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Put $w=1/z$ and $Y(w)=y(1/w)$. The equation becomesAt $w=0$ the regular-singular criterion requires $w(2/w)$ and $w^2(-w^{-n-4})$ to be analytic. This holds precisely for $n\le -2$. Thus the singularity at infinity is irregular, corresponding to essential exponential behaviour of solutions, exactly for $\boxed{n\ge -1}$. For $n\le -2$ it is regular singular, with powers and possibly logarithms rather than such essential exponentials.

For $y^{\prime \prime }=z^{3}y$, the Liouville–Green approximation uses $Q=z^3$, amplitude $Q^{-1/4}$ and phase integral $\int \sqrt{Q}dz=2z^{5/2}/5$. On a consistently chosen branch, the two leading solutions areThe derivative correction is smaller by order $z^{-5/2}$, and the two exponential signs give independent solutions.

Using the convention that Stokes lines have equal exponential phase, $\mathrm{Im}(z^{5/2})=0$, their rays are $\arg z=2\pi k/5$. The equal-magnitude, oscillatory rays, often called anti-Stokes lines, satisfy $\mathrm{Re}(z^{5/2})=0$, giving $\arg z=(2k+1)\pi /5$. If the alternative naming convention is used, these two names interchange; the stated equations specify the rays unambiguously.

To give the sector precisely, the solution recessive on the positive axis can be represented bywhere $K_{\nu }$ is the Modified Bessel function of the second kind. Substitution verifies the differential equation. Its asymptotic $K_{\nu }(w)\sim \sqrt{\pi /(2w)}{e}^{-w}$ is valid on closed subsectors of $|\arg w|<3\pi /2$ on the continued branch. ConsequentlyActual exponential decay occurs in the narrower sector $|\arg z|<\pi /5$; in the adjacent portions of the larger asymptotic sector the same continued exponential grows. This distinguishes the validity of the approximation to the positive-axis recessive solution from the directions where its modulus decays.

Write $c_1=X_1-Y$, $c_2=X_2+Y$, both positive, and $W_j=U_j^{\prime }(c_j)>0$. As usual the utility expectations must be meaningful; the improvement below is supported where both consumptions and derivatives are bounded, so it produces finite strictly positive expected gains without a tail differentiation problem.

Suppose $R=W_1/W_2$ is not almost surely constant. Choose $0<l<h$ and positive-probability sets $A$ and $B$ on which respectively $R\ge h$ and $R\le l$. By restricting these sets if necessary, keep both consumptions bounded and bounded away from zero, and their marginal utilities bounded. Put $a=\mathbb{E}(W_2{1}_A)>0$, $b=\mathbb{E}(W_2{1}_B)>0$, and choose $d$ strictly between $lb/(ha)$ and $b/a$. For $Z=1_B-d{1}_A$, we haveReplace $Y$ by $Y+\delta Z$ for small positive $\delta$. Agent one's utility has first derivative $-\mathbb{E}(W_{1}Z)>0$, and agent two's has derivative $\mathbb{E}(W_{2}Z)>0$. Positivity of consumptions is preserved on the bounded support, and uniform Taylor estimates there make both actual gains positive for sufficiently small $\delta$. Thus a Pareto-efficient transfer must satisfy

For given total wealth $s=X_1+X_2>0$, let $c$ be agent two's consumption. The ratio $R_s(c)=U_1^{\prime }(s-c)/U_2^{\prime }(c)$ is continuous and strictly increasing for $0<c<s$, by strict concavity. The endpoint marginal-utility assumptions give $R_s(c)\to 0$ as $c\downarrow 0$ and $R_s(c)\to \infty$ as $c\uparrow s$. Thus it has a unique solution $c=h_{\lambda }(s)$ to $R_s(c)=\lambda$. Continuous dependence on $s$ follows from strict monotonicity and continuity, so this solution is a measurable wealth-sharing rule. ThereforeFor a fixed weight, it is also the pointwise maximizer of $U_1(s-c)+\lambda U_2(c)$, whose strict concavity confirms uniqueness.

The Pontryagin maximum principle addresses optimization of an integral running reward, possibly with terminal reward, subject to a controlled differential equation, admissible control constraints and prescribed endpoint conditions. For a normal maximization problem $\dot{x}=f(t,x,u)$ with running reward $L$, the Hamiltonian is $H=L+\lambda f$; the optimal control maximizes it pointwise, and the adjoint satisfies $\dot{\lambda }=-H_x$ with the appropriate terminal transversality condition. State constraints require additional care when active.

Assume $p>0$ and $x(0)>0$. HereThe terminal stock is free and has no salvage reward, so an optimum with positive terminal stock has $\lambda (T)=0$. For $x>0$, maximization over $u\ge 0$ gives $u=x(p-\lambda )/2$ whenever $p-\lambda >0$. ThenSet $v=p-\lambda$, so $\dot{v}=v^2/4$ and $v(T)=p$. Integrating givesThe resulting $p-\lambda$ is positive throughout, validating the interior control formula. Integrating the state equation yieldsThe state remains positive, so its nonnegativity constraint is inactive and the terminal condition is consistent. The reward $pu-u^2/x$ is jointly concave in $(x,u)$ and the dynamics are linear, giving sufficiency of this solution. Alternatively the value function $V(t,x)=\lambda (t)x$ satisfies the Hamilton-Jacobi-Bellman equation with this maximizing control. At zero price the optimal control is zero, obtained by a continuous $p\downarrow 0$ limit; zero initial stock also gives zero extraction.

A random row vector has a multivariate normal distribution when every real linear combination of its coordinates is normally distributed, allowing a zero-variance constant. Its law is determined by its mean and covariance. For the conditional and testing formulas assume $\Sigma$ is positive definite, so both coordinate variances and the conditional variance are positive; the given ratio already requires ${\sigma }_{XX}>0$.

Put $Z=Y-\beta X$, with $\beta ={\sigma }_{XY}/{\sigma }_{XX}$. The pair $(X,Z)$ is jointly normal and $\mathrm{Cov}(X,Z)={\sigma }_{XY}-\beta {\sigma }_{XX}=0$, so $Z$ is independent of $X$. Its mean is ${\mu }_Y-\beta {\mu }_X$, and its variance is ${\sigma }_{YY}-{\sigma }_{XY}^2/{\sigma }_{XX}$. HenceStandardize each coordinate with its own positive standard deviation. The standardized pairs have mean zero and covariance $\left(\begin{array}{cc}1& \rho \\ \rho & 1\end{array}\right)$. Subtracting means and multiplying coordinates by positive constants leaves the sample correlation unchanged. Therefore the distribution of $r$ depends only on $\rho$.

For $n>2$, the stated statistic reduces algebraically toThis is strictly increasing from $(-1,1)$ onto $\mathbb{R}$, with derivative $\sqrt{n-2}(1-r^2{)}^{-3/2}$. Under $\rho =0$, change variables in the given Student density: $1+t^2/(n-2)=1/(1-r^2)$. HenceThe normalizing integral is $B(1/2,(n-2)/2)$. To test $\rho =0$ against a two-sided alternative at level $\alpha$, compute $t(r)$ and reject if $|t(r)|>t_{n-2,1-\alpha /2}$, equivalently ifA one-sided alternative uses the corresponding one-sided tail. Singular covariance gives $|r|=1$ almost surely when both variances are nonzero, and is outside the nondegenerate density calculation.

A geodesic is a curve with zero covariant acceleration in an affine parameter. For a unit-speed curve on an oriented surface with unit normal $N$, its geodesic curvature is $k_g=⟨{\gamma }^{\prime \prime },N\times {\gamma }^{\prime }⟩$: it measures the tangent-plane component of acceleration perpendicular to the tangent. A unit-speed curve is geodesic exactly when this component vanishes.

The exponential map ${\exp }_p(v)$ follows for unit time the geodesic starting at $p$ with initial velocity $v$, whenever it is defined. Since ${\exp }_p(tv)={\gamma }_v(t)$, its derivative at zero is $\boxed{d({\exp }_p{)}_0={\mathrm{id}}_{T_{p}S}}$. Choose an oriented orthonormal basis of $T_{p}S$ and let $v(\theta )$ be its unit circle. Geodesic polar coordinates are $F(r,\theta )={\exp }_p(rv(\theta ))$ in a normal neighborhood, with the usual angular singularity at $r=0$.

Put $T=F_r$ and $J=F_{\theta }$. Radial curves are unit-speed geodesics, so $E=⟨T,T⟩=1$ and ${\nabla }_{r}T=0$. The torsion-free connection gives ${\nabla }_{r}J={\nabla }_{\theta }T$, and thereforeSince $J(0,\theta )=0$, this proves $F=⟨T,J⟩=0$, the radial form of the Gauss lemma. Moreover $J(r,\theta )=r{v}^{\prime }(\theta )+O(r^2)$, with $|v^{\prime }|=1$, by the derivative of the exponential map. Hence $G=|J{|}^2=r^2+O(r^3)$, givingThe last derivative is the one-sided radial limit.

The global Gauss-Bonnet theorem for a compact oriented surface with piecewise smooth boundary isHere $K$ is Gaussian curvature, the boundary is positively oriented with the surface on its left, ${\epsilon }_j$ are signed exterior turning angles at corners, and $\chi$ is the Euler characteristic. Smooth geodesic boundary components contribute neither curvature nor corner terms.

A contractible simple closed geodesic on the cylinder would bound a disc, giving $\int KdA=2\pi$, impossible when $K<0$. Thus any such geodesic is an essential loop. If two distinct essential simple loops on an annulus intersect, the elementary bigon criterion gives two subarcs bounding an innermost disc: lifting the homotopic loops to the covering strip and choosing an innermost region supplies such a bigon. Distinct geodesics cannot have tangential intersections by uniqueness of the geodesic initial-value problem, so its two interior angles $\alpha ,\beta$ are positive. Gauss-Bonnet on this geodesic bigon gives $\int KdA=\alpha +\beta >0$, another contradiction. Therefore two distinct simple closed geodesics would be disjoint. They would then bound a compact annulus with geodesic boundary and Euler characteristic zero, forcing $\int KdA=0$, again impossible. There is at most one simple closed geodesic. Completeness of the noncompact surface is not needed for these compact-region arguments.

For a divisor $D$ on a smooth projective curve, let $L(D)=\{f\in k(X):\mathrm{div}(f)+D\ge 0\}\cup \{0\}$, and write $\ell (D)={\dim }_{k}L(D)$. The Riemann-Roch theorem stateswhere $g$ is the genus, $K_X$ is a canonical divisor given by a nonzero rational differential, and the degree of a divisor is the sum of its coefficients since $k$ is algebraically closed.

For genus one, a nonzero regular differential has a nonnegative divisor of degree $2g-2=0$, hence no zeros; therefore $K_X\sim 0$. Negative-degree divisors have no nonzero sections. Thus $\ell (n{P}_{\infty })=n$ for $n>0$, $\ell (0)=1$, and $L(n{P}_{\infty })=0$ for $n<0$.

Choose $x\in L(2P_{\infty })\setminus L(P_{\infty })$ and $y\in L(3P_{\infty })\setminus L(2P_{\infty })$. Their unique poles have orders two and three respectively. Distinct pole orders prove linear independence in the following complete list:Since $y^2$ also lies in the last six-dimensional space, it satisfiesThe coefficient $a$ cannot vanish because only $x^3$ among the displayed basis has a sixth-order pole available to cancel the leading pole of $y^2$. Homogenizing gives a plane cubic relation, valid also in characteristics two and three without completing squares or dividing by three.

It remains to prove that the map actually embeds the curve. For $D=3P_{\infty }$, Riemann-Roch gives $\ell (D)=3$, $\ell (D-Q)=2$ for every point $Q$, and $\ell (D-Q-R)=1$, including $R=Q$. The first equality makes the linear system base-point-free; the second separates distinct points, and the double-point case separates tangent directions. Therefore $D$ is very ample and ${\varphi }_D$ is a closed immersion into ${\mathbb{P}}^2$. Its pullback of a line has degree three, so its image has degree three. The displayed cubic equation consequently defines the image, which is smooth because it is isomorphic to $X$. Thus $\boxed{{\varphi }_{3P_{\infty }}:X\cong \text{a smooth plane cubic}}$. The affine expression $(1:x:y)$ extends across $P_{\infty }$ to $(0:0:1)$ by the pole orders.

Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal space admit a continuous $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. Using the usual convention, normal spaces are also $T_1$.

The Tietze extension theorem states that a continuous real-valued function on a closed subset of a normal space extends continuously to the whole space. If its range lies in a closed bounded interval, the extension may be chosen in that same interval.

Prove first the interval version for $f:S\to [-1,1]$. If a continuous residual $r$ on $S$ obeys $|r|\le M$, the sets $\{r\le -M/3\}$ and $\{r\ge M/3\}$ are disjoint closed subsets of $X$. Urysohn's lemma gives a continuous $h:X\to [-M/3,M/3]$ taking the endpoint values on those two sets. On each of the three ranges of $r$, it follows that $|r-h|\le 2M/3$ on $S$. Repeating with $M_j=(2/3{)}^j$ produces $h_j$ with $\Vert h_j{\Vert }_{\infty }\le M_j/3$ and a residual bounded by $M_{j+1}$. The series $\tilde{f}={\sum }_{j\ge 0}{h}_j$ converges uniformly, has absolute value at most $\sum M_j/3=1$, is continuous and restricts to $f$ as the residual tends to zero. Rescaling proves every closed bounded interval version.

For unbounded $f$, first extend $f/(1+|f|)$ to $g:X\to [-1,1]$. The closed set $B=g^{-1}(\{-1,1\})$ is disjoint from $S$. Urysohn's lemma gives $\psi :X\to [0,1]$ with $\psi {|}_S=1$ and $\psi {|}_B=0$. Then $h=\psi g$ takes values strictly between minus one and one everywhere and equals $f/(1+|f|)$ on $S$. Thus $h/(1-|h|)$ is a continuous real-valued extension of $f$, proving the full theorem.

Statement 1 is true: metric spaces are normal, and the bounded interval form applies. Statement 2 is true: compact Hausdorff spaces are normal, and the unrestricted real-valued form applies. Statement 3 is false. For a counterexample, let $X=\{a,b,c\}$ have open sets $\varnothing ,\{a\},\{a,b\},\{a,c\},X$. The subset $S=\{b,c\}$ is closed and discrete, so $f(b)=-1$, $f(c)=1$ is continuous on it. Every open neighborhood of $b$ or $c$ in $X$ contains $a$. A continuous extension would pull back disjoint neighborhoods of minus one and one to disjoint open sets containing $b$ and $c$, yet both would contain $a$. This is impossible.

For an open cover $X=A\cup B$, the Mayer–Vietoris theorem gives the long exact sequenceThe same sequence holds for a union of simplicial subcomplexes. All homology groups here have integer coefficients.

An explicit triangulation of $X_q$ is as follows. Take vertices $b_0,b_1,b_2$ forming a target circle, a circle of $3q$ vertices $a_0,\dots ,a_{3q-1}$, and a cone vertex $c$. With subscripts on $a$ interpreted modulo $3q$ and on $b$ modulo three, include trianglesfor every $j$, together with all their faces. The first two types triangulate the mapping cylinder of the degree-$q$ covering $a_j\mapsto b_j$; the last type fills the inner circle by a disc. A filled disc with an attached annular collar is still a disc, and identifying the outer collar boundary by this covering gives exactly the source quotient. Distinct inner vertices ensure this is a genuine simplicial complex, not just a one-cell presentation.

Let $A$ be the mapping cylinder and $B$ the coned disc. Then $A$ retracts to the target circle, $B$ is contractible and $A\cap B$ is the inner circle. Its inclusion into $A$ induces multiplication by $q$ on $H_1$. The exact sequence therefore reduces toConnectedness and the absence of higher-dimensional simplices giveThis is a two-dimensional Moore space with torsion in degree one.

The suspension of a connected complex is the union of two cones whose intersection retracts to the original complex. Reduced Mayer-Vietoris givesIn particular its $H_1$ vanishes. For the wedge of two connected polyhedra, choose neighborhoods retracting onto the two summands with contractible intersection at the wedge point. ThenDecompose each prescribed $G_i$ into cyclic summands. Realize each infinite cyclic summand in degree $i$ by $S^i$, and each finite nontrivial summand $\mathbb{Z}/q$ by $S^{i-1}{X}_q$, meaning $i-1$ successive suspensions. Take their finite wedge, or a point if there are no summands. The suspension formula shifts the torsion to degree $i$ and creates no other positive-degree homology, while the wedge formula adds summands. Thus the resulting connected polyhedron satisfies $\boxed{H_0(X)=\mathbb{Z},H_i(X)=G_i(1\le i\le n),H_i(X)=0(i>n)}$. The finite triangulations above, their simplicial suspensions and wedges provide an actual polyhedron.