I am trying to solve $f(n)=\frac{f(n-1)(1-(4f(n-1)-3)u)}{1-(1-2f(n-1))^2u}, n\in \mathrm{N}$, with $f(0)=1/2$ and for any parameter $u\in[0,1]$. Are there any pointers as to how I may approach finding an analytical form to this recurrence equation?
I have tried the following in Mathematica (which I am new to), but to no avail:
RSolve[f[n] == (f[n - 1] (1 - (4 f[n - 1] - 3) u))/(1 - (1 - 2 f[n - 1])^2 u) && f[0] == 1/2, f[n],n]
What would be the correct way to code it?
EDIT 1: To try to make things a bit more clear, parameter $u$ is not in the recursion part of the function, but I am looking for a formula in which that parameter would stay symbolic. The closed-form solution would be a function $g(n,u)$.
EDIT 2: Nov 7th: this is what I have attempted since posting the question. I have tried deriving with respect to $u$; this is because $u$ is the main "shape" parameter to this function, so it makes sense to look at changes of slope / concavity depending on $u$. As I have done some more reading since, please allow me to change notation to conform to what seems to be the common way of writing this (I am leaving the original part unedited because I do not want to create confusion with the current comments or answer). What above was written $f(n)$, without specifying $u$, I am now writing $f_n(x)$ (so also renaming $u$ into $x$):
- derivation (if I did not make an error as I fed multiple steps into Mathematica since I was not able to do this as one command): $$ \frac{\partial f_n(x)}{\partial x}=\frac{4f_{n-1}(x)(f_{n-1}(x)-1)^2+\big[(1-x)(4x(f_{n-1}(x)-2)f_{n-1}(x)+3x+1)\big]\frac{\partial f_{n-1}(x)}{\partial x}}{(4x(f_{n-1}(x)-1)f_{n-1}(x)+x-1)^2} $$
- then, solving the original recurrence for $f_{n-1}(x)$ yields two expressions (quadratic) as a function of $f_n(x)$ which I could theoretically plug into the above to obtain an equation which is both differential and linearly recursive, though the equation becomes very messy and I am not sure I am actually better off this way... $$ f_{n-1}(x)=\frac{4xf_n(x)-3x-1\pm\sqrt{p(x,f_n(x))}}{8(f_n(x)-x)} $$ where $$p(x,f_n(x))=16x(f_n(x))^2-8f_n(x)(x^2+3x)+9x^2+6x+1$$
- hence I could write: $$ \frac{\partial f_n(x)}{\partial x}=h(x,f_n(x),\frac{\partial f_{n-1}(x)}{\partial x}) $$ such that $h$ is linear in $\frac{\partial f_{n-1}(x)}{\partial x}$, but with $f_n(x)$ terms in various components making the whole thing too messy for my level of knowledge...
EDIT 3: Nov. 14th: the above function is in fact a simplification of the following linear combination: $$ f_n(x)=\frac{f_{n-1}(x) (\sqrt{x}+1)^2}{2+(4f_{n-1}(x)-2)\sqrt{x}} + \frac{ f_{n-1}(x)(\sqrt{x}-1)^2}{2-(4f_{n-1}(x)-2)\sqrt{x}} $$ My original thought was that I could solve each fraction more easily since they are not quadratic in $f_n$ anymore (similar to the idea suggested by @Somos, if I understood them correctly). Unforunately, and it now makes more sense to me (I think...), the combination of both closed forms, which are derived independently one from another, does not yield the closed-form of the combined recurrence, and I believe this is because each recurrence is non-linear.
Still, the nice symmetry in both terms might help find a solution, I will keep you updated...
You can have an outlook of this recurrence, by doing
$$ \left\{\frac{u+1}{2},\frac{u-1}{2 \left(u^2+u+1\right)}+1,\frac{u^3-1}{2 u (u+1) (u (u+3)+3)+2}+1\right\} $$