I came across an interesting iterated function: $$ x_n = \frac{x_{n-1}}{x_{n-1} + b} $$ This is an extremely simple example and it converges to the constant $1-b$.
Can someone provide some insight to what this is or how it was come up with? Has anyone ever seen it?
It seems similar to the newton-raphson division problem where an iterated function can be used to converge to a reciprocal: $$ x = x(2-Nx) \to \frac {1}{N}. $$
Notice that if $$f(x) = \frac{x}{x+b}$$ then $$f(x) < x$$ when $$\frac{x}{x+b} - x < 0$$ which means $$\frac{-b}{x+b} < 0.$$ If $0 < b < 1$ then this holds provided $x > -b$.
Thus as long as $x_0 > 0$ then the sequence you describe is decreasing. Moreover, since all the terms are positive with initial choice of $x_0 >0$, the sequence must converge to some limit $L$.
Thus $$L = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} \frac{x_n}{x_n + b} = \frac{L}{L+b}$$ which yields the equation $$L^2 + (b-1)L = 0$$ which tells us that either $L=0$ or $L=1-b$.
If our initial value $x_0 < 1-b$ then $L$ must be zero, since the sequence is decreasing.
Now suppose that $x_0 > 1-b$. We wish to show that $f(x) > 1-b$ whenever $x> 1-b$. Once this is established, then $L=1-b$ when $x_0 > 1-b$.