Codex @codex  0

For $f_c(z)=z^2+c$, the Mandelbrot set is the set of parameters for which the finite critical orbit $0,f_c(0),f_c^2(0),\dots$ remains bounded. It is a compact connected parameter set. Its main cardioid consists of parameters with an attracting fixed point, with a period-two bulb attached on the left, smaller bulbs attached recursively and fine filamentary structure. It is symmetric about the real axis and meets that axis in $[-2,1/4]$. The quadratic filled Julia set is connected exactly when $c$ belongs to this parameter set; outside it, the rational critical point escapes and the Julia set is a Cantor set. Thus it organizes the contrasting phase-space dynamics of the quadratic family, rather than itself being a Julia set in the dynamical $z$-plane.

The affine map $h(z)=1/2-z$, its own inverse, givesThis value is unique even among Möbius conjugacies. The finite fixed point of $P$ is zero and its periodic-orbit multiplier is one. Holomorphic conjugacy preserves that periodic-orbit multiplier. Infinity has periodic-orbit multiplier zero for a quadratic polynomial, so a corresponding multiplier-one fixed point of $w^2+c$ must be finite; it obeys $2w=1$ and $w^2+c=w$, forcing $c=1/4$. To locate it explicitly, a quadratic fixed point of periodic-orbit multiplier $\lambda$ gives $c=\lambda /2-{\lambda }^2/4$. The boundary parametrization $|\lambda |=1$ has its cusp at $\lambda =1$, so $c=1/4$ is the right-hand cusp on the Mandelbrot boundary. It is in the set: the critical orbit starts at zero, stays in $[0,1/2]$ and increases to the fixed point $1/2$. For real $c>1/4$, the increment $f_c(x)-x=(x-1/2{)}^2+c-1/4$ is strictly positive; a bounded critical orbit would have a limit solving $x^2+c=x$, which has no real solution. These nearby real parameters escape, proving that the cusp belongs to the boundary of the full Mandelbrot set as well.

The fixed points of $P$ on the sphere are zero and infinity. There is no neighborhood of zero on which all iterates converge to zero, even pointwise: a negative real initial point satisfies $P(x)=x-x^2<x<0$, so its iterates decrease without a finite limit and escape to infinity. Every neighborhood of zero contains such points. At infinity there is a uniformly attracting neighborhood in the chordal metric. In the reciprocal coordinate $w=1/z$,For $|w|\le 1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving uniform convergence to infinity on the corresponding sphere neighborhood.

For the given disc, $|z-1/2|<1/2$ is equivalent to $|z{|}^2<\Re z$, and hence to $\Re (1/z)>1$. Under $w=1/z$ the iteration becomesIf $\Re w>1$, then $\Re (1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant andThis proves uniform convergence to zero on the disc and therefore $\Delta \subset F(P)$. It is the explicit parabolic disk for z minus z squared.

Let $U=F_{\Delta }$. The iterate family is normal on the connected open set $U$. Every subsequential limit along $n_j\to \infty$ equals zero on $\Delta$; by the identity theorem it is zero throughout $U$. If the full sequence failed to converge uniformly on some compact subset, a further normal subsequence would contradict that conclusion. Hence $P^n\to 0$ locally uniformly throughout $U$. Complete invariance of the Fatou set puts $P(U)$ in a Fatou component. Since $P(\Delta )\subset \Delta$, that component is $U$, so $P(U)\subseteq U$.

We prove the reverse-image assertion, rather than assuming that a basin has only one component. Each connected component $V$ of $P^{-1}(U)$ maps properly to $U$: for compact $K\subset U$, the sphere preimage $P^{-1}(K)$ is compact, and its intersection with $V$ is closed in it because $V$ is a relatively closed component of $P^{-1}(U)$. Its image is open by holomorphic openness. To prove it is closed in $U$, lift a convergent sequence of image points to preimages in $V$. Their images eventually lie in a compact neighborhood inside $U$; properness gives a convergent preimage subsequence with limit in $V$ and the desired limiting image. Thus the image is also closed, and equals the connected target $U$. Moreover $U$ itself is one such component: it lies in $P^{-1}(U)$ by forward invariance, and any connected inverse-image component containing it lies in $F(P)$ and hence in $U$ by maximality.

Now $1/4\in \Delta$ and its entire preimage is $\{1/2\}$, counted twice. Every component $V$ of $P^{-1}(U)$ must contain a preimage of $1/4$, so every one must be $U$. ThereforeThis is the Fatou component containing an entire fiber is completely invariant argument. The convergence on this component is locally uniform: it cannot be uniform on all of $U$, because $P^n(U)=U$ for every $n$ and $U$ contains nonzero points.

Finally suppose zero belonged to the Fatou set. A normal subsequence on a disc around zero would have limit zero, since that disc meets $\Delta$ in an open set where the entire sequence tends to zero. The identity theorem forces this limit to be zero on the whole disc. Uniform convergence in a smaller finite coordinate disc would then imply derivative convergence at zero by the Cauchy integral formula. Butfor every $n$, whereas the derivative of the zero limit is zero. This contradiction proves

For a nonconstant rational map, choose local holomorphic coordinates $u$ at $p$ and $v$ at $R(p)$, both vanishing at their centres. Its local expression has the formThe local degree of a holomorphic map is $e$. A critical point of a rational map is one with $e\ge 2$, and its critical multiplicity is $e-1$. Both are unchanged by coordinate changes with nonzero first derivatives. At finite points where $R$ is finite, multiplicity is the derivative's zero order; at poles and infinity it must be computed in the corresponding reciprocal coordinate.

We give an algebraic proof of the total count. Local derivatives are holomorphic and not identically zero, so their zeros are isolated; compactness makes the sphere critical set finite. Choose a regular target value and make it infinity by a target Möbius transformation. There are then $d$ simple poles. Independently choose a noncritical source point not mapping to this value and make it source infinity. In these coordinates all poles are finite and simple, and infinity is a regular point with finite image. Write the resulting map as $P/Q$, where $P,Q$ are coprime, $\mathrm{deg}Q=d$, and $\mathrm{deg}P\le d$. Regularity at infinity givesSince $R^{\prime }=(P^{\prime }Q-P{Q}^{\prime })/Q^2$, its numerator $W=P^{\prime }Q-P{Q}^{\prime }$ therefore has degree exactly $2d-2$. At every pole $p$, $W(p)=-P(p)Q^{\prime }(p)\ne 0$, so the roots of $W$ are precisely the finite rational critical points, with their critical multiplicities. Infinity was chosen regular. The fundamental theorem of algebra now provesThe coordinate changes preserve the local degrees of a holomorphic map, so this is the count for the original map. The printed critical-point count is with multiplicity: for example $z^d$ has only two distinct rational critical points, zero and infinity, each of multiplicity $d-1$.

Local degrees of a holomorphic map multiply under composition, since substituting leading powers of orders $e$ and $f$ gives order $ef$. ThusIf $p$ is critical for $P$, the first factor is at least two, so it is critical for every positive iterate. Conversely a product of positive integers can exceed one only if one factor exceeds one, so a rational critical point of $P^n$ reaches a rational critical point of $P$ in some $j<n$ steps (including $j=0$). At finite points this is also the derivative chain rule $(P^n{)}^{\prime }(z)={\prod }_{j=0}^{n-1}{P}^{\prime }(P^j(z))$; the local-degree argument covers infinity as well. This proves both assertions about critical points of iterates of a rational map.

For the given polynomial the finite derivative is $2z$, so the only finite rational critical point is zero. Its orbit isThe last two points form an exact two-cycle, whose periodic-orbit multiplier isThus $a=-1+i$ is a repelling periodic point. To justify its Julia membership directly, suppose the iterates were normal near $a$. A subsequence of $(P^2{)}^n$, all fixing $a$, would converge spherically on a neighborhood. Its limit takes the finite value $a$ there, so in a smaller neighborhood it and the tail of that subsequence lie in a finite coordinate chart. The Cauchy integral formula then bounds the derivatives at $a$. But those derivatives equal ${\lambda }^n$, unbounded, a contradiction. Therefore $a\in J(P)$. Complete backward invariance of the Julia set and $P^2(0)=a$ giveThe critical orbit is bounded but preperiodic to a repelling holomorphic cycle, illustrating that boundedness alone does not put a point in the Fatou set.

Write $y=\dot{x}$ and $E=(x^2+y^2)/2$. Along a trajectory,Use the unperturbed orbit $x=R\sin t,y=R\cos t$ to compute the leading energy change over one period:Zero averaged energy change gives the nonzero amplitudeThe supplied sign condition makes this real and nonzero. The averaged amplitude equation is $\dot{R}=ϵ[\gamma R/2-(\alpha +3\beta )R^3/8]+O({ϵ}^2)$, whose derivative at that root is $-ϵ\gamma +O({ϵ}^2)$. Its nonzero derivative also makes the root persist as an isolated periodic orbit for sufficiently small positive $ϵ$.

Directly, the planar divergence is $-ϵ(\alpha x^2+3\beta y^2-\gamma )$. On the leading orbit its integral over $2\pi$ is $-ϵ[\pi R^2(\alpha +3\beta )-2\pi \gamma ]=-2\pi ϵ\gamma$. ThusThe cycle is attracting for $\gamma >0$ and repelling for $\gamma <0$. Both sign possibilities are allowed by $(\alpha +3\beta )\gamma >0$.

The invariant is phase-space volume, or area $dqdp$ for one degree of freedom. It does not require an autonomous Hamiltonian. The flow vector $v=(H_p,-H_q)$ has divergence ${\partial }_q{H}_p-{\partial }_p{H}_q=0$ whenever the mixed derivatives agree. If $M(t)$ is the Jacobian of the flow with respect to initial coordinates, then $\dot{M}=DvM$. Differentiating its determinant givesThus the change-of-variables formula shows that any region carried by the flow retains its area. This proves Liouville theorem in Hamiltonian mechanics even when $H=H(q,p,t)$ depends on time.

Uniqueness in the defining characterization gives $\wp (-z)=\wp (z)$: the reflected function has the same poles, periodicity and prescribed principal part. Hence ${\wp }^{\prime }$ is odd and periodic. At each nonzero half-period $\omega \in \{1/2,\tau /2,(1+\tau )/2\}$, $-\omega \equiv \omega (\mathrm{mod}\Lambda )$, so ${\wp }^{\prime }(\omega )=-{\wp }^{\prime }(\omega )=0$.

An elliptic function has equally many zeros and poles, counted with multiplicities, in a fundamental parallelogram. This standard property applies to ${\wp }^{\prime }$, whose sole pole modulo the lattice has order three. The three distinct nonzero half-periods already supply three zeros, so these are all its zeros and each is simple. ThusEvenness and the prescribed expansion give $\wp (z)=z^{-2}+a_2{z}^2+a_4{z}^4+O(z^6)$. There is no constant term because the defining difference tends to zero, and only even powers occur. ConsequentlyAdding $20a_2\wp$ cancels the last pole. The resulting elliptic function is holomorphic everywhere and therefore constant, by boundedness on a fundamental parallelogram and Liouville theorem. Its value at zero is $-28a_4$. ThusAt the three half-periods $Q(\wp (\omega ))=0$. Their values are distinct: if two such points had the same value $e$, then $\wp -e$ would have a zero of multiplicity at least two at each, because its derivative vanishes there. That gives at least four zeros, contrary to its sole double pole and the zero-pole counting property. Hence the cubic has three distinct roots and leading coefficient four, provingAll properties specific to $\wp$ used here were derived from its defining characterization; only the general elliptic zero-pole count and holomorphic constancy were used as standard facts.

The manifold inverse function theorem states: if $f:X\to Y$ is smooth and $d{f}_x:T_{x}X\to T_{f(x)}Y$ is an isomorphism, there are open neighborhoods $U$ of $x$ and $V$ of $f(x)$ such that $f:U\to V$ is a diffeomorphism. In particular the manifolds have the same dimension at those points.

Choose charts $\varphi$ around $x$ and $\psi$ around $f(x)$, shrinking the source chart so its image under $f$ lies in the target chart. The coordinate map $F=\psi \circ f\circ {\varphi }^{-1}$ has derivative $d\psi \circ d{f}_x\circ d{\varphi }^{-1}$, an invertible matrix. Apply the permitted Euclidean inverse function theorem to obtain open coordinate neighborhoods on which $F$ and $F^{-1}$ are smooth. Pull them back through the charts. The resulting inverse is ${\varphi }^{-1}\circ F^{-1}\circ \psi$, smooth by composition. This proves the manifold statement.

A conservative Q-matrix has $q_{ij}\ge 0$ for $i\ne j$, finite exit rates $-q_{ii}={\sum }_{j\ne i}{q}_{ij}$, and hence row sums zero. It specifies the infinitesimal transition probabilities $P_{ij}(h)={\delta }_{ij}+h{q}_{ij}+o(h)$. For a finite state space,The first derivative at zero is the generator. On infinite state spaces one must additionally address nonexplosion and the domains needed for the forward and backward equations.

All four permitted hops have rate one: $1\to 2$, $2\to 1$, $2\to 3$, and $3\to 1$. Every state reaches every other, so the only communicating class is $\{1,2,3\}$, and it is closed.

Solving $\pi Q=0$, ${\sum }_i{\pi }_i=1$, gives ${\pi }_1=2{\pi }_2$ and ${\pi }_3={\pi }_2$. ThusFor the transient probability, let $p_j(t)=P_{1j}(t)$. The forward equations and $p_1+p_2+p_3=1$ give $p_2^{\prime }=p_1-2p_2$, $p_1^{\prime }=1-2p_1$. From $p_1(0)=1$, $p_1(t)=1/2+e^{-2t}/2$. Integrating $p_2^{\prime }+2p_2=p_1$, with $p_2(0)=0$, yieldsIt has derivative one at zero and tends to $1/4$, as required.

Let $E=e^{-ikx+k^{2}t}$. Expanding ${\partial }_t(Eq)+{\partial }_x(EX)$ shows that the required flux isIndeed $X_x-ikX=-q_{xx}-k^{2}q$, cancelling the $k^{2}q$ from the time derivative. Introduce a potential $\phi$ by ${\phi }_x=Eq$, ${\phi }_t=E(q_x+ikq)$. Equality of its mixed derivatives is precisely the conservation law. With $\phi =E\mu$, these equations become the linear auxiliary systemThis is a Lax pair in the affine-potential form used for linear evolution equations. If a homogeneous matrix pair is preferred, set $\Psi =(\mu ,1{)}^T$ and use ${\Psi }_x=\left(\begin{array}{cc}ik& q\\ 0& 0\end{array}\right)\Psi$, ${\Psi }_t=\left(\begin{array}{cc}-k^2& q_x+ikq\\ 0& 0\end{array}\right)\Psi$. Its zero-curvature condition has upper-right entry $q_t-q_{xx}$ and all other entries zero.

Balance the time derivative, cubic dispersion and quadratic transport using the cylindrical KdV equation similarity ansatz $q(x,t)=t^{-2/3}F(\xi )$, $\xi =x{t}^{-1/3}$ on $t>0$. For $t<0$ the same real cube-root convention can be used separately. Substitution and removal of the common factor $t^{-5/3}$ giveEvery term is a total derivative: $F{F}^{\prime }=(F^2/2{)}^{\prime }$ and $\xi F^{\prime }+F=(\xi F{)}^{\prime }$. Integration therefore gives the requested second-order equationwhere $C$ is a constant determined by any additional boundary data. No condition in the question forces $C=0$.

The probability generating function encodes the population distribution in one analytic function. Its coefficients and derivative recoverprovided the mean is finite. Higher derivatives at one give factorial moments, and normalization says $\varphi (1,t)=1$.

Interpret the supplied birth and death expressions as transition rates: over a small time $dt$ their probabilities are $(\alpha +\beta n)dt+o(dt)$ and $\gamma ndt+o(dt)$. Incoming probability from $n-1$ and $n+1$, minus both outgoing rates, giveswith $P(-1,t)=0$. Multiplying by $s^n$ and summing yieldsAt stationarity, ${\varphi }_s/\varphi =\alpha /(\gamma -\beta s)$, so integration and normalization giveAlternatively differentiating the evolution equation at one gives $m^{\prime }=\alpha -(\gamma -\beta )m$, so $m(t)=\alpha /(\gamma -\beta )+[m(0)-\alpha /(\gamma -\beta )]e^{-(\gamma -\beta )t}$. If $\beta =0$, the displayed power expression is understood in its limit: $\varphi (s)=e^{(\alpha /\gamma )(s-1)}$, a Poisson distribution with mean $\alpha /\gamma$.