$P_{n}=rP_{n-1}$ $P_n \sim e^{(r-1)n}$

• If $r<1, \quad \text{then} \quad \lim P_n=0$.
• If $r>1, \quad \text{then} \quad \lim P_n= +\infty$.
• If $r=1, \quad \text{then} \quad \lim P_n=P_0$.

$P_{n}=P_{n-1}(r-bP_{n-1})$

If $x_n=\frac{b}{r}P_n$, then $x_{n}=rx_{n-1}(1-x_{n-1})$.

We define $f(x)=rx(1-x)$ and $F(x,r)=rx(1-x)$.

The legal value interval for $x$ is:

$0\le x\le 1$ (population size is non-negative), because

$(x_{n-1}>1 \Rightarrow x_{n}<0) \Rightarrow x \le 1$

and

$r > 4 \Rightarrow \frac{r}{4}=f\big(\frac{1}{2}\big)>1$

in other words an instance has been found where the population size violates the legal value interval.

Therefore, $0 < r\le 4$.

• http://physics.ntua.gr/ProgMech/