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The same pole-order argument provides a basisFor $n\ge 0$ the first set contains $\lfloor n/2\rfloor +1$ elements, and the second contains $\max (0,\lfloor (n-3)/2\rfloor +1)$. Distinct pole orders imply linear independence, and the coordinate-ring expression proves spanning. HenceFor $n<0$ a function in the space would have no poles anywhere and would vanish at infinity; a regular function on a projective integral curve is constant, so it must be zero. Equivalently the coordinate-ring pole argument first makes it constant, then the vanishing condition excludes nonzero constants.

A rational function with no poles away from $P_{\infty }$ is regular on the smooth affine curve. Its ring of regular functions is the coordinate ringso it has a unique expression $A(x)+yB(x)$ with polynomials $A,B$. This regular-function assertion uses smoothness, hence normality, of the affine curve, not Riemann-Roch theorem.

The nonconstant leading term of $A$ has even pole order $2\mathrm{deg}A$ at infinity, while $yB$ has odd pole order $3+2\mathrm{deg}B$. These different parities prevent cancellation. Thus no nonconstant such function has a pole of order at most one. Therefore

Let $P_j=({\lambda }_j,0)$, and let $Q_{\pm }=(0,\pm \sqrt{-{\lambda }_1{\lambda }_2{\lambda }_3})$. The latter are distinct because the product is nonzero and the characteristic is not two. At each $Q_{\pm }$, the derivative of the curve equation with respect to $y$ is nonzero, so $x$ is a local parameter and has a simple zero. At $P_j$, the derivative of the cubic in $x$ is nonzero, so $y$ is a local parameter with a simple zero; $x-{\lambda }_j$ has order two there.

The only poles occur at the point at infinity, and their orders were computed in part (i). Thus the principal divisors areOver a field where the $Q_{\pm }$ are not rational, their sum is interpreted as the degree-two zero divisor, or computed after scalar extension and then descended.

Use the usual hypothesis that the ground field has characteristic different from two, and work over its algebraic closure for the geometric points. This hypothesis is necessary for the smooth elliptic-curve calculation: in characteristic two, the displayed cubic model can be singular even when its three roots are distinct.

Homogenization givesAt $Z=0$ this says $X^3=0$, soTo compute the pole orders on a smooth cubic in Weierstrass form, work in the chart $Y=1$, writing $u=X/Y$ and $v=Z/Y$. The equation is $v={\prod }_j(u-{\lambda }_{j}v)$. Its derivative in $v$ at $(0,0)$ is one on the left after bringing the right side across, so $u$ is a local parameter. The equation gives $v=u^3+O(u^5)$. Consequently $x=u/v$ and $y=1/v$ have poles of orders two and three at $P_{\infty }$.

The previous permutation calculation shows that every automorphism either preserves $x^3,y^3$ separately or interchanges them. Their sum and product are therefore fixed by the Galois group and belong to $K$, proving the quadratic assertion.

For the depressed cubic, $\alpha +\beta +\gamma =0$, $\alpha \beta +\alpha \gamma +\beta \gamma =b$, and $\alpha \beta \gamma =-c$. Using $1+\zeta +{\zeta }^2=0$ givesand henceAlso $x^3{y}^3=-27b^3$. The desired polynomial isIts discriminant is $729c^2+108b^3=-27D$, where $D=-4b^3-27c^2$. Here $-27$ is a square in $K$, since $[3(\zeta -{\zeta }^2){]}^2=-27$.

One can justify the group criterion directly even when one resolvent vanishes. The nonzero root-difference producthas $d^2=D$ and changes sign exactly under odd root permutations. It is fixed by the entire group exactly when the group is contained in $A_3$. If $D$ is a square in $K$, then $d$ equals one of its two square roots in $K$; conversely, a fixed $d$ lies in $K$. ThereforeSeparability ensures $D\ne 0$, and the characteristic hypotheses ensure that all divisions and the sign argument are valid.

The irreducible cubic is separable, and its Galois group acts transitively on its three roots. The only transitive subgroups of $S_3$ areFor the cycle $\sigma :(\alpha ,\beta ,\gamma )\mapsto (\beta ,\gamma ,\alpha )$, the Lagrange resolvent for a cubic obeys $\sigma (x)={\zeta }^{2}x$. A transposition interchanging $\beta ,\gamma$ sends $x$ to $y=\alpha +\zeta \gamma +{\zeta }^2\beta$. Since $\zeta \in K$, these scalars are fixed by every automorphism. Thus the two possibilities for the image set areThese are descriptions of sets, so coincident values or a zero resolvent are not being asserted distinct.

For the ideal gas, $V=N{k}_{B}T/P$ gives ${\kappa }_T=1/P$. With $C_P/C_V=5/3$, the relation in part (ii) gives

Let ${\alpha }_V=V^{-1}(\partial V/\partial T{)}_P$. The Maxwell relations giveSetting $dS=0$ gives $(\partial T/\partial P{)}_S=TV{\alpha }_V/C_P$, hence ${\kappa }_S={\kappa }_T-TV{\alpha }_V^2/C_P$. Also $dS=(C_V/T)dT+(\partial P/\partial T{)}_{V}dV$, and the volume differential gives $(\partial P/\partial T{)}_V={\alpha }_V/{\kappa }_T$. Evaluating along constant pressure therefore yields $C_P-C_V=TV{\alpha }_V^2/{\kappa }_T$. Eliminate ${\alpha }_V$ to obtainThese derivatives are taken at fixed particle number, in a thermodynamically regular stable phase.

The one-particle partition function isThe classical indistinguishable noninteracting gas has $Z_N=Z_1^N/N!$; for labeled particles omit $N!$, which changes none of the requested derivatives. With $\beta =(k_{B}T{)}^{-1}$,The allowed equation of state gives $PV=N{k}_{B}T$, so the enthalpy is $E+PV=5N{k}_{B}T/2$ and $\boxed{C_P=5N{k}_B/2}$, at fixed particle number.

Set $G(x,y,t)=\Psi (x+y/2,t)\overline{\Psi }(x-y/2,t)$. The Schrödinger equation and its conjugate givesince ${\nabla }_x={\nabla }_{+}+{\nabla }_{-}$ and ${\nabla }_y=({\nabla }_{+}-{\nabla }_{-})/2$. In the Wigner transform, integration by parts turns ${\nabla }_y$ into $i\xi$, yielding the free transport equationAlong its characteristics $x(t)=x_0+t\xi$, the value is constant. ThusFor rapidly decreasing smooth data every integration by parts is justified; weaker data follow distributionally.

Use $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ with inverse factor $(2\pi {)}^{-n}$. The Fourier transform of the Schrödinger equation gives $i{\partial }_t\hat{\Psi }=|\xi {|}^2\hat{\Psi }/2$, hence $\hat{\Psi }(\xi ,t)=e^{-it|\xi {|}^2/2}\hat{f}(\xi )$. To invert the multiplier, first damp it by $e^{-\epsilon |\xi {|}^2}$. The Gaussian integral in each coordinate givesLet $\epsilon \downarrow 0$, with the branch continuous from positive real damping. The fundamental solution of a linear differential operator isFor Schwartz initial data this convolution and Fourier derivation are justified directly; the multiplier tends to one as $t\downarrow 0$ and has modulus one, giving the initial data and a unitary extension to $L^2$. The kernel itself is interpreted through the displayed Abel regularization.

If the polynomials converge uniformly on the full annulus to $g$, their limit is continuous and equals $f$ on the slit annulus by part (ii). That subset is dense in the full annulus, so $g=f$ everywhere. The Cauchy integral theorem and uniform convergence on the unit circle then give

Choose the argument in $(0,2\pi )$ on the slit annulus and letThese compact sets increase to the slit annulus, and each has connected complement: the omitted angular corridor joins the inner disk to the outside of the outer disk. The polynomial Runge theorem says that a function holomorphic near a compact set with connected complement can be approximated uniformly there by polynomials. Choose $p_j$ with ${\sup }_{K_j}|p_j-f|<1/j$. Every compact subset $K$ of the slit annulus has positive distance from the removed positive ray, so $K\subseteq K_j$ for all sufficiently large $j$. Hence $p_j\to f$ uniformly on every such compact subset.

Take the holomorphic function $f(z)=1/z$ on a neighborhood of the annulus. Every polynomial has integral zero around the unit circle by the Cauchy integral theorem, while ${\int }_{|z|=1}{z}^{-1}dz=2\pi i$. Uniform convergence on the annulus would imply convergence of these contour integrals, since their difference is bounded by $2\pi$ times the uniform error. Thus uniform polynomial approximation on the full annulus need not exist.

Let $K$ count second-fork arrivals. Its independent increment over $(s,s+t]$ is Poisson$(rt)$, andobtained by evaluating $\mathbb{E}(-1{)}^{K(s)}=e^{-2rs}$. If $K(s)$ is even, the next second-fork arrival goes to B, so no B arrival requires zero new arrivals. If it is odd, the next goes to C, so up to one new arrival is allowed. Reverse these alternatives for C. ThereforeAt $s=0$ these reduce to the distinct initial-delay survivors; as $s\to \infty$, both non-Poisson probabilities approach the equilibrium-delay survivor from part (b).

For a renewal holding time $T$ with mean $m$, the equilibrium residual-life distribution has survivor $m^{-1}{\int }_t^{\infty }\mathbb{P}(T>s)ds$. For C, and also the common long-run B/C renewal structure, $m=2/r$ and the Erlang survivor giveFor A the exponential law is already its own equilibrium delay.

Assume the first-fork choices are independent Bernoulli routings, as in Poisson thinning. Put $r=2\lambda /3$. Cars assigned to A form a Poisson process of rate $\lambda /3$, also an ordinary renewal process with independent exponential holding times of rate $\lambda /3$. The stream to the second fork is an independent Poisson process of rate $r$.

B receives its odd-numbered arrivals. Its initial delay is exponential of rate $r$, followed by independent Erlang distributions of shape two and rate $r$ between successive B arrivals. Thus B is a delayed renewal process, not Poisson. C receives the even-numbered arrivals: its first and all subsequent holding times are independent Erlang$(2,r)$, so C is an ordinary renewal process, not Poisson. Their holding-time density and survivor are $r^{2}t{e}^{-rt}$ and $e^{-rt}(1+rt)$. Both have asymptotic rate $r/2=\lambda /3$; equality of these mean rates with A does not make the processes Poisson.

At $p\in D$, each of the given ambient rotations fixes $p$, rotates its tangent plane and preserves the local surface. Naturality of the shape operator under that rotation makes $S_p$ commute with every planar rotation. A symmetric real two-by-two operator commuting with all rotations is scalar, so every point of $D$ is an umbilical point. The trace-free part of the shape operator is continuous. Density therefore makes it zero everywhere on $S$.

Locally write $S=\kappa I$. The Codazzi equation gives $(X\kappa )Y=(Y\kappa )X$ for tangent fields $X,Y$; choosing independent fields proves $d\kappa =0$. Thus the surface is locally a plane if $\kappa =0$, or has constant spherical centersince its differential is $I-S/\kappa =0$. A change of normal reverses both $N$ and $\kappa$, leaving this center unchanged, so these local descriptions patch across the connected surface. In the flat case the whole surface lies in one plane, where a regular surface without boundary is open; a nonempty such set cannot also be compact and closed. Hence the flat case is excluded. In the other case $S$ lies on one sphere of radius $1/|\kappa |$. Its inclusion in that sphere is locally open, since both have dimension two, and its compactness makes the image closed. Connectedness of the sphere forces the image to be the entire sphere. Therefore $S$ is a sphere.

The shape operator is the self-adjoint tangent map $S_p=-d{N}_p$, using a locally chosen unit normal. Its eigenvalues ${\kappa }_1,{\kappa }_2$ are the principal curvatures, and their eigenvectors give the principal directions. At an umbilical point they coincide, so $S_p=\kappa I$ and every tangent direction is principal. Reversing the normal changes both curvature signs but does not change the principal directions or umbilicity.

The Theorema Egregium states that Gaussian curvature is determined by the first fundamental form, and therefore preserved by local isometries. For a local surface parametrization $r$, write $g_{ij}=r_i\cdot r_j$, choose a unit normal $N$, and set $b_{ij}=r_{ij}\cdot N$. Decomposing the second derivatives into tangent and normal parts givesCommute the third partial derivatives of $r$ and take tangent components. The terms involving derivatives of $b$ are normal; the tangent terms give the Gauss equationwhere $R$ is computed from derivatives and products of the displayed Christoffel symbols, using $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. For example the tangent term from differentiating $b_{ij}N$ is $-b_{ij}{b}_k{}^{\ell }{r}_{\ell }$, which produces exactly the difference of the two products on the right. ConsequentlyThe last expression depends only on $g$ and its derivatives. Pulling back $g$ under a local isometry therefore preserves $K$, proving the theorem.