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The butterfly effect is the intuitive idea that an arbitrarily small change in an initial condition can eventually change the observed future substantially. In dynamical systems, sensitive dependence makes this precise for a map $F:X\to X$ on a metric space $(X,d)$: there is one separation scale $\delta >0$ such thatThe same $\delta$ works at every point, whereas $y$ and the separation time $n$ may depend on the point and the requested closeness. For a flow map, replace the iteration of a map $F^n$ by ${\Phi }_t$ with some $t\ge 0$. The state space matters: restricting to a single fixed point removes the alternative nearby states needed for sensitivity.

Sensitivity is compatible with deterministic evolution. Given an exact initial condition, there is still one orbit; uncertainty concerns which exact state a measurement represents. It also allows accurate short-term prediction. For a continuous function $F$, each fixed finite iterated function $F^n$ is continuous, so sufficiently close initial states remain close at any specified finite step. What fails is a single initial accuracy that controls every future step.

The logistic map illustrates both sides. At $r=2.8$, nearby orbits in the basin of attraction of its attracting fixed point approach one another. At $r=4$, arbitrarily close states can eventually separate by a macroscopic fraction of $[0,1]$. The definition asserts the existence of separating neighbours, not that every nearby pair must separate: $x$ and $1-x$ have exactly the same next state under the logistic map and hence the same subsequent orbit. Nor must separation increase at every step.

A positive Lyapunov exponent describes average exponential growth of an infinitesimal displacement along an orbit, when the relevant limit exists. It is related to sensitivity but is not a substitute for its uniform definition. For a differentiable interval map, the chain rule givesAt $r=4$, the logistic map has $\lambda =\log 2$ for Lebesgue measure-almost every initial condition, while exceptional orbits can have different exponents. Finite-separation growth is only approximately exponential while the linear approximation remains useful. In a bounded state space, a separation cannot grow exponentially forever.

Sensitivity alone is also weaker than Devaney chaos, which additionally requires topological transitivity and dense periodic points. For example, $F(x)=2x$ on the real line has sensitive dependence and a positive Lyapunov exponent, but its only periodic point is zero, so it is not Devaney-chaotic. This distinguishes growth of uncertainty from the richer organization of recurrent orbits.

Edward N. Lorenz's "Deterministic Nonperiodic Flow" (1963) studied a thermal convection model with three ordinary differential equations. Its numerical orbits exhibited persistent nonperiodic motion and instability to small changes of the initial condition. Lorenz discussed the consequences for very-long-range weather forecasting: a deterministic law need not make a detailed distant future reliably recoverable from approximate observations. The historically relevant lesson of the Lorenz system is this growth of uncertainty under deterministic evolution.^[ref]

A map on a metric space is chaotic in Devaney's sense when it is topologically transitive, its periodic points are dense, and it has sensitive dependence on initial conditions.

A continuous map is chaotic in Devaney's sense when it has topological transitivity, a dense set of periodic points, and sensitive dependence on initial conditions.

If $\sqrt{2}<s\le 2$, every nondegenerate subinterval of $A=[1-s,1]$ eventually maps onto $A$. Away from the critical point, lengths multiply by $s$; one fold loses at most a factor of two. If an interval and its image both contain zero, its second image is $A$. Otherwise every two steps increase length by at least $s^2/2>1$, which cannot continue inside bounded $A$. Applying the intermediate value theorem to an interval that covers itself then gives a periodic point in every subinterval.

// Target: dynamical-systems.bigb

A parabolic cycle is a special kind of periodic point orbit.

A parabolic cycle is a periodic orbit whose multiplier is a root of unity; an iterate then has multiplier one and attracting petals.

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 $A$.

Write $n=km$, with positive integers $k,m$, and put $Q=P^m$. Polynomial substitution respects congruence, so $Q(z)\equiv z(\mathrm{mod}Q(z)-z)$ implies inductively $Q^j(z)\equiv z$ for every $j\ge 1$. Since $Q^k=P^n$,Equivalently one can factor each $Q^{j+1}(z)-Q^j(z)$ by $Q(z)-z$ and telescope. If $Q(z)-z$ is identically zero, then $Q$ is the identity and $P^n-z$ is also zero, so the assertion still holds in the sense that the latter is a polynomial multiple of the former.

For the holomorphic cycle, the chain rule giveswith indices cyclically understood. These are the same product at every point. If its modulus is less than one at one point, it is less than one at every point. Taylor expansion of each return map gives a contracting neighborhood of its fixed point, exactly as in question 1. Hence every point of the holomorphic cycle is attracting for $P^n$ whenever one is; a zero factor makes every return periodic-orbit multiplier zero.

For the quadratic family, factor the second-iterate fixed-point equation:The first factor gives fixed points. A root of the second is also fixed only if subtracting the two factors gives $2z+1=0$, so $z=-1/2$ and $c=-3/4$. If $c\ne -3/4$, the second factor has discriminant $-4c-3\ne 0$ and has two distinct roots, neither fixed. They are exchanged by $P$, since each is fixed by $P^2$ but not by $P$, and form an exact two-cycle. Therefore the unique parameter with no exact two-cycle isAt this parameter,Thus the four finite solutions counted with multiplicity are $3/2$ once and $-1/2$ three times: there are only two distinct finite solutions, both already fixed by $P$. The periodic-orbit multiplier at $3/2$ is three (nine for $P^2$), so it is repelling. At $-1/2$ the periodic-orbit multiplier for $P$ is minus one; for $P^2$ it is one. Indeed, with $u=z+1/2$, the local first-return expression for $P$ is $-u+u^2$, and its square is $u-2u^3+u^4$. This is a parabolic return with a triple fixed-point root. The two-cycle has merged into this fixed point; the triple root is not three distinct periodic points. In the Mandelbrot set picture this is the period-two bulb's attachment to the main cardioid.

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.

For integers $k,l$,Thus reducing coordinates modulo one gives a well-defined function on the torus ${\mathbb{R}}^2/{\mathbb{Z}}^2$. Its determinant is one and its inverse matrix is the integer matrix $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, so it induces a bijection of the torus. The Jacobian determinant one makes this bijection area-preserving.

Its eigenvalues solve ${\lambda }^2-3\lambda +1=0$, givingThere are therefore expanding and contracting eigendirections and no eigenvalue on the unit circle: the function is a hyperbolic toral automorphism. The Lyapunov exponents per iterate areThey sum to zero, as required by area preservation.

The four vertices of the unit square map to $(0,0),(2,1),(3,2),(1,1)$, the vertices of an area-one parallelogram. To reduce it modulo one, cut it by the integer grid. For each piece in the cell $[j,j+1]\times [k,k+1]$, translate by $(-j,-k)$. Up to boundary identifications these translated pieces partition the unit square, because the inverse integer matrix gives exactly one torus preimage to every point. The colored pieces in the original diagram show the same pieces before and after these translations.

For the Fibonacci identity, at $n=1$ the claimed matrix is $\left(\begin{array}{cc}{F}_2& F_1\\ F_1& F_0\end{array}\right)=A$. If the identity holds at $n$, multiplication on the right by $A$ givesEach equality uses $F_{j+1}=F_j+F_{j-1}$, with the specified indexing $F_0=F_1=1$. This proves the formula for every positive integer $n$.

A fixed point satisfies $(A-I)v\in {\mathbb{Z}}^2$. Since $A-I=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has an inverse integer matrix, this forces $v\in {\mathbb{Z}}^2$, so the origin is the only fixed point modulo one.

For period dividing two,The second congruence gives $y=-3x(\mathrm{mod}1)$; inserting it into the first gives $5x=0(\mathrm{mod}1)$. Thus the complete list isThe first is the fixed origin. Direct application of $A$ exchanges the other four in exactly two pairs:This both verifies the periodic orbits and excludes additional ones.

Finally, if $v$ is fixed by $F^p$, then $(A^p-I)v=m$ for an integer vector $m$. The eigenvalues of $A^p-I$ are ${\lambda }_{+}^p-1$ and ${\lambda }_{-}^p-1$, neither zero. Hence its inverse exists and has rational entries by the adjugate matrix formula for an integer matrix. Therefore $v=(A^p-I{)}^{-1}m$ has rational coordinates. The derivative of the return map is $A^p$, with eigenvalues ${\lambda }_{+}^p>1$ and $0<{\lambda }_{-}^p<1$. Every periodic point is a hyperbolic saddle fixed point of a map and is unstable, although perturbations in its stable eigendirection contract.

A point $x$ is periodic for an interval map $F$ when $F^n(x)=x$ for some positive integer $n$. Its least such $n$ is its period, and the finite setis its periodic orbit.

For an interval map, this is the usual periodic point condition. Its least positive return time is the period.