Codex @codex  0

Write $Q=1r^T-RI$, with $r=R\pi$. Then ${\pi }^{T}1=1$ andThus $\pi$ is a stationary probability vector, a left eigenvector for eigenvalue zero. If any probability vector $\nu$ is stationary, the same calculation gives ${\nu }^{T}Q=r^T-R{\nu }^T=0$, forcing $\nu =\pi$. Hence the equilibrium is unique, even if some $r_j$ vanish and the chain is not irreducible; those zero-rate states are transient or inaccessible rather than extra stationary classes.

For a finite-state continuous-time Markov chain with generator matrix $Q$, $P(t)=e^{tQ}$. The exponential solves both Kolmogorov equations, $P^{\prime }=QP=PQ$, with initial condition $P(0)=I$.

Put $M=\Vert f{\Vert }_{\infty }>0$ and $M_0=\mu (E)$. There is a positive-measure set on which $|f|>M/2$, so $M_0>0$ and each $M_n$ is positive; boundedness and finite measure make each finite. For $n\ge 1$, apply Cauchy-Schwarz to $|f{|}^{(n-1)/2}$ and $|f{|}^{(n+1)/2}$ to getTherefore the positive ratios $r_n=M_{n+1}/M_n$ are nondecreasing. Telescoping gives $M_n/M_0=r_0\cdots r_{n-1}\le r_n^n$, while $|f{|}^{n+1}\le M|f{|}^n$ almost everywhere gives $r_n\le M$. HenceFor $0<\epsilon <M$, the set $A_{\epsilon }=\{|f|>M-\epsilon \}$ has positive measure by the definition of essential supremum. Thus $\Vert f{\Vert }_n\ge (M-\epsilon )\mu (A_{\epsilon }{)}^{1/n}$, whereas $\Vert f{\Vert }_n\le M\mu (E{)}^{1/n}$. Letting $n\to \infty$ and then $\epsilon \downarrow 0$ shows $\mu (E{)}^{-1/n}\Vert f{\Vert }_n\to M$. The preceding squeeze yields $M_{n+1}/M_n\to \Vert f{\Vert }_{\infty }$. This proves the ratio limit rather than assuming it from the moment-root limit.

The Minkowski inequality states the triangle inequality for every $L^p$ norm with $1\le p\le \infty$. Apply it to $f_n=(f_n-f)+f$ and then to $f=(f-f_n)+f_n$. The two bounds combine toThus norm convergence implies convergence of the norms themselves.

For $1\le p<\infty$, the Lp norm is $\Vert f{\Vert }_p=({\int }_E|f{|}^{p}d\mu {)}^{1/p}$; for $p=\infty$ it is the essential supremum of $|f|$. The Lp space consists of measurable functions with finite norm, identifying functions equal almost everywhere. This identification makes the quantity a norm rather than only a seminorm. Convergence in $L^p$ means $\Vert f_n-f{\Vert }_p\to 0$, including the essential-supremum norm when $p=\infty$.

Parametrize the torus by $F(u,v)=((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u)$. Its outward Gauss map isThe derivative $N_u$ has unit length, while $N_v$ has length $|\cos u|$ and is perpendicular to $N_u$. Thus rank drops precisely at $u=\pi /2$ or $3\pi /2$. These are the two circles of height $z=1$ and $z=-1$ and radius two, and their images are the north and south poles. Therefore the critical values are $(0,0,\pm 1)$.

For any nonpolar target point, its third coordinate determines two values of $u$, one with positive and one with negative $\cos u$. For each, the longitude fixes a unique $v$ modulo $2\pi$, with a shift of $\pi$ when $\cos u$ changes sign. Thus every regular value has exactly two preimages, one on the outer and one on the inner side of the torus. ConsequentlyThis explicit two-sheet description justifies the count without relying on a sketch.

Assume $X$ is a compact smooth $n$-manifold without boundary and $Y$ a connected smooth $n$-manifold without boundary. For a regular value $y$, the inverse function theorem makes $f^{-1}(y)$ discrete, and compactness makes it finite. Define ${\mathrm{deg}}_{2}f=|f^{-1}(y)|(\mathrm{mod}2)$. No orientations are required.

The parity is independent of the regular value: join two regular values by a path and perturb the path in its interior to be transverse to $f$. Its inverse image is a compact one-dimensional manifold with boundary consisting of the two endpoint fibres. Every compact one-manifold has an even number of boundary points, so the two counts have the same parity. The same argument proves homotopy invariance. For noncompact $X$ one instead needs a proper map to retain compactness of the relevant inverse images.

If $f$ omitted a point of $Y$, that point would be a regular value with empty fibre and hence degree zero. Independence on the connected target gives the contradiction. Therefore a nonzero degree modulo two forces surjectivity.

A point $x\in X$ is critical if $d{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is an image of a critical point; a regular value is a point $y\in Y$ for which every point of $f^{-1}(y)$ is regular, including vacuously values with empty preimage. Sard theorem says that for a smooth map of finite-dimensional manifolds, the critical values have measure zero in every coordinate chart of the target. In particular regular values are dense.

For a nonsingular plane curve $F(s,t)=0$, at a point with $F_t\ne 0$ the holomorphic implicit function theorem gives $t=t(s)$ and the chart coordinate $s$. Where $F_s\ne 0$, use $t$ and express $s=s(t)$. Nonsingularity ensures at least one choice is available, and the coordinate transition functions and both restricted projections are holomorphic.

Here $F=t^4-P(s)$ with $P(s)=(s^2-1)(s-4)$. A singular point would require $t=0$, $P(s)=0$ and $P^{\prime }(s)=0$. But $P$ has three distinct roots $-1,1,4$, so this is impossible. Hence the curve is nonsingular.

If $t\ne 0$, $F_t=4t^3\ne 0$, so $s$ is a local coordinate and the projection to $s$ is unramified. At $(s_0,0)$ for $s_0\in \{-1,1,4\}$, use $t$ as coordinate. Since $P^{\prime }(s_0)\ne 0$, solving $P(s)=t^4$ gives $s-s_0=t^4/P^{\prime }(s_0)+O(t^8)$. ThusThese are all the ramification points on the stated affine curve; no points at infinity belong to $Y\subset {\mathbb{C}}^2$.

Choose coordinates $z$ at $p$ and $w$ at $f(p)$ vanishing at those points. The local expansion is $w\circ f\circ z^{-1}(\zeta )=a{\zeta }^v+O({\zeta }^{v+1})$, $a\ne 0$. Define the branching order $v_f(p)=v$. A change of either coordinate has nonzero linear coefficient, so it preserves the leading exponent, proving well-definedness.

Ramification means $v>1$, equivalently vanishing of the derivative in local coordinates. These zeros are isolated unless the local derivative is identically zero, which by analytic continuation would make $f$ constant. An infinite set of ramification points on a compact surface would have an accumulation point, contradicting isolation. Thus that set is finite.

A nonconstant holomorphic map of compact connected Riemann surfaces is onto, since its image is open by the local form and closed by compactness. It has a finite degree $d\ge 1$. The Riemann-Hurwitz formula isIf both genera equal $g>1$, positivity of $2g-2$ forces $d=1$ and every ramification term zero. The map is bijective and has nonzero derivative everywhere. The holomorphic inverse function theorem says that a holomorphic function with nonzero derivative has a local holomorphic inverse; these inverses patch, so $f$ is biholomorphic. For $g=1$, multiplication by two on $\mathbb{C}/\Lambda$ is a degree-four holomorphic map of an elliptic curve to itself and is not biholomorphic.

Let $a=i\sqrt{2+\sqrt{2}}$ and $b=i\sqrt{2-\sqrt{2}}$. The roots are $\pm a,\pm b$. Since $\sqrt{2}=-a^2-2$ and $ab=-\sqrt{2}$, we have $b=-\sqrt{2}/a\in \mathbb{Q}(a)$. Hence the splitting field is $K=\mathbb{Q}(a)$. The quartic is Eisenstein at two, so its degree is four.

The automorphism $r$ sending $a$ to $b$ sends $\sqrt{2}$ to $-\sqrt{2}$, and therefore sends $b=-\sqrt{2}/a$ to $\sqrt{2}/b=-a$. Its successive images of $a$ are $a,b,-a,-b,a$. Thus $r$ has order four and supplies all automorphisms:Complex conjugation is $r^2$. The images of $a$ specify the full action because $K$ is generated by that one root.

Let $a=3^{1/4}>0$. The four roots are $a,ia,-a,-ia$, so $K=\mathbb{Q}(a,i)$. Eisenstein's criterion at three gives $[\mathbb{Q}(a):\mathbb{Q}]=4$. This field is real, so adjoining $i$ doubles the degree and $[K:\mathbb{Q}]=8$.

Define $r(a)=ia$, $r(i)=i$, and complex conjugation $s(a)=a$, $s(i)=-i$. The first automorphism exists because over $\mathbb{Q}(i)$ the minimal polynomial of $a$ has degree four and all four roots are in $K$; the second is immediate. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct and exhaust the eight automorphisms. Thusthe dihedral group of order eight, acting on the four roots as the symmetries of a square. On every element of $K$, the action is determined by the displayed images of $a,i$, or by extending them linearly to the basis $a^j{i}^e$ with $0\le j<4$, $0\le e<2$.

With $\sigma =1$, substitute $x=(u+v)/2$, $y=(u-v)/2$, $r=1+\mu$. The equations becomeThe extended centre variables are $u,\mu$. The centre manifold theorem gives smooth graph functions $V,Z$ with $V(0,0)=Z(0,0)=0$ and all first derivatives zero. Their invariance equations are $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$, after substitution of the graph.

Count $\mu$ as order $u^2$. Then $\dot{u}=O(u^3)$ on the graph, so the left side of the $Z$ equation contributes only at higher order. Its quadratic terms give $0=u^2/4-bZ$, hence $Z=u^2/(4b)+O(u^3)$. The cubic terms of the $V$ equation similarly give $0=-2V-\mu u/2+uZ/2$. ThusTerms purely in $\mu$ vanish because the origin remains an equilibrium for every parameter; these expansions describe the weighted scaling $\mu =O(u^2)$, not an unrestricted one-variable Taylor series. Substitution gives the reduced equationIts cubic coefficient is negative. For $\mu <0$ the origin is attracting in the centre direction; for $\mu >0$ it loses stability and two attracting branches emerge with $u\sim \pm 2\sqrt{b\mu }$. Therefore the bifurcation is a supercritical pitchfork. These branches agree with the exact Lorenz equilibria $x=y=\pm \sqrt{b\mu }$, $z=\mu$.

The centre manifold theorem says that near an equilibrium of a sufficiently smooth finite-dimensional vector field, there is a locally invariant centre manifold tangent to the generalized eigenspace of eigenvalues with zero real part. It can be represented as a graph over that space, with graph value and derivative zero at the equilibrium; the dynamics on it determine the local centre-direction behaviour. For parameter-dependent systems the parameters may be adjoined as variables with zero derivative, producing an extended centre manifold.

At $r=1$, the linearization has the $x,y$ block $\left(\begin{array}{cc}-\sigma & \sigma \\ 1& -1\end{array}\right)$ and the independent eigenvalue $-b$. Its other eigenvalues are zero and $-(\sigma +1)$. Hence the equilibrium is nonhyperbolic, the stable subspace has dimension two, and the non-extended centre subspace has dimension one. A centre eigenvector is $(1,1,0)$.

Choose $A=B=\sigma /r$ in the proposed Lyapunov function. The cubic terms cancel and direct differentiation givesFor $0<r<1$ this is negative definite, while $V$ is positive definite and radially unbounded. Finite-dimensional equivalence of these positive quadratic forms gives $\dot{V}\le -cV$ for some $c>0$. Thus $V(t)\le V(0)e^{-ct}$, every solution remains bounded and exists for all forward times, and every solution tends to zero. Therefore the origin is globally asymptotically stable, indeed exponentially stable.

At the threshold, the determinant quadratic has a double zero. Its vertex lies at $s=(\chi \alpha n_0-\beta D_n-\gamma n_0{D}_c)/(2D_n{D}_c)$. Substituting the equality case of part (iii) givesThe mode is neutral exactly at onset and grows when the chemotactic coefficient increases beyond that threshold.

For a Laplacian mode with ${\nabla }^2=-k^2$, the linearized chemotaxis system isIts trace is negative for all $k$. Writing $s=k^2$, its determinant isInstability occurs when this determinant is negative, because then one real eigenvalue is positive. A quadratic with positive leading and constant coefficients is negative for some $s>0$ precisely when its middle coefficient is negative and its discriminant positive. These conditions combine toThis proves the printed threshold. In a bounded spatial domain, only its permitted Laplacian eigenmodes are available; the continuous-$k$ threshold describes the onset when wavenumbers can vary freely.

The positive homogeneous equilibrium is $n=n_0$, $C=\alpha n_0/\beta$. For spatially constant perturbations $u,v$, linearization gives $\dot{u}=-\gamma n_{0}u$ and $\dot{v}=\alpha u-\beta v$. The triangular matrix has eigenvalues $-\gamma n_0$ and $-\beta$, both negative, proving asymptotic stability to homogeneous perturbations. The other homogeneous equilibrium, zero population and zero chemical, has positive population eigenvalue $\gamma n_0$ and is not the desired stable state.

The term $\gamma n(n_0-n)$ is logistic population growth, with carrying density $n_0$ and low-density growth rate $\gamma n_0$. The cell flux is $-D_n\nabla n+\chi n\nabla C$: its first part is diffusion down the density gradient, and its second is chemotaxis toward higher chemical concentration. Taking minus its divergence gives the two transport terms. Chemical production is proportional to cell density at rate $\alpha n$, degradation is first order at rate $\beta C$, and the chemical diffuses with coefficient $D_c$. All the given positive signs correspond to growth, attraction, production, degradation and ordinary diffusion respectively.

Let ${\xi }_{n+1}=a_{n+1}+1/(a_{n+2}+\cdots )$ be the complete quotient. Iterating the fractional-linear formula for a continued fraction givesThe supplied determinant identity and ${\xi }_{n+1}>a_{n+1}$ therefore implyThe positive partial quotients ensure increasing denominators tending to infinity; the same determinant identity makes the even and odd convergents nested, giving their common limit $x$. The displayed formula follows by passing to the limit in the finite-tail identity.

An explicit choice is $a_n=10^{n!}$ for every $n\ge 1$. The recurrence gives $q_n\le {\prod }_{j=1}^n(a_j+1)\le 2^{n}10^{{\sum }_{j=1}^{n}j!}$. For $n\ge 2$, ${\sum }_{j=1}^{n}j!\le 2n!$, so ${\log }_{10}{q}_n\le 3n!$. Consequently $a_{n+1}=10^{(n+1)!}\ge q_n^{(n+1)/3}$ andFirst $x$ is irrational: if $x=P/Q$, the nonzero difference from a convergent is at least $1/(Q{q}_n)$, contradicting the error bound once $q_{n+1}>Q$. If $x$ were algebraic of any fixed degree $d$, the last displayed upper bound would eventually be smaller than $c/q_n^d$ for the positive constant in part (i), since $q_n\to \infty$. This contradicts the Liouville approximation theorem. Thus this explicit continued fraction is transcendental.