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An invariant interval for a scalar map is an interval whose points remain in that interval under every forward iterate. Here invariance means forward invariance and does not assert surjectivity onto the interval.

The contraction mapping theorem states that a self-map $T$ of a nonempty complete metric space with a uniform contraction constant $q<1$ has a unique fixed point $p$. For every starting point, the iteration $x_{n+1}=T{x}_n$ converges to $p$, with $d(x_n,p)\le q^{n}d(x_1,x_0)/(1-q)$.

For the cosine fixed-point iteration, any real $x_0$ gives $x_1\in [-1,1]$. This closed interval is complete, and cosine maps it into $[\cos 1,1]\subseteq [-1,1]$. The mean value theorem giveswith $\sin (1)<1$. Apply the contraction mapping theorem to the tail starting at $x_1$. Therefore $\boxed{x_n\to p,p=\cos p}$ for every real initial value. Numerically $p\approx 0.739085$. The global real-line derivative bound is only one; entering the invariant interval supplies the strict uniform bound needed by the theorem.

First establish a uniform upper bound. Direct subtraction givesbecause the quadratic polynomial has positive leading coefficient and discriminant $49-56=-7$. The assumed positivity of $F(N_0)$ therefore places $N_1$ in $(0,3/2)$.

This interval is forward invariant. On $[0,1/2]$ and on $[1,3/2]$, the three factors in $P(N)$ have nonnegative product, hence $F(N)\ge 1/2$. On $[1/2,1]$ the product is nonpositive, butso $F(N)\ge 1/8>0$. Together with the upper bound this proves $F((0,3/2))\subset (0,3/2)$. Induction now givesThe hypothesis on the first image is essential for initial values outside that invariant interval.

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.