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For $1<s\le \sqrt{2}$, set $h_s(x)=-(s-1)x$ and $J_s=[-(s-1),s-1]$. The tent map sends $J_s$ to $[1+s-s^2,1]$, with disjoint interiors, and $T_s^2(J_s)\subseteq J_s$. Directly,Successive squaring eventually brings every $s>1$ into $(\sqrt{2},2]$. A horseshoe for an interval map of an iterate in the renormalized coordinate lifts to a horseshoe of an iterate of the original map.

// Target: dynamical-systems.bigb

Given a finite measurable partition of an interval, the itinerary of $x$ records which partition element contains each iterate $T^n(x)$. Prescribing a finite initial word defines an itinerary cylinder. For the full tent map, every length-$N$ binary cylinder has Lebesgue measure $2^{-N}$ up to endpoint conventions, so its itinerary process is a fair i.i.d. Bernoulli process.

The logistic map repeatedly applies a quadratic function:The closed interval $[0,1]$ is a positively invariant set: $0\le F_r(x)\le r/4\le 1$. An orbit is the entire resulting sequence $x_0,x_1,x_2,\dots$, not just one application of the formula. For example, with $r=2$ and $x_0=0.2$, the first states are $0.2,0.32,0.4352,0.49160192,\dots$, approaching the fixed point $1/2$.

As a simple model of population dynamics, $x_n$ is a scaled abundance in generation $n$. When abundance is small, $x_{n+1}/x_n=r(1-x_n)$ is close to $r$; increasing abundance reduces this per-capita multiplier. The falling side of the quadratic function represents sufficiently severe crowding reducing the next generation. The scale $x=1$ is not generally the carrying capacity: the positive fixed point is $1-1/r$. The logistic map is a discrete-generation model; the continuous-time logistic growth equation has a different evolution rule and does not produce its period-doubling cascade.

The practical question is how abundance behaves after the initial adjustment: does it settle, alternate, or continue varying? The same formula permits all these possibilities. The following figure shows three orbits, their cobweb plots, and a bifurcation diagram. In a cobweb plot, moving vertically to $y=F_r(x)$ calculates the next state, and moving horizontally to $y=x$ makes that value the input for the next step.

The fixed points solve $F_r(x)=x$:The second fixed point is in $[0,1]$ for $r\ge 1$ and coincides with zero at $r=1$. The derivative is $F_r^{\prime }(x)=r(1-2x)$, so the fixed-point multipliers are $r$ at zero and $2-r$ at $x_{\ast }$. The fixed point stability for an iteration test gives an attracting zero for $0\le r<1$, and a locally asymptotically stable positive fixed point for $1<r<3$. A multiplier of modulus one needs a nonlinear argument; the strict derivative test alone settles neither $r=1$ nor $r=3$.

This derivative test has a direct interpretation. A small error $e_n=x_n-x_{\ast }$ obeysAt $r=2.8$, the leading multiplier is $-0.8$: the error changes sign, explaining the alternating approach, while its magnitude eventually shrinks by approximately a factor $0.8$ per step. At $r=3.2$, the multiplier is $-1.2$, and the positive fixed point repels nearby states. However, a different outcome has asymptotic stability: the logistic map two-cycle attracts nearby orbits.

For $3<r<1+\sqrt{6}$, the attracting period-two orbit has the two valuesAt $r=3.2$ they are approximately $0.513045$ and $0.799455$, with $F_r(x_{-})=x_{+}$ and $F_r(x_{+})=x_{-}$. Their periodic-orbit multiplier is $4+2r-r^2=0.16$, so a small displacement shrinks over two steps even though the fixed point is unstable. The brief algebraic justification is to factor $F_r^2(x)-x$, remove the fixed point factors, and multiply the two derivatives using the chain rule.

At $r=3$, the positive fixed-point multiplier passes through $-1$ and a period-doubling bifurcation creates the period-two orbit. At $r=1+\sqrt{6}\approx 3.44949$, its periodic-orbit multiplier passes through $-1$ and an attracting period-four periodic orbit emerges. At $r=3.5$, the period-four values are approximately $0.382820,0.826941,0.500884,0.874997$ in that order. Further period-doubling bifurcations accumulate near $r\approx 3.56995$. Beyond that value there are also parameter windows with attracting periodic orbits, including a period-three window near $r=3.83$. Neither a crowded numerical plot nor a parameter beyond the accumulation value proves sensitive dependence on initial conditions on the attracting set.

The endpoint $r=4$ supplies an analytically transparent example. The homeomorphism $h(u)={\sin }^2(\pi u/2)$ satisfieswhere $T$ is the full tent map. Thus these maps have topological conjugacy; stretching and folding in the tent map explain sensitive dependence on initial conditions and Devaney chaos for $F_4$ on $[0,1]$. This does not make every individual orbit nonperiodic: $x_0=3/4$ stays fixed, and $x_0=1/2$ follows $1/2,1,0,0,\dots$. Sensitivity means that these exceptional orbits also have arbitrarily close initial states that eventually depart substantially.

Robert M. May's "Simple mathematical models with very complicated dynamics" (1976), Nature 261, pp. 459–467, reviewed how simple deterministic recurrence relations can pass from stable abundances through period-doubling cascades to irregular behaviour. May presented the logistic map as an especially accessible example and discussed the relevance of such models to population dynamics. The paper explicitly describes itself as a review of results found by several researchers, bringing their findings together for a broad scientific audience.^[ref]

The image of the full interval under the tent map is $T_s([-1,1])=[1-s,1]$. Since $s>1$, this image contains the critical point zero and the right endpoint one; its next image is again $[1-s,1]$. Moreover, any interval $B$ with $T_s(B)=B$ lies inside the image of $[-1,1]$. Thus the core interval of an expanding tent map isEvery point outside $A$ enters $A$ after one iterate and then stays there. For $s=2$ there are no points of the domain outside $A$.

To prove interval exactness of a tent-map core, first take a nondegenerate closed interval $L\subseteq A$ and put $L_n=T_s^n(L)$. Each $L_n$ is an interval. If $L_n$ does not straddle zero, the corresponding branch is affine with slope of magnitude $s$, so $|L_{n+1}|=s|L_n|$. If it straddles zero, its image has right endpoint one andIf both $L_n$ and $L_{n+1}$ contain zero, then $L_{n+1}$ contains $[0,1]$, whence $L_{n+2}=A$. Otherwise at most one of these two steps folds across the critical point. Consequently, unless the full core has already been reached,Since $s^2/2>1$, indefinite failure to reach $A$ would give intervals of unbounded length inside an interval of length $s$. This is impossible, proving $\boxed{T_s^k(L)=A\text{ for some }k}$. For an interval with open endpoints, apply the argument to any nondegenerate closed subinterval inside it; its image already covers $A$, while the image of the original interval cannot leave $A$.

Now let $U$ be any nonempty relatively open subset of $A$ and choose a closed nondegenerate $J=[u,v]\subset U$. Some $F=T_s^k$ maps $J$ onto $A$, so there are points $p,q\in J$ with $F(p)=1-s$ and $F(q)=1$. Therefore $F(p)-p\le 0$ and $F(q)-q\ge 0$. The intermediate value theorem yields $z\in J$ with $F(z)=z$. This is a periodic point of $T_s$, of period dividing $k$. Every such $U$ contains one, so the periodic points are dense in the whole core $A$.

A horseshoe for an interval map consists of an open interval $J$ containing two disjoint open subintervals each mapped onto $J$. Glendinning chaos means that some positive iterate has a horseshoe.

Solving on the appropriate branches of the tent map givesHere $F(x_{-2})=x_{-1}$ and $F(x_{-1})=x_0$. On $J=[x_{-1},x_0]$ the second iterate isBoth endpoints map to $x_0$, and the minimum is $v=\mu -{\mu }^2/2$. Both monotone branches cover $J$ when $v\le x_{-1}$, equivalently $(\mu -1)({\mu }^2-2)\ge 0$. Restricting them to preimages of the interior proves a horseshoe for $F^2$ when $\mu \ge \sqrt{2}$.

For $1<\mu \le \sqrt{2}$, $G(J)\subseteq J$. Set $d=(\mu -1)/(\mu +1)$ and $h(x)=(x_0-x)/d$. This maps $J$ onto $[0,1]$; substitution in each linear piece gives $hG{h}^{-1}=F_{{\mu }^2}$. The renormalization of the tent map near its fixed point therefore makes $F^4$ conjugate on $J$ to $F_{{\mu }^2}^2$. Applying the preceding threshold proves its horseshoe when $2^{1/4}\le \mu <\sqrt{2}$.

For $\mu \ge \sqrt{2}$, the inverse branch of $G$ fixing $x_0$ contracts distance to $x_0$ by ${\mu }^{-2}$. Compose $m$ copies of it with the other inverse branch. This contraction has a fixed point with a periodic itinerary, distinct from $x_0$ and at distance $O({\mu }^{-2m})$. Hence nontrivial periodic points accumulate at $x_0$.

For $\mu <\sqrt{2}$, the conjugate tent parameter $\nu ={\mu }^2<2$ has every nonzero periodic point in $[\nu -{\nu }^2/2,\nu /2]$. This interval is forward invariant; below it, a nonzero point increases until entering it. The positive lower bound separates those periodic points from zero, which corresponds to $x_0$. Thus no nontrivial periodic points are sufficiently close to $x_0$ from the left. A point sufficiently close on the right maps to such a left-hand point; periodicity would pass to its image and is impossible. Thus $x_0$ is isolated among periodic points, even though a fourth-iterate horseshoe exists elsewhere.

Yes. Let $A$ be $T$-invariant. Since the itinerary variables generate $\mathcal{A}$ modulo null sets, $A$ can be regarded as an event in the canonical sequence model. For every $N$, invariance and $X_n\circ T=X_{n+1}$ implyThus $A$ belongs to the tail sigma-algebra of the i.i.d. sequence $(X_n)$. By the Kolmogorov zero-one law, $m(A)$ is zero or one. Hence the tent map is an ergodic measure-preserving transformation.

For a Borel set $A\subseteq [0,1]$, the two inverse branches of the tent map are $y\mapsto y/2$ and $y\mapsto 1-y/2$. Apart from their immaterial common endpoint,Each branch scales Lebesgue measure by $1/2$, and their images lie in opposite half-intervals. HenceAlso $m([0,1])=1$, so $m$ is an invariant probability measure.

For $0<\mu <1$, all orbits converge to one fixed point, and for $\mu =1$ every orbit is eventually fixed or period two, so no iterate has a horseshoe. For the symmetric tent map with slope magnitude $\mu >1$, the topological entropy is $\log \mu >0$. The interval-map positive-entropy horseshoe theorem says that a continuous interval map has positive topological entropy exactly when some iterate has a horseshoe. HenceFor $\mu \ge \sqrt{2}$ this already follows from part (iii); for $1<\mu <\sqrt{2}$, a higher iterate supplies the horseshoe.

For $1<\mu \le \sqrt{2}$, the interval $J=[1/(\mu +1),\mu /(\mu +1)]$ is invariant under the second iterate of the tent map. The affine coordinate $h(x)=(\mu /(\mu +1)-x)/((\mu -1)/(\mu +1))$ conjugates that restriction to the tent map with parameter ${\mu }^2$. Its two linear pieces give the identity directly. This turns the two-iterate horseshoe threshold into a four-iterate threshold and explains why chaotic invariant sets need not have periodic points near the original nonzero fixed point.