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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]

For $1<s\le \sqrt{2}$, use centered tent-map period-doubling renormalization. LetThe image $T_s(J_s)=[1+s-s^2,1]$ lies to the right of $J_s$, with disjoint interiors, because $1+s-s^2\ge s-1$. It lies in the positive branch. Consequently,and its range $[-c,(s^2-1)c]$ is contained in $J_s$. The orientation-reversing affine map $h_s$ gives the exact topological conjugacyThus renormalization squares the slope and halves the return period. Since $s>1$, there is a least $n\ge 0$ for which $s^{2^n}>\sqrt{2}$. Minimality implies $s^{2^n}\le 2$, and all previous slopes are in the renormalizable range. Successive affine maps therefore conjugate the restriction of $T_s^{2^n}$ on a nested central interval to $T_{s^{2^n}}$.

By the preceding part, an iterate $T_{s^{2^n}}^m$ has a horseshoe for an interval map. Pull its two branch intervals and its range interval back through the composed conjugacy. They give a horseshoe for $T_s^{2^{n}m}$. ThereforeAt the endpoint $s=\sqrt{2}$, one renormalization gives the full slope-two tent map, so the argument includes that endpoint without assuming the strict inequality from part (a).

The normalization fixes the critical point at zero and its image at one. It removes arbitrary height and spatial scales, so successive renormalized return maps can be compared in one function space rather than drifting through rescaled copies of the same dynamics. The second iterate near the critical point has a minimum; a negative rescaling reverses it back to a maximum.

Because $f(0)=1$, the period-doubling renormalization operator satisfiesPreservation of normalization therefore forcesThe scale is a functional of $f$, not an independently adjustable constant. In the renormalizable regime it lies in $(-1,0)$. If $h(x)=ax$, then $\mathcal{T}(f)=h^{-1}\circ f^2\circ h$ on the central restrictive interval. Thus topological conjugacy relates one iteration of the renormalized map to two iterations of the original map, and the new critical value is again one. For $f_{\mu }$, $a=1-\mu$, so this regime starts beyond the superstable two-cycle at $\mu =1$.

Normalization alone does not make every normalized unimodal map renormalizable: the central interval and its image must form the appropriate period-two restrictive pair, so that the rescaled second iterate is again an even unimodal map into $[-1,1]$. The operator is considered on the neighbourhood where these additional conditions hold.

Let $H:X\to Y$ be a homeomorphism with $H\circ f=g\circ H$. This is topological conjugacy. Choose a compatible metric $e$ on $Y$ and pull it back to $X$:Because $H$ is a homeomorphism, $d_H$ is a compatible metric. The conjugacy identity implies $H{f}^k=g^{k}H$ for every $k\ge 0$, so the corresponding Bowen metrics satisfyThe bijection $H$ therefore transports separated sets in either direction without changing their separation, and $s_n(d_H,ϵ)=s_n(e,ϵ)$. Their exponential growth rates, and then their limits as $ϵ\downarrow 0$, agree. Part (a) lets us replace $d_H$ by any compatible metric on $X$. Hence

The set $I^n$ consists precisely of points whose first $n$ positions $x,f(x),\dots ,f^{n-1}(x)$ all lie in $L\cup R$. HenceDeleting the first position preserves this condition, proving $f(\Lambda )\subseteq \Lambda$. For the reverse inclusion, if $y\in \Lambda$, take $x=g_L(y)$. Then $x\in L$ and $f^k(x)=f^{k-1}(y)\in L\cup R$ for every $k\ge 1$. Thus $x\in \Lambda$ and $f(x)=y$. In fact both inverse branches give preimages in $\Lambda$. Thereforewhich proves full invariance, not merely forward invariance.

For any infinite symbolic sequence $s$, the itinerary cylinders $I_{s_1\cdots s_n}$ are nonempty compact nested intervals. The nested interval theorem gives at least one point in their intersection. Every $x\in \Lambda$ has a unique itinerary because $L\cap R=\varnothing$. To justify that the itinerary also determines a unique point, it suffices to show the interval lengths shrink. The fact that logistic inverse branches contract in an angular coordinate works for every $\mu >4$; a direct claim that $|g_L^{\prime }|,|g_R^{\prime }|<1$ in the original coordinate would fail for some $\mu$ just above four.

Use the homeomorphism $h(u)={\sin }^2(\pi u/2)$ of $[0,1]$. The transformed inverse branches areTheir derivatives satisfyThe inequality follows by squaring and using $4/\mu <1$. Thus every length-$n$ transformed itinerary cylinder has length at most ${\mu }^{-n/2}$. Since $|h^{\prime }|\le \pi /2$, its original interval has length at most $(\pi /2){\mu }^{-n/2}$ and shrinks to a point. This gives the claimed one-to-one identification with the one-sided full shift.

Finally part (ii) givesThus $f$ acts by the left shift. The identification is a topological conjugacy: matching long prefixes forces closeness by the shrinking estimate, while distinct finite itinerary cylinders are disjoint compact sets and so their inverse coding is continuous. In particular $f$ need not be one-to-one on $\Lambda$, just as a one-sided left shift has two possible first symbols before any given tail.

A fixed point $x_{\ast }$ satisfies $f(x_{\ast },\mu )=x_{\ast }$. A perturbation obeys $u_{n+1}=m{u}_n+O(u_n^2)$, where the periodic-orbit multiplier is $m=f_x(x_{\ast },\mu )$. It is linearly attracting for $|m|<1$ and repelling for $|m|>1$; at $m=\pm 1$ nonlinear terms decide Lyapunov stability. A crossing through $+1$ can produce a saddle-node, transcritical or pitchfork bifurcation, depending on the mathematical model parameter dependence and symmetry. A crossing through $-1$ generically gives a period-doubling bifurcation. A real one-dimensional function cannot have the complex conjugate pair needed for a Neimark–Sacker bifurcation.

For a genuine period-two orbit, study the return map $f\circ f$ at either point. By the chain rule its periodic-orbit multiplier isIts modulus determines linear stability by the same criterion, now per two iterates.

For the real quadratic iteration map, the fixed-point equation givesThe derivative is $-2x$. The lower branch $x_{-}$ is repelling whenever $\mu >-1/4$. The upper branch satisfies $|-2x_{+}|<1$ exactly forAt $\mu =-1/4$ the two fixed points merge in a saddle-node bifurcation. With $u=x+1/2$ at that mathematical model parameter, $u^{\prime }=u-u^2$, so the point attracts from the right and is repelling from the left. At $\mu =3/4$, $x_{+}=1/2$ has periodic-orbit multiplier $-1$ and undergoes a period-doubling bifurcation. At the threshold itself $u^{\prime }=-u-u^2$ and the two-step return is $u^{\prime \prime }=u-2u^3-u^4$, proving weak asymptotic stability there despite the inconclusive linear test.

Factor the second-iterate equation:Removing the fixed-point factor yields the periodic orbit pointsThey are distinct and exchanged by $f$ exactly when $\mu >3/4$; at equality they coincide with the fixed point. Their product is $1-\mu$, hence their periodic-orbit multiplier is $4(1-\mu )$. ThusIt is repelling for $\mu >5/4$ and reaches its next period-doubling bifurcation threshold at $\mu =5/4$. At that exact endpoint it is weakly attracting: if $f^2$ is expanded about either periodic orbit point as $v^{\prime }=-v+A{v}^2+B{v}^3+O(v^4)$, then $A=2\mu -6x_i^2$, $B=-4x_i$, and $A^2+B=20\pm 10\sqrt{2}>0$. Its second return is $v^{\prime \prime }=v-2(A^2+B)v^3+O(v^4)$.

At $\mu =2$ put $h(y)=4{\sin }^2(\pi y)-2=-2\cos (2\pi y)$. The double-angle identity givesThis verifies the printed relation to the Bernoulli shift. Precisely, it is a semiconjugacy: $h(y)=h(1-y)$, so it is not invertible on the circle. To obtain a genuine interval topological conjugacy, use the homeomorphism $H(u)=-2\cos (\pi u)$ from $[0,1]$ onto $[-2,2]$, and the full tent mapThe same identity proves $f(H(u),2)=H(T(u))$.

The full tent map has dense periodic points, is topologically mixing and has sensitive dependence on initial conditions. To see these properties directly, its $n$th iterate has $2^n$ monotone branches on dyadic intervals of length $2^{-n}$, each mapping onto $[0,1]$. Every nonempty open interval contains such a branch domain, and a later iterate covers all of $[0,1]$, proving topological mixing. On any branch, the function graph of the iterate has a fixed point by the intermediate value theorem, so periodic points occur in every open interval. Given any initial point and any neighborhood, some iterate of that neighborhood covers the interval; one can choose a nearby point whose image differs from the initial point's image by at least a fixed positive distance. This proves sensitive dependence on initial conditions. Uniform continuity of the inverse homeomorphism transfers these properties to $f$ on $[-2,2]$.

Consequently the real quadratic iteration map is chaotic on its invariant interval in the Devaney sense: topological transitivity, dense periodic points and sensitive dependence on initial conditions, with all these orbits bounded. There are also aperiodic orbits: periodic points form a countable union of finite sets of roots of an equation of the equations $f^n(x)=x$, whereas the interval is uncountable. The semiconjugacy alone would not justify treating $h$ as invertible; the tent-map argument supplies the missing rigorous transfer.

The family $x_{n+1}=\mu -x_n^2$ has fixed points $(-1\pm \sqrt{1+4\mu })/2$. Its upper branch attracts linearly for $-1/4<\mu <3/4$ and loses stability in a period-doubling bifurcation. The resulting two-cycle has points $(1\pm \sqrt{4\mu -3})/2$ and multiplier $4(1-\mu )$, attracting for $3/4<\mu <5/4$. At $\mu =2$, the map is conjugate on $[-2,2]$ to the full tent map.