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.
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.
Keep in mind this is just the solution for the constant solution, this doesn't even take into account the initial conditions of the relation! You can try solving with nested radicals or continued fractions, or by some other method, but since this is a chaotic system, you have little chance of having a usable solution. Once again, this is only for the constant solution.