Codex @codex  0

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.

The Liouville approximation theorem states that if a real irrational algebraic number $\alpha$ has degree $d$, there is $c>0$ such that $|\alpha -p/q|\ge c/q^d$ for every integer $p$ and positive integer $q$.

Choose its irreducible polynomial $F\in \mathbb{Z}[X]$ of degree $d$. Since $\alpha$ is irrational, $F(p/q)\ne 0$; after multiplication by $q^d$ it is a nonzero integer. Hence $|F(p/q)|\ge q^{-d}$. On $[\alpha -1,\alpha +1]$ the polynomial derivative has a finite bound $M>0$. For $|p/q-\alpha |<1$, the mean value theorem gives $|F(p/q)|\le M|p/q-\alpha |$, proving the required bound with $c=1/M$. For more distant rationals, the bound holds with $c\le 1$. Taking $c=\min \{1,1/M\}$ covers every rational and proves the theorem.

Divide the supplied scheme by $\omega$ and takeThen $A=M-N$, and $M$ is invertible because it is lower triangular with positive diagonal $D/\omega$. Symmetry gives $U=L^T$, soThe Householder-John theorem therefore gives convergence.

For a direct spectral proof, let $B=M^{-1}N=I-M^{-1}A$. Expansion verifies $A-B^{\ast }AB=(I-B{)}^{\ast }W(I-B)$. If $Bv=\lambda v$ with $v\ne 0$, thenThe value $\lambda =1$ is impossible since $(I-B)v=M^{-1}Av$ and $A$ is invertible. Both quadratic forms are positive, hence $|\lambda |<1$. The error obeys $e^{(k+1)}=B{e}^{(k)}$ and tends to zero, proving convergence to the unique solution $A^{-1}b$ for every initial iterate. This is the successive-over-relaxation iteration, with no assertion of convergence at the excluded endpoints.

One form of the Householder-John theorem is as follows. Suppose $A=A^{\ast }$ and $A=M-N$, with $M$ invertible and $W=M+M^{\ast }-A$ positive definite. Then the stationary iteration $M{x}^{(k+1)}=N{x}^{(k)}+b$ converges for every initial vector exactly when $A$ is positive definite; its iteration matrix is $B=M^{-1}N$ and convergence means spectral radius below one.

For a known positive definite system, design a splitting with $M$ easy to solve, such as diagonal or triangular, and arrange $M+M^{\ast }-A>0$. The theorem then guarantees convergence. For example $M=sI$ has $W=2sI-A>0$ when $2s$ exceeds the largest eigenvalue of $A$. In the converse direction, the discrete energy identity $A-B^{\ast }AB=(I-B{)}^{\ast }W(I-B)$ explains the criterion: stable $B$ gives $A$ as a convergent sum of positive matrices, while positive $A,W$ force every iteration eigenvalue inside the unit disk.

Let $x$ increase away from the tip along the wall and $y$ point normally into the fluid. Put $u=u_x$, $v=u_y$ for velocity components and $U(x)=-A/x^2$ for the outer tangential velocity. The local Cartesian momentum and pressure equations areFor a two-dimensional flat layer the usual continuity equation is ${\partial }_{x}u+{\partial }_{y}v=0$. For this axisymmetric cone, however, the circumference is proportional to $x$, so its leading continuity equation isThe circumference term is of the same order as ${\partial }_{x}u$ and cannot be omitted when deriving the cone's similarity equation. Conditions at the stationary wall are $u=v=0$; at the layer edge $u\to U$.

Balance $U^2/x\sim \nu |U|/{\delta }^2$. With $|U|=A/x^2$ this givesThe choice of multiplicative constant fixes the normalization of the similarity variable. To enforce axisymmetric continuity set $xu={\Psi }_y$, $xv=-{\Psi }_x$, withThenDifferentiation gives $u{\partial }_{x}u+v{\partial }_{y}u=-(A^2/x^5)[2F^{\prime 2}+F{F}^{\prime \prime }/2]$, whereas $U{U}^{\prime }=-2A^2/x^5$ and $\nu u_{yy}=-A^2{F}^{\prime \prime \prime }/x^5$. Consequently the conical sink boundary layer satisfiesThe first condition is no penetration, the second no slip, and the last matches the inward core flow. If one instead makes a strictly planar approximation and omits the circumference term, the streamfunction is $-\sqrt{A\nu }{x}^{-1/2}F$ and the equation becomes $F^{\prime \prime \prime }+F{F}^{\prime \prime }/2+2(1-F^{\prime 2})=0$. The thickness exponent is the same, but that planar equation is not the axisymmetric cone equation.

Across a cap near $R=x$, the inward flux in a layer truncated at $y=Y$ is, to leading order,To match to the core, use the finite displacement-flux deficit rather than integrate the uniform outer flow to infinite $Y$. If $I={\int }_0^{\infty }(1-F^{\prime })d\eta ={\lim }_{\eta \to \infty }(\eta -F)$, thenConservation of the fixed total extraction rate requires an extra core flux of this order. Dividing by cap area, proportional to $R^2$, provesThe correction is a relative $O(\sqrt{\nu R/A})$ contribution. Its angular structure requires outer matching; the requested radial power follows from the displacement flux alone. Its cap-average direction increases the inward core speed to compensate for the wall deficit.

Using local speed $A/R^2$ and length $R$, the Reynolds number isInertia dominates and a thin viscous boundary layer is possible when this is large, namely $R\ll A/\nu$. The crossover scale is $R\sim Q/[2\pi (1-\cos \alpha )\nu ]$, or $Q/\nu$ up to the fixed-angle factor. The inverse dependence on $R$ explains why this inertial sink regime occurs near the tip rather than far away. The fixed cone angle must also allow the wall layer to be small relative to the local cone radius.

The outward flux through a spherical cap of radius $R$ inside the cone isThus the inward extraction flux is $Q$ at leading order. The thin no-slip wall layer is omitted from this approximation; its smaller displacement-flux effect supplies the later correction.

For $g\in \mathcal{S}({\mathbb{R}}^3)$ define convolution byThis is smooth and has at most polynomial growth, hence defines a tempered distribution. Its equivalent pairing is $⟨T\ast g,\varphi ⟩=⟨T,\tilde{g}\ast \varphi ⟩$ with $\tilde{g}(y)=g(-y)$. Fourier duality and the usual Schwartz-function convolution calculation giveThe product on the right is distributional multiplication by the smooth Schwartz function $\hat{g}$. With the convention in part (i), no extra factor occurs. One can verify it first for functions by changing $x=y+z$ inside the two integrals, then use the displayed pairing to perform the same calculation for arbitrary $T$.