due to your sign convention). – flonk Apr 9 '15 at 22:14
• I learned that there might be a chance for finding the radical, but the recurrence approach is the wrong choice. What I (falsely) wanted is to first find $f(x)$ and then insert $x=0$ to get the radical $f(0)$, however, its solution would just yield $f(x)$ as a function of given $f(0)$. This is conceptually wrong (even if the dynamics where not chaotic) but it became obvious for me from your answer, which made me finally think about the initial conditions. – flonk Apr 10 '15 at 8:30