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The nonzero fixed point of the logistic map is $x_{\ast }=(\mu -1)/\mu$, with multiplier $F^{\prime }(x_{\ast })=2-\mu$. It is stable for $1<\mu <3$, and its multiplier crosses $-1$ at $\mu =3$. To identify the emerging orbit explicitly, factorThe remaining two roots, interchanged by $F$, areThey coincide with $x_{\ast }$ at three and are distinct for $\mu >3$. Their cycle multiplier isIt lies between $-1$ and $1$ for $3<\mu <1+\sqrt{6}$. This establishes the first period-doubling bifurcation and the stable bifurcating two-cycle.

At ${\mu }_2=1+\sqrt{6}$, ${\Lambda }_2=-1$ and $d{\Lambda }_2/d\mu =-2\sqrt{6}\ne 0$. The second flip is nondegenerate: for $H=F^2$ at either cycle point, the cubic flip coefficient is $c=(H^{\prime \prime \prime }/6)+(H^{\prime \prime }/2{)}^2=-\mathcal{S}H/6$ when $H^{\prime }=-1$. Here $\mathcal{S}$ denotes the Schwarzian derivative. Since $\mathcal{S}F=-6/(1-2x{)}^2<0$ and $\mathcal{S}(F^2)=(\mathcal{S}F\circ F)(F^{\prime }{)}^2+\mathcal{S}F$, this coefficient is positive. Locally, after following the fixed point of $H$, its second iterate has the form $H^2(y)-y=2a(\mu -{\mu }_2)y-2c{y}^3+\cdots$, with $a=2\sqrt{6}>0$. Nonzero roots therefore appear with $y^2\sim a(\mu -{\mu }_2)/c$ on the increasing-parameter side. These are a four-cycle of $F$, proving the second period-doubling bifurcation occurs at $\mu =1+\sqrt{6}$.

Let $F(x)=\sqrt{2+\sqrt{x}}$ for $x\ge 0$. This is an increasing continuous function. Starting at $a_1=\sqrt{2}$, induction gives $\sqrt{2}\le a_n\le 2$: if $a_n\le 2$, then $a_{n+1}\le \sqrt{2+\sqrt{2}}<2$, and every iterate is at least $\sqrt{2}$.

The first step has $a_2>\sqrt{2}=a_1$. Since $F$ is increasing, $a_n\ge a_{n-1}$ implies $a_{n+1}=F(a_n)\ge F(a_{n-1})=a_n$. Thus this sequence is increasing and bounded above. Let $L_{-}={\sup }_n{a}_n$. For every $\epsilon >0$, a term exceeds $L_{-}-\epsilon$; all later terms lie between it and $L_{-}$. This proves convergence, the elementary bounded monotone sequence theorem.

For the second starting value, write the iterates as $b_1=4$ and $b_{n+1}=F(b_n)$. Here $b_2=2<4$. Monotonicity of $F$ propagates this inequality, so $b_n$ is decreasing. It is bounded below by $\sqrt{2}$ because every positive iterate is at least that value. Hence it converges to its infimum $L_{+}$. From $b_2=2$ onward all terms lie in $[\sqrt{2},2]$.

Continuity of $F$ permits passing to each limit in the recurrence, giving $L_{\pm }^2=2+\sqrt{L_{\pm }}$. To prove that these fixed points agree, suppose $r,s\ge \sqrt{2}$ both satisfy that equality. Subtracting the two equations and rationalizing givesThe second factor is positive: $r+s\ge 2\sqrt{2}$, whereas $(\sqrt{r}+\sqrt{s}{)}^{-1}\le 1/(2\sqrt{\sqrt{2}})<2\sqrt{2}$. Thus $r=s$. ThereforeExistence has been supplied by the two convergent sequences, and the subtraction argument gives the required uniqueness in their common interval.

The constant solutions are fixed points of $G(u)=u^2(1+u)/2$. Solving $G(u)=u$ gives $u(u+2)(u-1)=0$, hence $u_{\ast }=0,-2,1$. For a discrete dynamical system, a small perturbation satisfies ${\eta }_{n+1}=G^{\prime }(u_{\ast }){\eta }_n+O({\eta }_n^2)$, whereA multiplier of magnitude less than one contracts perturbations; a multiplier greater than one expands them. Therefore $u=0$ is asymptotically stable, while $u=-2$ and $u=1$ are unstable. In particular, if $|u|\le 1/2$, then $|G(u)|\le (3/4)|u{|}^2\le (3/8)|u|$, proving convergence to zero without relying solely on the vanishing linear multiplier.

The contraction mapping theorem says that if $(X,d)$ is a nonempty complete metric space and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one fixed point.

For the proof, start at $x_0\in X$ and set $x_n=T^n{x}_0$. Iterating the contraction bound gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality then gives, for $m>n$,Thus $(x_n)$ is a Cauchy sequence and has a limit $x_{\ast }$ by completeness. Since the contraction is continuous, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves existence and uniqueness.

For the integral map, the fundamental theorem of calculus shows that $Tf$ is continuous whenever $f$ is continuous. In the supremum norm,Hence $\Vert Tf-Tg{\Vert }_{\infty }\le \frac{3}{2}\max (a^2,b^2)\Vert f-g{\Vert }_{\infty }$. We can chooseThe space $C[a,b]$ with this supremum norm is complete: a uniformly Cauchy sequence has pointwise real limits, the Cauchy bound passes to those limits to give uniform convergence, and a uniform limit of continuous functions is continuous. The contraction mapping theorem therefore gives a unique continuous $y$ with $y(x)=1+{\int }_0^{x}3ty(t)dt$. The fundamental theorem of calculus makes this $y$ differentiable, with $y^{\prime }=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the integral equation, so uniqueness transfers to the ordinary differential equation. Explicitly $y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$, as direct differentiation confirms.

Let $J$ be a finite collection of resources with positive integer capacities $C_j$, and let $R$ be a finite collection of nonempty routes. In the standard unit-resource loss network, a call on route $r$ needs one unit simultaneously at each resource in $r$; fixed routing means that these requirements do not change when a request is rejected. Suppose route $r$ has an independent Poisson process of arrivals of rate ${\nu }_r$ and independent holding times of mean $1/{\mu }_r$. Its offered traffic is ${\alpha }_r={\nu }_r/{\mu }_r$. The Erlang fixed point approximation approximates resource blocking events as independent, rather than claiming this independence for the exact loss network.

Write $b_j$ for the approximate blocking probability at resource $j$. The reduced-load approximation screens a call by every other resource that it needs, so the offered traffic seen by resource $j$ isFor a single resource with offered traffic $a$ and capacity $C$, the Erlang loss formula isThus the Erlang fixed point approximation is the systemThe resulting route acceptance probability is ${\prod }_{j\in r}(1-b_j)$. In particular, the screening product for $a_j$ excludes $j$ itself; including it would describe a different approximation.

For existence, define $T_j(b)=E(a_j(b),C_j)$ on $[0,1{]}^J$. Each $a_j$ is a continuous polynomial, and the denominator of the Erlang loss formula is at least one. Consequently $T$ is continuous and maps this nonempty compact convex set into itself. The Brouwer fixed-point theorem supplies a fixed point. At any such point $b_j<1$, since finite offered traffic and positive capacity give $E(a,C)<1$.

For uniqueness, we construct a strictly convex function whose minimizer is equivalent to an Erlang fixed point. Put $y_j=-\log (1-b_j)$. First establish the single-resource facts needed for this change of coordinates. For $a>0$, the inverse blocking probability has the expressionThe terms with $k<C$ decrease strictly with $a$, so $E(a,C)$ increases strictly from zero to one. Let $a_C(y)$ be its inverse parametrization defined by $E(a_C(y),C)=1-e^{-y}$, with $a_C(0)=0$. If $K$ has the truncated Poisson distribution with weights proportional to $a^k/k!$, then shifting the sum gives its mean, the carried load of an Erlang loss resource, asDifferentiating the normalized weights showsThe variance is positive because both $K=0$ and $K=1$ have positive probability. Moreover, $m_C(a)\to C$ as $a\to \infty$, since the largest term in the truncated Poisson distribution dominates. Henceis continuous, strictly increasing from zero to $C$.

Now minimize the following convex potential for the Erlang fixed point over the nonnegative orthant:Each exponential term is convex, and each integral is a strictly convex function of its coordinate because its derivative $U_{C_j}$ is strictly increasing. Their sum is therefore strictly convex. It is also a coercive function: once $z$ is large enough, $U_{C_j}(z)\ge C_j/2$, so each integral grows at least linearly in its coordinate, while the exponential terms are nonnegative. Thus $F$ attains exactly one minimum.

The coordinate derivative isAt an interior coordinate of a minimizer it vanishes. At $y_j=0$, it is nonpositive, whereas the one-sided minimum condition requires it to be nonnegative; it therefore vanishes there too. Conversely, a zero gradient is a global minimizer of this convex function. Multiplying ${\partial }_{j}F=0$ by $e^{y_j}$ givesApplying the Erlang loss formula converts this exactly to $b_j=E(a_j(b),C_j)$. The change of coordinates is one-to-one, proving uniqueness under fixed routing. A zero-capacity resource can instead be fixed as blocked with probability one, with all routes using it removed before applying the proof; an unused positive-capacity resource has blocking probability zero.

For alternative routing, consider a symmetric three-node loss network. Each of its three links has capacity $1000$; every node pair offers traffic $950$, tries its direct link first, and, if that link is blocked, tries the two-link path through the third node. Let all approximate link blocking probabilities equal $b$. Each link receives its direct offered traffic and the overflow from the other two node pairs. An overflow is generated with probability $b$ and reaches the link under consideration only when its other required link accepts, with probability $1-b$. The symmetric reduced-load approximation therefore givesFor $g(b)=E(950[1+2b(1-b)],1000)-b$, evaluate the Erlang loss formula by the stable recursion $E(a,0)=1$, $E(a,k)=aE(a,k-1)/(k+aE(a,k-1))$. It givesThese signs can also be certified with exact rational arithmetic, since all four loads are rational. Continuity and the intermediate value theorem give three distinct roots, one in each intervening interval. Thus the alternative-routing approximation need not be unique. This is multiplicity of approximate fixed points, not multiplicity of stationary distributions for the exact finite irreducible Markov chain.

The associated even map is simply $g(x)=\mu -x^2$. For nonzero $x$, the odd quadratic Lorenz map has the two branches $f_L(x)=x^2-\mu$ on the positive side and $f_L(x)=\mu -x^2$ on the negative side. Their limiting values at zero are $-\mu$ and $+\mu$, while the prescribed point value is $f_L(0)=-\mu$. Thus the branches join continuously only at $\mu =0$, where the origin is a fixed point and $f_L(x)=x|x|$.

The small fixed point of $g$ isFor small negative $\mu$, $r<0$ and the odd quadratic Lorenz map has two stable fixed points $r,-r$. Their periodic-orbit multiplier is $2|r|<1$. As $\mu \uparrow 0$ these fixed points collide at zero. For small positive $\mu$, $r>0$ andso a stable symmetric two-cycle is present, with periodic-orbit multiplier $4r^2<1$. Hence the global transition isThe large fixed points $\pm (1+r)$ remain unstable near this transition and are not created or destroyed there. The parabola $g$ itself is smooth as $\mu$ passes zero; its fixed point crosses the critical point and changes the sign of its periodic-orbit multiplier. The change of orbit type occurs in the sign-lifted discrete dynamical system, whose discontinuity switches orientation.

The positive fixed point $r=(-1+\sqrt{1+4\mu })/2$ of $g$ has periodic-orbit multiplier $-2r$. At $\mu =3/4$, $r=1/2$ and the periodic-orbit multiplier crosses $-1$. Factoring the second-iterate fixed-point equation givesThe second factor yields a two-cycleIt exists for $\mu >3/4$ and has periodic-orbit multiplier $4pq=4(1-\mu )$. Its stability range is $3/4<\mu <5/4$, so $g$ undergoes a supercritical period-doubling bifurcation at $\mu =3/4$. At $\mu =1$ the two-cycle is $\{0,1\}$ and its periodic-orbit multiplier is zero, making it superstable.

For positive $\mu$ and $0<|x|<\sqrt{\mu }$, direct composition gives the local second iterateThe one-sided limits at zero are opposite, and agree with zero exactly when $\mu =1$ in this neighborhood. There $F_1(x)=\mathrm{sgn}(x)(2x^2-x^4)$ is continuous at zero and fixes it. Thus the local return branch has the same discontinuity-closing transition as the map in part (a), with effective parameter ${\mu }^2-\mu$ crossing zero.

For $3/4<\mu <1$, both $p,q$ are positive and the period transfer from a quadratic map to its Lorenz sign lift gives the two stable cyclesFor $1<\mu <5/4$, $p>0$ and $q<0$, so the periodic-orbit multiplier of the $g$ cycle is negative and the lifted orbit isIt is one stable four-cycle. Hence the gluing of cycles in an iterated Lorenz map givesEquivalently, $F_{\mu }$ changes from two small stable fixed points $\pm q$ to a stable small two-cycle $\pm (-q)$. At the critical value itself, the actual convention gives $f_L(0)=-1$, $f_L(-1)=0$, $f_L(1)=0$: the surviving critical two-cycle is $\{0,-1\}$, and $+1$ is preperiodic. This critical itinerary must be evaluated directly, not with part (b)'s noncritical sign product.

There is also a precise qualification to the printed use of “continuous”. The whole second iterate is not continuous everywhere at $\mu =1$: $F_1(1)=-1$, while ${\lim }_{x\to 1^{-}}{F}_1(x)=1$ and ${\lim }_{x\to 1^{+}}{F}_1(x)=-1$. Jumps persist at the nonzero preimages of the original discontinuity. Therefore the stated global-bifurcation criterion holds for the return map restricted to a neighborhood of zero, as in the sketches; if continuity on all of $\mathbb{R}$ is demanded literally, the claim about $f_L^2$ needs that restriction. The orbit creation and destruction above is unaffected.

A complete lattice has every join, hence every directed and finite join. Conversely, if a partially ordered set is directed-complete and has finite joins including the empty join, then for any subset $A$ the set of joins of its finite subsets is nonempty and directed. Its least upper bound is $\bigvee A$. Including the empty join supplies a bottom element and handles $A=\varnothing$.

For a directed set of partial functions from $A$ to $B$, their union is a function: any two functions have a common extension, so they agree wherever their domains overlap. This union is exactly their least upper bound in the extension order, proving directed completeness.

Write $C_x$ for the intersection of all sets containing $x$ and closed under $f$ and directed joins. Then $xRy$ means $y\in C_x$. Reflexivity is immediate, and transitivity follows because any closed set containing $x$ must contain $y$ and then every $z$ with $yRz$. The upper set $[x,\infty )$ is closed: $f$ is inflationary, and a directed join of elements above $x$ is above $x$. Thus $xRy$ implies $x\le y$, proving antisymmetry.

If $h,k\in H$, then $xRk(x)Rh(k(x))$, so $h\circ k\in H$. Moreover $h(k(x))\ge h(x)$ by monotonicity, and $h(k(x))\ge k(x)$ by inflationarity. This proves that $Hx$ is directed. Define $h_0(x)=\bigvee Hx$. Pointwise monotonicity of every $h$ implies monotonicity of $h_0$. Since $Hx\subseteq C_x$ and $C_x$ is closed under directed joins, $xR{h}_0(x)$; hence $h_0\in H$.

The original map $f$ belongs to $H$ by the definition of closedness, so $f\circ h_0\in H$ and $f(h_0(x))\le h_0(x)$ by the definition of the join. Inflationarity gives the reverse inequality. Finally, if $p=f(p)$ and $p\ge x$, the lower set $(-\infty ,p]$ is closed by monotonicity of $f$ and the join property. Thus every $h(x)$ is at most $p$, and so is $h_0(x)$. This proves the least fixed point from a directed family of maps: $h_0(x)$ is the least fixed point above $x$.

The permutation representation on the complex vector space with basis $X$ has permutation character ${\pi }_X(g)=|{\mathrm{Fix}}_X(g)|$. The multiplicity of the trivial character is $⟨{\pi }_X,1_G⟩=|G{|}^{-1}{\sum }_g|\mathrm{Fix}(g)|$. Count pairs $(g,x)$ with $gx=x$: each orbit contributes $|G|$, by the orbit-stabilizer theorem. Hence this multiplicity equals the number of orbits.

For a transitive action the average number of fixed points is one. The identity fixes $|X|>1$ points, so not every other element can fix at least one; otherwise the average would exceed one. Thus there is a derangement $g$ with ${\pi }_X(g)=0$.

If ${\pi }_X=1_G+m\chi$, evaluating at that derangement gives $\chi (g)=-1/m$. Character values are algebraic integers, since a finite-order representation matrix has roots of unity as eigenvalues, and the sum of algebraic integers is integral. A rational algebraic integer is an integer. Therefore $-1/m$ is an integer, forcing $m=1$.

The diagonal action on $X^2$ is $g(x,y)=(gx,gy)$. Its permutation character is ${\pi }_X(g{)}^2$, since both coordinates must be fixed. The orbit-count result and the real-valued character ${\pi }_X$ giveIf any $m_j\ge 2$ and another nontrivial constituent occurs, this is at least $1+4+1=6$. If just one nontrivial constituent occurs, the previous result forces its multiplicity to one. Thus $r\le 5$ implies every multiplicity is one, and the formula then gives $k=r$.

For the cubic population iteration, put $g(u)=\lambda u(1-u^2)$. Its fixed points solveThus $u_{\ast }=0$ always exists. For $\lambda \ne 0$ there are alsowhenever the expression under the square root is positive, namely $\lambda <0$ or $\lambda >1$. At $\lambda =1$ these coincide with zero; at $\lambda =0$ only zero exists.

The criterion for fixed point stability for an iteration is $|g^{\prime }(u_{\ast })|<1$. To justify it, continuity of $g^{\prime }$ gives a small interval around the fixed point where $|g^{\prime }|\le q<1$. The mean value theorem then gives $|g(u)-u_{\ast }|\le q|u-u_{\ast }|$, so this interval is invariant and the error decreases geometrically. This proves local asymptotic stability rather than merely linear boundedness.

Here $g^{\prime }(u)=\lambda (1-3u^2)$. At zero its value is $\lambda$, so zero is locally asymptotically stable for $-1<\lambda <1$. At a nonzero fixed point it isThe strict inequality $|3-2\lambda |<1$ is equivalent to $1<\lambda <2$. Both nonzero real fixed points are therefore locally asymptotically stable on that interval. This proves the two requested existence ranges. The strict derivative criterion alone makes no conclusion at the endpoint multipliers of modulus one.