Problem Statement
This problem is inspired by another problem I worked on recently. Consider a sequence of positive numbers $x_{m}$ defined by its first value $0<x_{1}<1$ and the following recursive relation for $p,q\geq 1$:
$$ x_{p+q}=x_{p}+x_{q}-2x_{p}x_{q} $$
The problem then asks for a closed form of $x_{m}$.
My Solution
The recursive relation can be rewritten as the following:
$$ x_{p+q}=x_{p}\cdot\left(1-x_{q}\right)+\left(1-x_{p}\right)\cdot x_{q} $$
From this relation, it's quite trivial to see that $0<x_{m}<1$ for all $m\geq 1$. Looking at the inequality, the first thing that came to my mind was to treat $x_{m}$ as a probability:
Consider a (potentially unfair) coin with probability $x_{1}$ to land head. One can think of $x_{m}$ as the probability that there are an odd number of heads when the coin is tossed $m$ times. This fit the recursive relation, because to have an odd number of heads in $p+q$ tosses, either we have an odd number of heads in the first $p$ tosses & an even number of heads in the remaining $q$ tosses, or the other way around. Thus, a closed form expression is possible:
$$ \begin{aligned} x_{m}&=\sum_{\text{odd }j}\binom{m}{j}\cdot x_{1}^{j}\left(1-x_{1}\right)^{m-j} \\ &=\frac{1}{2}\left[1-\left(1-2x_{1}\right)^{m}\right] \end{aligned} $$
Remarks
This is a solution using probability / combinatorial argument. I am wondering if there are other alternatives e.g. pure algebraic solutions or something else.
It's not really a recursion, it's a functional equation. Write the equation as $$ 1 - 2 x_{p+q} = (1 - 2 x_p) (1 - 2 x_q) $$ Thus if $y = 1 - 2 x$ we have $y_{p+q} = y_p y_q$, which is easy to solve: $y_p = c^p$, i.e. $x_p = (1-c^p)/2$, for arbitrary $c$.