= Solution
In the <squared-denominator population map>, $r$ is the low-density per-generation multiplication factor, while $b$ controls crowding: $1/b$ is the characteristic population scale. The nonzero <fixed point> for $r>1$ and its multiplier are
$$
\boxed{N_*=(\sqrt r-1)/b,\qquad F'(N_*)=2/\sqrt r-1.}
$$
Since this multiplier lies strictly between minus one and one, $N_*$ is locally asymptotically stable. This holds for every finite $r>1$, even when the map is locally decreasing at its <fixed point>.
For the <post-initial invariant interval for a squared-denominator population map>, put $M=r/(4b)$ and $L=4r^2/[b(r+4)^2]$. The derivative $F'(N)=r(1-bN)/(1+bN)^3$ shows that the global maximum is $M$ at $N=1/b$. The first two iterates of the supplied initial condition are consequently
$$
N_2=M,\qquad N_3=F(M)=L.
$$
For $r>4$, $1/b<L\leq M$. Hence the map decreases throughout $[L,M]$, with minimum $F(M)=L$ there, and its global maximum bound ensures $F(L)\leq M$. Thus $F([L,M])\subseteq[L,M]$, and induction proves
$$
\boxed{L\leq N_t\leq M\quad\text{for every }t\geq2.}
$$
Both endpoints are attained at $t=2,3$. The lower bound printed without a time qualification is false at $t=1$, because the given value $N_1=1/b$ is strictly below $L$ when $r>4$. The invariant bounds apply after that initial step, not to it.
The <cobweb plot> below uses $n=bN$ and $r=9$. The first vertical move reaches the maximum, the next reaches its image, and subsequent vertical/horizontal steps remain in the invariant interval and approach the stable <fixed point>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-1-population-cobweb.png]
{title=Cobweb of the squared-denominator population map, showing the invariant interval from the second iterate}
Back to article page
