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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 areSince this multiplier lies strictly between minus one and one, $N_{\ast }$ 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 $F^{\prime }(N)=r(1-bN)/(1+bN{)}^3$ shows that the global maximum is $M$ at $N=1/b$. The first two iterates of the supplied initial condition are consequentlyFor $r>4$, $1/b<L\le M$. Hence the map decreases throughout $[L,M]$, with minimum $F(M)=L$ there, and its global maximum bound ensures $F(L)\le M$. Thus $F([L,M])\subseteq [L,M]$, and induction provesBoth 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.

Starting at $N_1=1/b$ with $r>4$, the next two iterates are the maximum $M=r/(4b)$ and its image $L=4r^2/[b(r+4{)}^2]$. Since $1/b<L\le M$ and the squared-denominator population map decreases on $[L,M]$, its image there lies in $[L,M]$. Thus the displayed interval is invariant after the initial step, and both bounds are attained. The initial value itself is below $L$, so it must not be included in the lower-bound assertion.