Beverton-Holt population dynamic model

Question

I have the model:

$$x_{n+1} =\frac{rx_n}{1+x_n}$$

where $r$ is a positive constant. Using the transformation $y=\frac{1}{x_n}$ show that $y_{n+1}$ is a linear function of $y_n$ and find $x_n$ in terms of $n$.

Can I use that $y_{n+1}=\frac{1}{x_{n+1}}$?

Answer 1

$$ y_{n+1} = \frac{1}{r}(y_n+1) $$ has a simple closed form: set $y_n=z_n+\frac{1}{r-1}$. The recurrence takes the form $z_{n+1}=\frac{1}{r}z_n$, so: $$ z_n = \frac{z_0}{r^n},\qquad y_n=\frac{z_0}{r^n}+\frac{1}{r-1},\qquad x_n=\frac{1}{\frac{z_0}{r^n}+\frac{1}{r-1}} $$ and at last

$$ x_n = \color{red}{\frac{(r-1)r^n x_0}{(r-1)+(r^n-1)x_0}}.$$