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Let $\mathcal{A}$ be the closure of the polynomials in $C(K)$ for the uniform norm. It is a closed algebra, since products of uniformly convergent sequences converge uniformly on the compact set $K$.

Choose a point $W$ with $|W|>{\sup }_{z\in K}|z|$. The uniformly convergent geometric seriesshows that this reciprocal belongs to $\mathcal{A}$. Join $w$ to $W$ by a continuous path in $\mathbb{C}\setminus K$. Its image is compact and disjoint from $K$, so their distance $d$ is positive. Partition the path into finitely many steps shorter than $d/2$.

Suppose $1/(b-z)\in \mathcal{A}$ and $|a-b|<d/2$, with $b$ on the path. Thenconverges uniformly on $K$, since $|b-z|\ge d$. Every partial sum belongs to $\mathcal{A}$, by its algebra property, and so does the limit. Working backwards along the partition transfers the pole from $W$ to $w$. Thus $1/(w-z)\in \mathcal{A}$: a polynomial $P$ exists withThis pole-moving polynomial approximation is the elementary mechanism behind the polynomial Runge theorem.

Solve $\dot{Q}=-AQ$, $Q(0)=I$. Since $A^T=-A$, differentiation gives $Q^{T}Q=I$. To see why the Lax equation preserves the spectrum, direct differentiation givesThus $L(t)=Q(t)L(0)Q(t{)}^T$ is an orthogonal similarity, so all eigenvalues and all coefficients of its characteristic polynomial are first integrals.

A $2n$-dimensional Hamiltonian system is Liouville integrable when it has $n$ functionally independent first integrals, including the Hamiltonian, whose pairwise Poisson brackets vanish, on an open dense regular region.

For the three-particle periodic Toda lattice, take $p_i={\dot{q}}_i$. The proposed Hamiltonian haswhich are exactly the equations of motion. Differentiating the new variables givesThe transformation is not canonical. For examplerather than canonical coordinate brackets. Moreover $a_1{a}_2{a}_3=1/8$, so the map forgets the common translation of all the $q_i$ and is not an invertible coordinate change on the six-dimensional phase space.

Multiplication of the displayed matrices gives diagonal entries $2(a_i^2-a_{i-1}^2)$ and off-diagonal entries $(b_{i+1}-b_i)a_i$ in $LA-AL$. Therefore they form the required Lax pair. The three characteristic coefficients areThey are independent first integrals, as permitted without proof. Equivalently one may takeIndeed $C_1=-P/2$, $C_2=P^2/8-H/4$ and $C_3=1/4-J/8$. Their mutual Poisson commutation also follows directly: $P$ generates the common translation, and $\{H,J\}=0$ because $J$ is conserved.

There are axis equilibria $E_{\pm }=(\pm \sqrt{\mu },0)$ when $\mu \ge 0$, and $E_a=(a,a^2-\mu )$ for every $\mu$. At an axis equilibrium $(x_{\ast },0)$ the eigenvalues of the Jacobian matrix are $2x_{\ast }$ and $a-x_{\ast }$. Thus a negative axis root is a stable node when $a<x_{\ast }$ and a saddle when $a>x_{\ast }$. A positive axis root is an unstable node when $a>x_{\ast }$ and a saddle when $a<x_{\ast }$. Equalities give a zero eigenvalue.

At $E_a$,It is a saddle for $\mu <a^2$. For $\mu >a^2$, it is stable if $a<0$, unstable if $a>0$, with real node eigenvalues for $a^2<\mu \le 2a^2$ and a focus for $\mu >2a^2$. At $\mu =2a^2$ the node-focus transition is not a bifurcation of hyperbolicity when $a\ne 0$. For $a=0$, $E_a$ has imaginary eigenvalues when $\mu >0$ and the linearized system is a centre. For $a\ne 0$, the candidate local bifurcation values are $\boxed{\mu =0and\mu =a^2}$. When $a=0$ these zero-eigenvalue events coincide at $\mu =0$, while the off-axis equilibrium also has zero-real-part imaginary eigenvalues for every $\mu >0$.

The centre manifold theorem reduces local dynamics near zero-real-part eigenvalues to an invariant manifold tangent to their eigenspace; transverse directions with nonzero real parts retain their stable or unstable character. For a varying parameter, adjoin $\dot{\mu }=0$ and use an extended centre manifold for a parameter.

Take $a=1$. At $\mu =0$ near the origin, $y=0$ is an exact extended centre manifold. Its reduced equation isThis is a saddle-node bifurcation: no nearby axis equilibria for $\mu <0$, one at zero, and two for $\mu >0$. The transverse eigenvalue is positive, so the negative root is a saddle and the positive root an unstable node for small positive $\mu$.

At $\mu =1$, put $u=x-1$, $v=y$, $\nu =\mu -1$. Then $\dot{u}=2u+u^2-v-\nu$, $\dot{v}=-uv$, $\dot{\nu }=0$. Write the extended centre manifold as $u=h(v,\nu )$. Its invariance equation is $h_v(-hv)=2h+h^2-v-\nu$. Matching degrees givesHence the reduced dynamics areThe exact equilibrium branches are $v=0$ and $v=-\nu$, crossing at zero and exchanging stability in the centre direction. This is a transcritical bifurcation; the transverse eigenvalue remains positive. For $\nu <0$ the axis branch is an unstable node and the off-axis branch a saddle; for $\nu >0$ their roles reverse locally.

When $a=0$, the two parameter values merge and the origin at $\mu =0$ has a double zero eigenvalue with nonzero nilpotent linear part. The centre manifold is then two-dimensional, so the preceding one-dimensional classifications cannot simply be reused. Also the off-axis branch has imaginary eigenvalues for every $\mu >0$, with identically zero trace: this is not a generic Hopf bifurcation obtained by varying $\mu$. These degeneracies already follow without computing a new normal form.

Use the entire function $e^{ik{z}^3}$. The steepest descent contour starting at $1$ iswith the cube-root branch starting at $1$. Along it $ik{z}^3=ik-ku$, so the real part decreases and the imaginary part stays constant. It approaches the ray $\arg z=\pi /6$, on which $z=e^{i\pi /6}{u}^{1/3}$ and $ik{z}^3=-ku$.

Close a contour using the interval $[0,1]$, these two outgoing paths truncated at parameter $R$, and a short connecting arc. The two endpoints differ by $O(R^{-2/3})$, and the integrand on the connector decays exponentially in $R$. Cauchy integral theorem therefore gives ${\int }_0^1{e}^{ik{t}^3}dt=I_1-I_2$.

The specified substitutions giveandExpand $(1+iu{)}^{-2/3}=1-2iu/3+O(u^2)$ near zero, and apply Watson lemma, or split the integral at a fixed small $u$ and integrate the Taylor remainder. This gives the first two asymptotic termsKeeping the next term gives $-2e^{ik}/(9k^2)+O(k^{-3})$. The fractional power comes from the degenerate stationary endpoint at zero; the oscillatory integer-power terms come from the ordinary endpoint at one.

Let initial wealth be $w$ and let ${\theta }_i$ be the number of units held of risky asset $i$; the bank investment is $w-{\theta }^T{S}_0$. Put $b=\mu -(1+r)S_0$. Terminal wealth is normal, with mean $(1+r)w+{\theta }^{T}b$ and variance ${\theta }^{T}V\theta$. The expectation of the exponential utility isFor $\gamma >0$ and positive definite covariance matrix $V$, maximizing this is equivalent to maximizing the certainty equivalent ${\theta }^{T}b-\gamma {\theta }^{T}V\theta /2$. Thus the optimal investment isAll investors have proportional risky holdings. A market portfolio can therefore be represented by ${\theta }_M=V^{-1}b$, with any positive scalar multiple giving the same return. Suppose its initial value $P={\theta }_M^T{S}_0$ is positive and $b\ne 0$. Set $B=b^T{V}^{-1}b$. For returns $R_i=S_1^i/S_0^i-1$ and $R_M={\theta }_M^T{S}_1/P-1$,The beta of an asset is its covariance with the market return divided by the market variance. ConsequentlyThis is the capital asset pricing model. Since ${\beta }_i={\rho }_i{\sigma }_i/{\sigma }_M$, dividing the excess-return identity by ${\sigma }_i$ gives the Sharpe ratio identityThese ratio statements require nonzero variances and nonzero initial portfolio value. If $V$ is singular and $b$ is in its range, replace $V^{-1}$ by its pseudoinverse; any kernel holding can be added to an optimizer. If $b$ has a component in the kernel, it gives a deterministic excess gain that can be scaled without bound, so there is no finite optimal portfolio. If $b=0$, there is no distinguished risky market portfolio and its beta ratios are undefined.

Assume nonnegative wealth and rewards, and initially $0<p_i\le 1$. Set $F_0(x)=x$, and let $R_i$ take the values $0,2a_i$ with equal probability conditional on not being caught. For the retirement threshold with multiplicative capture risk, the dynamic programming equation is the displayed maximum; capture contributes zero. Write $t={\max }_i{a}_i{q}_i/p_i$.

Inductively, if $x\ge t$, all future successful wealth levels also lie above $t$, so $F_{s-1}(x+R_i)=x+R_i$. Continuing in town $i$ is then worth $q_i(x+a_i)\le x$, and retirement is optimal. If $x<t$, choose $i$ with $a_i{q}_i/p_i>x$. Continuing for just one night and then retiring is worth $q_i(x+a_i)>x$, so immediate retirement cannot be optimal for any $s\ge 1$. This proves an optimal policy continues exactly when $\boxed{x<t}$, with retirement chosen at the threshold.

Now suppose $q_1>q_2$ and $q_1{a}_1>q_2{a}_2$. Since also $p_1<p_2$, town $1$ has the larger threshold, so $t=a_1{q}_1/p_1$. The functions $F_s$ are nondecreasing in wealth and in $s$, and $F_s(x)\ge x$; these properties follow by induction from the dynamic programming equation.

Prove by induction on the remaining horizon that town $2$ is never optimal when continuation is worthwhile. For $s=1$, the difference between the town values is $(q_1-q_2)x+q_1{a}_1-q_2{a}_2>0$. If $a_2\le a_1$, monotonicity giveswith strictness because the first bracket is positive under the hypotheses.

For $a_2>a_1$, first suppose $x+2a_2\ge t$. The second town's successful large-reward branch retires, so its value is $q_2[F_{s-1}(x)+x+2a_2]/2$. Town $1$ is worth at least $q_1[F_{s-1}(x)+x+2a_1]/2$. Their difference is at leastIf instead $x+2a_2<t$, all states after either first-night reward remain below the threshold. By induction a policy starting in town $2$ then chooses town $1$ on its second night. Its value isInterchanging the first two towns gives exactly the same value, because capture probabilities multiply and independent rewards add. After starting in town $1$, however, the induction hypothesis says town $2$ is strictly inferior on the second night at every possible surviving wealth level. Replacing that second choice by the optimal one strictly improves the value. This proves town $1$ dominates throughout every finite horizon.

If a town has $p_i=0$ and $a_i>0$, its threshold is interpreted as infinity and retirement is not optimal before the horizon ends. A town with $q_i{a}_i=0$ offers no gain; the natural threshold is zero. These conventions handle the degenerate capture probabilities rather than dividing by zero.

An estimator ${\hat{\theta }}_n$ is consistent if, for every parameter value $\theta$ and every $ϵ>0$, ${\mathbb{P}}_{\theta }(|{\hat{\theta }}_n-\theta |>ϵ)\to 0$.

For the proposed root criterion, fix $ϵ>0$. Convergence in probability at the two points givesOn this event the intermediate value theorem gives a zero between the two points. Since the zero is unique, it is ${\hat{\theta }}_n$. Therefore $\boxed{{\hat{\theta }}_n\stackrel{P}{\to }{\theta }_0}$, without needing uniform convergence of $S_n$.

For independent $N({\theta }_0,1)$ observations, the log-likelihood has derivative $n({\bar{X}}_n-\theta )$ and negative second derivative $-n$. Its unique maximizer is ${\hat{\theta }}_n={\bar{X}}_n$. Apply the criterion to $S_n(\theta )=\theta -{\bar{X}}_n$: the strong law of large numbers gives the limit $S(\theta )=\theta -{\theta }_0$, with the required signs. Hence the maximum-likelihood estimator is consistent.

In the bivariate model the log-likelihood, apart from constants, isMinimizing each squared pair gives $\boxed{{\hat{\mu }}_i=(X_{1i}+X_{2i})/2}$. Maximizing the resulting likelihood in the variance givesEach difference is $N(0,2{\sigma }^2)$, independently of the others and independently of its unknown mean parameter. Thus the strong law of large numbers gives ${\hat{\sigma }}^2\to {\sigma }^2/2$ in probability, not ${\sigma }^2$. This is the Neyman-Scott incidental parameter problem: the number of nuisance means grows with the sample size, but each mean is estimated from only two observations. The usual fixed-dimensional likelihood consistency theory therefore does not apply. The corrected estimator $2{\hat{\sigma }}^2$ is consistent.

A parametrization is $X(\theta ,z)=(\cosh z\cos \theta ,\cosh z\sin \theta ,z)$. It gives a homeomorphism $S^1\times \mathbb{R}\to S$, with inverse obtained from the height and angular position. Its first fundamental form isFor a surface of revolution with radius $r(z)$, the Gaussian curvature is $-r^{\prime \prime }/[r(1+r^{\prime 2}{)}^2]$. Thus $\boxed{K=-{\mathrm{sech}}^{4}z<0}$. The waist $z=0$ is a closed geodesic: ${\Gamma }_{\theta \theta }^z=-\tanh z$ vanishes there, and $\theta$ can be parametrized at constant speed.

For the uniqueness of a closed geodesic on a negatively curved cylinder, use Gauss-Bonnet theorem with piecewise geodesic boundary:A geodesic disk with smooth boundary would require curvature integral $2\pi$. A disk bounded by a geodesic monogon would require $\pi +\alpha >0$, and a geodesic bigon would require $\alpha +\beta >0$. All are impossible when $K<0$.

Here is the topological reduction needed to apply this observation without assuming simplicity. Lift a closed geodesic to the universal covering plane of the cylinder. A self-intersection of its lift supplies an innermost monogon by the Jordan curve theorem, so the lift is embedded. A contractible closed geodesic would similarly supply a geodesic disk or monogon, so its winding number is nonzero. Its lift is consequently a proper line, invariant under a nonzero power of the deck translation. Two distinct translates of that line cannot intersect: periodicity would repeat an intersection, and innermost arcs between intersections would bound a bigon. The disjoint translates are ordered transversely, with the deck translation preserving that order. A finite cyclic permutation of this ordered family is impossible, so the lift is invariant under the primitive deck translation too. Thus the original geodesic has a simple image and is at most a repeated traversal of it.

Likewise two distinct simple essential closed geodesics cannot intersect: their periodic lifts would produce a bigon. If disjoint, the Jordan curve theorem on the cylinder says they bound a compact annulus. Its two boundaries have zero geodesic curvature and $\chi =0$, so Gauss-Bonnet gives $\int KdA=0$, again impossible. Therefore there is at most one closed geodesic image. Reparametrizations, reversed orientation and multiple traversals are understood to describe the same geometric geodesic.

A reduced positive definite binary quadratic form $a{x}^2+bxy+c{y}^2$ satisfies $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. These inequalities select representatives under proper equivalence of binary quadratic forms; the changes of variables have integer entries and determinant $1$.

For the first binary quadratic form, put $x=u-v$, $y=v$. For the second, put $x=-u-3v$, $y=u+2v$. Both changes preserve coprime integers, and giveProper representation means using $\gcd (u,v)=1$. For $f$, $|v|\ge 2$ gives a value greater than $5$, while $v=0$ forces $u=\pm 1$. For $|v|=1$, the smallest values are $4$ and $5$. For $g$, $v=0$ again gives only $1$ primitively; $|v|=1$ gives $11,12,15,\dots$, and $|v|\ge 2$ gives at least $44$. Thus the three smallest properly represented values areFor example the original variables can be chosen as $(1,0),(-1,1),(-2,1)$ for $f$, and $(-1,1),(-3,2),(-4,3)$ for $g$.

Every positive definite binary quadratic form can be reduced: integer shears first make $|b|\le a$, and interchanging the variables when $c<a$ decreases the positive leading coefficient, so this procedure terminates. To enumerate the reduced forms of discriminant minus forty-four, reduction gives $3a^2\le 44$, hence $1\le a\le 3$. Checking $b^2-4ac=-44$ with the reduced inequalities gives exactlyThe second represents only even integers. The last two have the same represented values, by $u\mapsto -u$. The first and third are the reduced versions of $g$ and $f$. Therefore every represented odd integer occurs in at least one of $f,g$, whether or not its original representation was primitive.

On the complex torus $\mathbb{C}/\Lambda$, the Weierstrass elliptic function $\wp$ is a degree-two map to the Riemann sphere, and its fibres are the pairs $\{z,-z\}$, with coalescence at the four branch points. An even elliptic function is constant on every fibre, so away from those branch points it descends to a meromorphic function $Q$ of $\wp$.

The descent also holds at a branch point. At a half-period $h$, evenness and periodicity give $f(h+u)=f(h-u)$, so its local Laurent series uses only even powers of $u$. Also $\wp (h+u)-\wp (h)$ is a local coordinate proportional to $u^2$. At $0$, use $1/\wp (z)\sim z^2$ instead. Thus $Q$ extends meromorphically across every branch point and infinity. Every meromorphic function on the Riemann sphere is a rational function, proving $f=Q\circ \wp$.

For a nonconstant rational function $Q=p/q$ in coprime form, $\mathrm{deg}Q=\max (\mathrm{deg}p,\mathrm{deg}q)$: a generic equation $p-wq=0$ has that many roots on the sphere. Degrees of holomorphic maps multiply under composition. Hence the degree isConstant functions have degree zero. Degree two therefore occurs exactly when $Q$ is a Möbius transformation:Conversely every such function is even, elliptic and of degree two.

The Baire category theorem in the needed version states that a nonempty complete metric space cannot be a countable union of closed sets with empty interior. To prove it, suppose $F_1,F_2,\dots$ are such sets and start with a nonempty open set $U$. Inductively choose closed balls ${\overline{B}}_j$ of positive radius at most $2^{-j}$, with ${\overline{B}}_1\subset U\setminus F_1$ and ${\overline{B}}_{j+1}\subset B_j\setminus F_{j+1}$. This is possible because each complement is open and dense. The centres form a Cauchy sequence; completeness supplies a limit in every closed ball. It lies in $U$ and outside every $F_j$, proving the theorem.

Here the sequence spaces are over the real scalars, as required by the Euclidean-plane assertion. Suppose an isometry $\phi :{\mathbb{R}}^2\to c_0$ existed. On the compact complete space $S^1\times S^1$, the setsare closed. They cover that space: for a nonzero sequence in $c_0$, the supremum norm is attained at a coordinate; for $x=y$ every coordinate works. By the Baire category theorem, one $F_n$ contains a product of open arcs. Shrink the arcs to be disjoint. The real coordinate difference is then continuous and nonzero, so its sign is constant on the product. Thus the chord distance has the separated formIn particular its alternating sum on every four-point rectangle is zero. Parametrize the arcs by angles $s,t$ with $0<t-s<2\pi$. Their chord distance is $D(s,t)=2\sin ((t-s)/2)$, whose mixed derivative isThe zero rectangle identity would force this mixed derivative to vanish, a contradiction. Therefore no such isometry exists.

For a real vector $x\in {\ell }_1^n$, list its signed sums over all $2^n$ choices $\epsilon \in \{-1,1{\}}^n$, followed by infinitely many zero coordinates:This is linear, has finite support and lies in $c_0$. Choosing the signs of the components gives ${\max }_{\epsilon }|\epsilon \cdot x|={\sum }_j|x_j|$. Thus $\boxed{\Vert Tx{\Vert }_{\infty }=\Vert x{\Vert }_1}$, the required linear isometry.

A simplicial approximation to $f:|K|\to |L|$ is a simplicial map $g:K\to L$ satisfying the open-star conditionfor every vertex $v$. Equivalently, whenever $f(x)$ lies in the interior of a simplex of $L$, $|g|(x)$ lies in that simplex. In particular $f$ and $|g|$ are joined by the simplexwise straight-line homotopy.

The Simplicial approximation theorem says that, for finite $K$, every continuous $f$ has such an approximation after sufficiently many barycentric subdivisions of $K$. For general complexes one uses a subdivision adapted to the open-star cover; a single uniform number of barycentric subdivisions is the compact-domain version.

Given triangulations $X\cong |K|$, $Y\cong |L|$, choose a simplicial approximation $g$ on a suitable subdivision $K^{\prime }$. Its chain map sends an oriented simplex $[v_0,\dots ,v_r]$ to $[g(v_0),\dots ,g(v_r)]$, with repeated-vertex simplices sent to zero. It commutes with the boundary operator and induces $g_{\ast }$. Compose this with the subdivision isomorphism $H_r(K)\to H_r(K^{\prime })$ and the triangulation identifications to define $f_{\ast }:H_r(X)\to H_r(Y)$.

For $z\mapsto z^n$, a positively oriented fundamental loop winds $n$ times around the target circle. With integer coefficients,All higher homology groups vanish. The $H_1$ result can also be read from a subdivision into $n$ copies of a target fundamental cycle.