I want to solve the recurrence relation


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.

  • $\begingroup$ 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... $\endgroup$ – Start wearing purple Jul 14 '14 at 14:19
  • $\begingroup$ Just realized how old this question was, but I felt it need closure. $\endgroup$ – 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.

Mandelbrot Set

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.

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  • $\begingroup$ 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). $\endgroup$ – flonk Apr 9 '15 at 22:14
  • $\begingroup$ @flonk you can't solve that way. When you manipulate the equation back to square root form, you end up with f(x) relying on f(x+1) which means a self similar solution would have to be done inversely, which brings us back to the mandlebrot set. The flaw is in having the present value namely f(0) rely on all of future existent values namely f(1),f(2),f(3)...interesting on paper, chaotic in practice. $\endgroup$ – Zach466920 Apr 10 '15 at 0:21
  • $\begingroup$ 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. $\endgroup$ – flonk Apr 10 '15 at 8:30
  • $\begingroup$ @flonk an infinite radical can give points of the Julia set, which in turn is quite related to the Mandelbrot set. The Julia set is that equation for the Mandelbrot set, but instead of having variation of parameters you vary the initial conditions. That can be solved in the form of infinite radicals. But one infinite radical only gives one point in the set. $\endgroup$ – Zach466920 Apr 10 '15 at 14:12

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