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The regular level set theorem says that a regular level of a smooth map from an $m$-manifold to an $r$-manifold is an $(m-r)$-manifold. The corresponding sublevel-set theorem says that $\{F\le c\}$ is a manifold with boundary $F^{-1}(c)$ when $c$ is a regular value of a smooth real-valued function, with the interior $F<c$.

Apply this to $F(x)=|x{|}^2$ on ${\mathbb{R}}^{n+1}$. On $F^{-1}(1)$ its gradient $2x$ is nonzero. Thus $S^n$ is an $n$-manifold and $B^{n+1}$ is an $(n+1)$-manifold with boundary $\boxed{\partial B^{n+1}=S^n}$. Locally the implicit function theorem supplies $F-1$ as one coordinate; the ball occupies its nonpositive side, which is a half-space after reversing that coordinate. This also covers $n=0$: the sphere consists of two points and the ball is the interval with those endpoints.

A regular value $y$ of a smooth map $f:X\to Y$ is one for which the differential $d{f}_x:T_{x}X\to T_{y}Y$ is surjective at every $x\in f^{-1}(y)$. A value outside the image is regular vacuously. The condition concerns all preimages, not just one.

An embedded smooth $n$-dimensional manifold $X\subseteq {\mathbb{R}}^N$ has, around every point, a parametrization by an open subset of ${\mathbb{R}}^n$ which is a homeomorphism onto a relative neighbourhood, has rank $n$, and has smooth coordinate changes. Equivalently, after smooth ambient coordinates it is locally a coordinate $n$-plane. For a manifold with boundary use relatively open subsets of ${\mathbb{H}}^n=\{u_n\ge 0\}$ instead, with parametrizations and transitions smoothly extendible across the boundary. Points corresponding to $u_n=0$ form the boundary. Interior charts are ordinary manifold charts; a boundary point is not a corner.

A singular point would satisfy $p=0$, $p_w=2w=0$ and $p_z=-3z^2+2z+1=0$. Thus $w=0$ and $z=1$ or $-1/3$. But $p(1,0)=1$ and $p(-1/3,0)=-5/27$, both nonzero. The curve is smooth everywhere.

The complex implicit function theorem says that near a zero of a holomorphic function with a nonzero partial derivative, the corresponding variable is a uniquely defined holomorphic function of the others. Where $p_w\ne 0$, use $z$ as a local coordinate with $w=w(z)$; where $p_z\ne 0$, use $w$ with $z=z(w)$. On overlaps the coordinate transitions are holomorphic with nonzero derivatives. These charts give $Z$ a Riemann surface structure, and both coordinate projections are holomorphic in them.

A ramification point of a nonconstant holomorphic map is a point where in local coordinates the map has the form $u\mapsto u^e$ with $e\ge 2$; its ramification, or branching, order is $e-1$. For $g=w$, take $(z,w)=(1,i)$. Here $p_w=2i\ne 0$, so $z$ is a local coordinate. Differentiating $w(z{)}^2=z^3-z^2-z$ gives $w^{\prime }(1)=0$ and $2i{w}^{\prime \prime }(1)=4$. ThereforeThe nonzero quadratic coefficient gives local degree $\boxed{2}$ and branching order $\boxed{1}$. If “branching order” denotes local degree rather than its excess in a different convention, the corresponding value is two.

A rational function is a quotient $r=P/Q$ of polynomials with $Q\ne 0$. Cancel common factors. It determines a map on the Riemann sphere by assigning infinity at poles and assigning at infinity the limit in the sphere. Equivalently homogenize $P,Q$ to the same degree $d=\max (\mathrm{deg}P,\mathrm{deg}Q)$ and set $f([Z:W])=[P_h(Z,W):Q_h(Z,W)]$. They have no common projective zero: finite ones would contradict coprimality, and at infinity at least one has a nonzero leading coefficient. This also proves the map is holomorphic.

The degree of the nonconstant rational function is $d$. The degree of a holomorphic map of compact Riemann surfaces is the number of preimages of any target point, counted with local mapping multiplicity. For target $[a:b]$, the homogeneous equation $b{P}_h-a{Q}_h=0$ is a nonzero homogeneous polynomial of degree $d$. Over $\mathbb{C}$ it factors into $d$ linear factors, with multiplicity, giving exactly $d$ projective zeros. Locally its zero order is precisely the map's multiplicity, since the other target-coordinate denominator is nonzero. The two degrees are equal.

There is a Möbius transformation sending any ordered triple of distinct points to $0,1,\infty$. When all are finite one can use $T_A(z)=(z-a_1)(a_2-a_3)/[(z-a_3)(a_2-a_1)]$; the formulas for infinity are its limits. Define $T_B$ in the same way. Then $T_B^{-1}\circ T_A$ sends $A$ to $B$ and restricts to a biholomorphism $X\setminus A\to X\setminus B$.

Let $P_m$ be the linear subspace of polynomials of degree at most $m$. It has finite dimension $m+1$, is proper in $P$, and $P={\bigcup }_{m\ge 0}{P}_m$. For any norm on $P$, the restricted norm on $P_m$ is equivalent to its coefficient Euclidean norm: on the coefficient unit sphere the continuous norm has a positive minimum and a finite maximum. Thus $P_m$ is complete. A complete subspace of a metric space is closed, since any sequence converging to an ambient limit is Cauchy and has the same limit in the subspace.

Part (b) makes each $P_m$ a closed set with empty interior, hence nowhere dense. If $P$ were a Banach space, this countable union would contradict the Baire category theorem. No norm makes the space of all polynomials complete.

If a linear subspace $V$ of a normed space contains an open ball $B(v,r)$, then $v\in V$ and subtraction gives $B(0,r)\subseteq V$. For any $x\in X$, choose a positive scalar $t$ with $\Vert tx\Vert <r$. Then $tx\in V$ and hence $x\in V$. Thus $V=X$. Every proper linear subspace has empty interior.

The Baire category theorem states that a complete metric space is a Baire space: the intersection of countably many open dense sets is dense. Equivalently, a nonempty open subset cannot be a countable union of nowhere dense sets.

Let $U_n$ be open dense and $O$ any nonempty open set. Choose a closed ball $B_1$ of positive radius at most $1/2$ inside $O\cap U_1$. Inductively choose $B_{n+1}$ inside the interior of $B_n$ intersected with $U_{n+1}$, with positive radius at most $2^{-(n+1)}$. Density makes this intersection nonempty, and openness supplies a sufficiently small closed ball. The centres form a Cauchy sequence because all later centres lie in $B_n$, whose diameter tends to zero. Completeness gives a limit $x$. Since each ball is closed and contains all later centres, $x\in B_n$ for every $n$. Thus $x\in O\cap {\bigcap }_n{U}_n$, proving density. Applying this to complements of the closures of nowhere dense sets gives the equivalent formulation.

Write $f=X^2+bX+c$ and $g=X^2+BX+C$. Since their common square-free discriminant $d$ is congruent to zero or one modulo four, it must be one modulo four; thus $b,B$ are odd. It is not a rational square because a square-free integer square is only zero or one. Both polynomials are irreducible over $\mathbb{Q}$, and their roots generate the same quadratic extension $\mathbb{Q}(\sqrt{d})$. Their integral roots are $(-b+\sqrt{d})/2$ and $(-B+\sqrt{d})/2$, respectively. Each generated order isthe usual ring of integers for square-free $d\equiv 1(\mathrm{mod}4)$. This identifies both quotients with the same ring. Explicitly, if $\xi$ denotes the image of $X$ modulo $g$, send the image of $X$ modulo $f$ to $\xi +(B-b)/2$. Equal discriminants give $f(\xi +(B-b)/2)=g(\xi )=0$, and the opposite translation gives its inverse. The two quotient rings are isomorphic.

The cubic has no rational root: its only possible integer roots are $\pm 1$, and neither works. It is therefore irreducible. Let $\alpha$ be its image in $K$; it is an algebraic integer, so $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_K$. The power-basis discriminant isThe index-discriminant formula from part (i) says that the square of $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ divides 59. It must be one. Hence $\boxed{{\mathcal{O}}_K=\mathbb{Z}[\alpha ]}$, with integral basis $1,\alpha ,{\alpha }^2$ and field discriminant $-59$.

For a basis of a module $m_1,\dots ,m_d$ of the full-rank module $M$, define its discriminant of elements of a number field byThe field discriminant is this determinant for an integral basis of ${\mathcal{O}}_K$. A change of integral basis with determinant $\pm 1$ does not change it. Write $m_j={\sum }_i{b}_{ij}{e}_i$ in an integral basis and let $B=(b_{ij})$. The two trace matrices satisfy $T_M=B^{\mathsf{T}}{T}_{{\mathcal{O}}_K}B$. The index of the lattice is $r=|\det B|$, as follows for example from Smith normal form. ThereforeMultiplicative closure makes $M$ an order when it contains one, but the determinant argument itself requires only full finite index, so it also covers the stated module hypothesis without that extra condition.

Take a maximal matching in a graph with $m$ edges and let $S$ be its $2m$ endpoints. Maximality makes its complement an independent set, since an edge wholly outside $S$ could be added. Every vertex outside $S$ therefore sends all its $k$ edges into $S$, giving $k(n-2m)$ crossing edges. Every vertex in $S$ already uses at least one incident edge inside $S$, its matched edge, so it sends at most $k-1$ outside. HenceThe maximum matching in a graph is at least this large. For $k=2$, take any disjoint union of $r$ triangles. It is 2-regular, has $n=3r$ vertices and matching number $r=n/3$, since each triangle contributes exactly one edge. As $r$ varies this gives infinitely many equality examples.

Let $A,B$ be the two parts of the bipartite graph. Counting edges gives $k|A|=k|B|$, hence $|A|=|B|=n/2$. For $S\subseteq A$, the $k|S|$ edges incident to $S$ all end in $N(S)$, whose vertices have total capacity $k|N(S)|$. Thus $|N(S)|\ge |S|$. Hall's marriage theorem gives a perfect matching, and no matching in a graph can have more than $n/2$ edges. The matching number is $\boxed{n/2}$.

Hall's marriage theorem states that a finite bipartite graph with parts $A,B$ has a matching in a graph saturating $A$ exactly when $|N(S)|\ge |S|$ for every $S\subseteq A$. Necessity follows because the matched neighbours of $S$ are distinct.

For sufficiency use induction on $|A|$. The empty case is immediate. If a nonempty proper subset $S$ has $|N(S)|=|S|$, the induced graph on $S,N(S)$ satisfies Hall's condition and has a saturating matching in a graph by induction. The remaining graph on $A\setminus S,B\setminus N(S)$ also satisfies it: for $T\subseteq A\setminus S$,Apply induction again and unite the two matchings in a graph. If there is no such tight proper subset, every nonempty proper $S$ has $|N(S)|\ge |S|+1$. Choose a vertex $a$ and a neighbour $b$, which exists by Hall's condition. Delete both. Every remaining subset loses at most one neighbour and still satisfies Hall's condition. Induction supplies a matching in a graph of the remainder; add $ab$. For $|A|=1$ this last step directly proves the assertion.

For constant equation-of-state parameter $w$, the cosmological continuity equation integrates to $\rho ={\rho }_0{a}^{-3(1+w)}$ using today's normalization. In conformal time, $\mathcal{H}=a^{\prime }/a=\dot{a}$, so the Friedmann equation becomesSince ${\mathcal{H}}^{\prime }=a\ddot{a}$, the Friedmann acceleration equation givesFor $k{c}^2=1$, put $q=(1+3w)/2$. When $q\ne 0$, integration gives $\mathcal{H}=\cot (q(\tau -{\tau }_0))$. Integrating $a^{\prime }/a=\mathcal{H}$ then yields the closed constant-equation-of-state Friedmann scale factorChoose a connected interval with positive sine; the origin of conformal time can absorb ${\tau }_0$. For $w>-1/3$ this is the expansion and recollapse branch between consecutive zeros of the sine. Negative $q$ instead gives the corresponding minimum-radius branch on its positive-sine interval. The printed power formula excludes $w=-1/3$: in that exceptional case $\mathcal{H}=H_0$ is constant and $a=A{e}^{H_0\tau }$, with $H_0^2+1=8\pi G{\rho }_0/3$. A real solution requires the latter constant to be at least one; equality permits a static solution.

For pressure-free matter, $w=0$, so $\beta =1/2$. Set the big-bang origins to zero. The closed matter-dominated Friedmann solution follows by ${\sin }^2(\beta \tau )=(1-\cos 2\beta \tau )/2$ and integrating $dt=ad\tau$:The entire cycle is $0\le 2\beta \tau \le 2\pi$; maximum expansion has $a=\alpha$ at $2\beta \tau =\pi$, and the big crunch occurs at $t=\alpha \pi /(2\beta )$.

Multiply the Friedmann equation by $a^2$ and differentiate:Insert the cosmological continuity equation to obtainOn a nonstatic branch divide by $2a\dot{a}$; continuity extends the result across isolated turning points. The resulting Friedmann acceleration equation isAn everywhere static branch needs the independent spatial Einstein field equations: the two differentiated constraints alone cannot determine the pressure when $\dot{a}\equiv 0$. This is the qualification to deriving the acceleration equation solely by division.

By Gibbs inequality,with equality exactly when $\boxed{a_j=p_j}$. Put $q_j=u_j/U$, a probability distribution. The optimal log-optimal multiplicative betting rate isFor $U<1$ this is strictly positive by Gibbs inequality; taking $0<\epsilon <g_{\ast }$ in the preceding strong law of large numbers bound proves growth with probability tending to one.

For $U>1$ the casino can set $u_j=U{p}_j$. Every optimal return multiplier is then exactly $1/U$, so the fortune decreases deterministically as $f{U}^{-m}$. More generally $g_{\ast }=-\log U<0$ for that choice. An arbitrary choice with $U>1$ need not cause decline. For example, take $p_1=p_2=1/2$, $u_1=0.01$, $u_2=1.99$. Then $U=2$ but $g_{\ast }=(1/2)\log [0.25/(0.01\cdot 1.99)]>0$. For any fixed positive $p$, making one $u_j$ sufficiently small while preserving $U$ also makes the optimal rate positive.

Let $J_1,J_2,\dots$ be independent rolls with the specified probability distribution, and let $F_m$ be the fortune after $m$ games. Its multiplicative evolution givesby the strong law of large numbers. The summands have finitely many finite values, so its hypotheses hold. For any $\epsilon >0$, the probability that $f{e}^{m(g-\epsilon )}\le F_m\le f{e}^{m(g+\epsilon )}$ tends to one. Thus maximizing $g(a)$ maximizes the typical exponential growth rate, the objective of log-optimal multiplicative betting. Maximizing expected fortune would be a different objective and would not control the typical repeated-game outcome.

For probability distributions $p,q$ with positive entries, Gibbs inequality iswith equality exactly when $p=q$. Indeed, $\log x\le x-1$, with equality only at $x=1$, givesThis also proves the equality condition. With zero entries the same Kullback-Leibler divergence statement holds using $0\log (0/q)=0$ and $p\log (p/0)=+\infty$ for $p>0$, or by restricting to the support and taking limits.

The applied pressure changes the dynamic condition to ${\varphi }_t+g\eta =-p/\rho$. For its specified spatial mode, the two amplitude equations becomeWhen $\Omega \ne \omega$, the corresponding particular oscillation isFree oscillations with amplitudes $C\cos \omega t+D\sin \omega t$ may be added. If the mode starts from rest at $t=0$, its total amplitude instead isNear resonance, the response grows too large for the linearization. At exact resonance the rest-start limit is $A=(\epsilon h/2)(\omega t\cos \omega t-\sin \omega t)$, with secular growth. Dissipation or nonlinear effects would be required to limit it.