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Use the standard ideal-fluid assumptions underlying the stated relation: both fluids are incompressible, inviscid and unbounded away from the interface, with no surface tension. An irrotational flow has $u_j=\nabla {\varphi }_j$, and incompressibility gives $\Delta {\varphi }_j=0$ in each fluid. The unsteady Bernoulli equation iswhere the arbitrary time functions can be absorbed into the potentials. On $z=\zeta (x,t)$ the material-interface and normal-stress conditions areDisturbance velocities decay as $z\to \pm \infty$. The equilibrium pressures are $p_{\ast }-{\rho }_{j}gz$.

Linearize at the resting flat interface. The equations become $\Delta {\varphi }_j=0$, ${\zeta }_t={\varphi }_{j,z}$ on $z=0$, andFor $k>0$, write $\zeta =\eta e^{i(kx-\omega t)}$. Decay selects ${\varphi }_1=A_1{e}^{-kz}{e}^{i(kx-\omega t)}$ above and ${\varphi }_2=A_2{e}^{kz}{e}^{i(kx-\omega t)}$ below. The kinematic conditions give $A_1=i\omega \eta /k$ and $A_2=-i\omega \eta /k$. Substituting into the dynamic condition givesFor a signed wave number replace $k$ by $|k|$. A heavier lower fluid gives stable interfacial gravity waves; a heavier upper fluid gives ${\omega }^2<0$ and Rayleigh-Taylor instability. The density ordering is not silently assumed positive in the derivation.

Assume homogeneous space and time, isotropic space, inertial frames related linearly, the relativity principle and a common isotropic speed $c$ for light. Take coincident origins at zero and parallel axes, with $S^{\prime }$ moving at $v$ along $x$. The primed origin requires $x^{\prime }=A(x-vt)$. Write $t^{\prime }=Bx+Dt$; mapping each light ray $x=\pm ct$ into $x^{\prime }=\pm c{t}^{\prime }$ gives $D=A$ and $B=-Av/c^2$. Rotational symmetry about the boost axis gives a common transverse scale $B(v)$; reversing the boost direction by spatial isotropy gives $B(v)=B(-v)$. Reciprocity then gives $B(v{)}^2=1$, and continuity and the common orientation select $B(v)=1$. Thus transverse directions are unchanged. Reciprocity requires the inverse to have the same scale $A$ with $v$ replaced by $-v$. Composing gives $A^2(1-v^2/c^2)=1$; continuity from $A=1$ at $v=0$ selects the positive root. Hence the Lorentz transformation isThe printed wavelength formula uses the angle of the line of sight from observer towards star, opposite to photon propagation. Choose the transverse axis so the ray lies in the $xy$ plane. With that directed-angle convention the photon four-momentum isThe inverse coordinate boost takes a rest-frame four-vector to $S$, giving $p^0=\gamma (p^{\prime 0}+\beta p^{\prime 1})$ and $p^1=\gamma (p^{\prime 1}+\beta p^{\prime 0})$, where $\beta =v/c$. Thus $\nu =\gamma {\nu }^{\prime }(1-\beta \cos {\theta }^{\prime })$, and the aberration of light relation is $\cos \theta =(\cos {\theta }^{\prime }-\beta )/(1-\beta \cos {\theta }^{\prime })$. More directly, the boost from $S$ to $S^{\prime }$ gives $p^{\prime 0}=\gamma (p^0-\beta p^1)$, hence ${\nu }^{\prime }=\gamma \nu (1+\beta \cos \theta )$. Since wavelength is $c/\nu$, the relativistic Doppler effect givesIf the angle were instead measured along the actual photon momentum, both cosine signs would reverse and the same physical relation would have a minus sign. Specifying this orientation avoids confusing the two versions.

For $\cos \theta =1$, $\lambda /{\lambda }^{\prime }=\sqrt{(1+\beta )/(1-\beta )}$: a receding star is redshifted. For $\cos \theta =-1$, the ratio is $\sqrt{(1-\beta )/(1+\beta )}$: an approaching star is blueshifted. For $\cos \theta =0$, $\boxed{\lambda /{\lambda }^{\prime }=\gamma }$: the transverse redshift is the time-dilation effect. This transverse condition refers to the observer's angle, not ${\theta }^{\prime }=\pi /2$ in the source frame.

An isotropic tensor has unchanged components under every proper orthogonal change of basis: $T_{i_1\dots i_n}=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\dots j_n}$ for every $R\in SO(3)$. Orthogonality gives $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$, so the Kronecker delta is isotropic. The determinant identity gives $R_{ia}{R}_{jb}{R}_{kc}{ϵ}_{abc}=(\det R){ϵ}_{ijk}={ϵ}_{ijk}$, so the Levi-Civita symbol is isotropic too. If reflections are included, this last object is a pseudotensor: it changes sign. The question's isotropy convention therefore refers to proper rotations.

For the sphere moment $T_{i_1\dots i_n}={\int }_{S^2}{\hat{x}}_{i_1}\cdots {\hat{x}}_{i_n}dS$, change variables to $R\hat{x}$. Rotations preserve the sphere and its area measure; inserting the component transformation proves that $T$ is an isotropic tensor. For the second moment, half-turns about coordinate axes kill the off-diagonal components, while rotations interchanging axes make the diagonal entries equal. Thus it is a multiple $a{\delta }_{ij}$. Contracting the indices gives $3a={\int }_{S^2}|\hat{x}{|}^{2}dS=4\pi$. Odd moments vanish by $\hat{x}\mapsto -\hat{x}$, including the third moment. The supplied general fourth-rank form and symmetry under every permutation of the four indices make its three coefficients equal. A double contraction then gives $4\pi =b(9+3+3)$. ThusTo explain the weighted volume integrals, put $v=(1,i,0)$; its bilinear square is $v\cdot v=1+i^2=0$, without complex conjugation. Separating radial and angular factors givesFor $n=2$ the angular factor is $(4\pi /3)v\cdot v=0$. For $n=3$ it is zero by oddness. For $n=4$ it is $(4\pi /15)3(v\cdot v{)}^2=0$. Therefore all three requested integrals vanish, assuming the radial integrals exist. More generally the null-vector cancellation of radial moments follows from rotation about the third axis, which multiplies the integral by $e^{in\varphi }$ without otherwise changing it.

The function is odd, so its Fourier series has no constant or cosine terms. If $\lambda =m\in \mathbb{Z}$, it is already a Fourier basis function: for $m>0$ the only sine coefficient is $b_m=1$; for $m<0$ it is $b_{|m|}=-1$; for $m=0$ the function is zero.

For real noninteger $\lambda$, the product-to-sum identity givesSince both numerator sines equal $(-1{)}^n\sin (\pi \lambda )$, the Fourier series of a noninteger-frequency sine isIt equals the function for $-\pi <x<\pi$. At either periodic endpoint it equals the average of the two one-sided limits, namely zero; the noninteger-frequency function has a jump in its periodic extension there. Equality also holds in the square-integral sense on the full interval.

For $\lambda =1/2$, $b_n=8(-1{)}^{n+1}n/[\pi (4n^2-1)]$ and ${\int }_{-\pi }^{\pi }{\sin }^2(x/2)dx=\pi$. The Parseval identity for a real sine series is $(1/\pi ){\int }_{-\pi }^{\pi }{f}^2={\sum }_{n\ge 1}{b}_n^2$. Hence

For nonempty $F$, the triangle inequality gives $d(x,z)\le d(x,y)+d(y,z)$ for every $z\in F$. Taking the infimum yields $d(x,F)\le d(x,y)+d(y,F)$. Interchanging $x,y$ provesThus the distance to a set is a Lipschitz function, hence continuous. If $x\notin F$ and $F$ is closed, its open complement contains a ball $B(x,ϵ)$; every point of $F$ is at least $ϵ$ away. Hence $\boxed{d(x,F)>0}$ for $x\notin F$.

For two nonempty disjoint closed sets, let $h(x)=d(x,F_1)-d(x,F_2)$. It is continuous, negative on $F_1$ and positive on $F_2$. The disjoint open sets $U_1=\{h<0\}$ and $U_2=\{h>0\}$ contain them. Empty sets cause no difficulty, using an empty neighborhood for the empty set. This proves every metric space is a normal topological space.

In any normal space, a closed $F$ contained in an open $U$ can be separated from $X\setminus U$: choose disjoint open $W\supset F$ and $V\supset X\setminus U$. Because $X\setminus V$ is closed and contains $W$, $\overline{W}\subset X\setminus V\subset U$. First choose disjoint open $U_1,U_2$ around $F_1,F_2$, then apply this argument separately to each pair $F_i\subset U_i$. It gives ${\overline{W}}_i\subset U_i$, and therefore $\boxed{{\overline{W}}_1\cap {\overline{W}}_2=\varnothing }$. This is closed-neighborhood shrinking in a normal space.

For the two specified hyperbola branches, explicit choices areThey are open and contain the branches with products $-1$ and $1$. Their closures lie respectively in $\{x\le 0,xy\le -1/2\}$ and $\{x\ge 0,xy\ge 1/2\}$. Any common point would have $x=0$, making both product inequalities impossible. Thus their closures are disjoint even though the original branches can approach one another arbitrarily closely at large height.

The fundamental theorem of calculus shows that the differential initial-value problem is equivalent toIn one direction integrate $x^{\prime }=F(t,x(t))$ from zero. In the other, continuity of the integrand makes the right side continuously differentiable with precisely that derivative and initial value.

Consider the closed subset $\mathcal{X}=\{x\in C([-b,b]):\Vert x-x_0{\Vert }_{\infty }\le r\}$ of the complete space with the uniform norm. It is complete: a uniformly Cauchy sequence has a uniform continuous limit, and the closed range constraint passes to the limit. DefineThe bound $|F|\le C$ gives $\Vert Tx-x_0{\Vert }_{\infty }\le bC<r$, so $T$ maps $\mathcal{X}$ into itself. The uniform Lipschitz condition givesHere negative $t$ changes only the orientation of the integral; its length is still at most $b$.

Construct $x^{(0)}(t)=x_0$ and $x^{(m+1)}=T{x}^{(m)}$. Induction gives $\Vert x^{(m+1)}-x^{(m)}{\Vert }_{\infty }\le q^m\Vert x^{(1)}-x^{(0)}{\Vert }_{\infty }$. Summing this geometric bound proves uniform Cauchy convergence to some $x\in \mathcal{X}$. The same contraction estimate shows $T{x}^{(m)}\to Tx$, so $x=Tx$. The integral equation now makes $x$ a $C^1$ solution. If another solution $y$ existed, it would also be fixed, and $\Vert x-y{\Vert }_{\infty }\le q\Vert x-y{\Vert }_{\infty }$ would force equality. This proves existence and uniqueness on the entire stated interval, implementing the contraction mapping theorem rather than just invoking it.

If $C=0$, the vector field is zero and the solution is constant. If $K=0$, the vector field is independent of the space variable and the integral solves the problem directly. The divisions by zero in the printed interval bound are naturally interpreted as imposing no restriction from the corresponding term.

At $P=\sigma (u,v)$, regularity makes ${\sigma }_u,{\sigma }_v$ linearly independent. The tangent space is $T_{P}U=\mathrm{span}\{{\sigma }_u,{\sigma }_v\}$, and an oriented unit normal vector isReversing its sign chooses the opposite orientation. The first fundamental form is the squared length of a tangent displacement:where $E={\sigma }_u\cdot {\sigma }_u$, $F={\sigma }_u\cdot {\sigma }_v$, $G={\sigma }_v\cdot {\sigma }_v$. Its determinant $EG-F^2=|{\sigma }_u\times {\sigma }_v{|}^2$ is positive.

Use $N_{\mathrm{II}}={\sigma }_{vv}\cdot \mathbf{n}$ to distinguish the scalar coefficient of the second fundamental form from the normal vector. Since $\mathbf{n}\cdot \mathbf{n}=1$, its derivatives are perpendicular to it and hence tangent. Write ${\mathbf{n}}_u=a{\sigma }_u+b{\sigma }_v$ and ${\mathbf{n}}_v=c{\sigma }_u+d{\sigma }_v$. Differentiating $\mathbf{n}\cdot {\sigma }_u=\mathbf{n}\cdot {\sigma }_v=0$ givesSubstitution of the tangent expansions provesThe shape operator is $-d\mathbf{n}$; its matrix in the parametrized tangent basis is the negative transpose of the coefficient matrix shown. Its determinant, the product of the principal curvatures, is thereforethe Gaussian curvature. Transposition and the two minus signs do not change this determinant.

Let $\pi$ be a prime element of the Gaussian integers. Its norm $\pi \overline{\pi }$ is a positive integer greater than one, so $\pi$ divides some ordinary positive prime number in its integer prime factorization. It cannot divide two distinct such primes $p,q$: an integer Bézout identity $up+vq=1$ would imply $\pi \mid 1$, contrary to its being a nonunit. This proves existence and uniqueness of the rational prime lying below $\pi$.

Up to associates, the complete list of Gaussian prime elements is: $1+i$; the positive rational primes $p\equiv 3(\mathrm{mod}4)$; and, for every rational prime $p\equiv 1(\mathrm{mod}4)$, the two nonassociate elements $a+bi$ and $a-bi$, where $a,b>0$ and $p=a^2+b^2$. Representatives related by multiplication by one of $\pm 1,\pm i$ are associates. The last pair accounts for both factors of a split prime.

Every residue class modulo $p$ has a unique representative $a+bi$ with $0\le a,b<p$, since $p\mid (a-a^{\prime })+(b-b^{\prime })i$ means both coordinate differences are divisible by $p$. Hence $\boxed{|R/pR|=p^2}$ andFor $p=2$, the class of $1+i$ is nonzero but its square is $2i=0$; thus the quotient ring is not a field. For $p\equiv 3(\mathrm{mod}4)$, a root $r^2=-1$ would imply $r^{p-1}=(-1{)}^{(p-1)/2}=-1$, contradicting Fermat's little theorem. The quadratic $X^2+1$ is therefore irreducible and its quotient is a field with $p^2$ elements.

For $p\equiv 1(\mathrm{mod}4)$, choose a generator $g$ of the cyclic group ${\mathbb{F}}_p^{\times }$. Then $r=g^{(p-1)/4}$ satisfies $r^2=-1$. The two nonzero classes $i-r$ and $i+r$ have product zero in $R/pR$. ThusFor $p\equiv 1(\mathrm{mod}4)$ it is instead isomorphic to ${\mathbb{F}}_p\times {\mathbb{F}}_p$, by the distinct linear factors and the Chinese remainder theorem.

For a state $i$, let $N_i=\{n\ge 1:P_{ii}^n>0\}$ and $d_i=\gcd N_i$. If distinct states $i,j$ communicate, choose positive-probability paths $i\to j$ and $j\to i$ of lengths $r,s$. Then $r+s$ is a return time to each. For every $n\in N_i$, concatenate $j\to i$, the $i$-return loop, and $i\to j$ to see that $s+n+r\in N_j$. Since $d_j$ divides this and $s+r$, it divides $n$. Thus $d_j\mid d_i$, and interchanging the states gives $d_i\mid d_j$. This proves period is constant on a communicating class.

Read the transition matrix from the original PDF; the converted TeX loses the location of the second row's final entry. The communicating classes areWithin $\{1,3,4\}$ the only internal cycle is $1\to 3\to 4\to 1$. Return lengths are multiples of three, and length three has positive probability, so its period is three. This class is open because it has exits to other classes. The class $\{2,7\}$ is the closed deterministic two-cycle, of period two. The absorbing class $\{6\}$ is closed and has period one.

State 5 has outgoing transitions but no incoming transition at all, so it is an open singleton class and $N_5=\varnothing$. For this return-free state in a Markov chain, the convention $\gcd \varnothing =0$ gives period zero; if period is defined only when positive return times exist, it is undefined instead. In particular, this singleton is not aperiodic: aperiodicity would require period one.

Write $Q={\sum }_i(X_i-\bar{X}{)}^2$ and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal-sample likelihood isWithout the mean restriction, its maximum has $\hat{\mu }=\bar{X}$ and ${\hat{\sigma }}^2=Q/n$. Under the null, the maximizing variance is $Q_0/n$. Substituting them gives the generalized likelihood-ratio test statisticwhere $S^2=Q/(n-1)$ and $T=\sqrt{n}(\bar{X}-{\mu }_0)/S$. Assume $n\ge 2$; $Q>0$ almost surely for positive population variance.

The ratio decreases strictly as $|T|$ increases, so rejecting for small $\Lambda$ is equivalent to rejecting for large $|T|$. Under $H_0$, $T$ has Student's t-distribution with $n-1$ degrees of freedom. Therefore the exact size-$\alpha$ rule isThis is the two-sided Student's t-test. Equivalently reject when $\Lambda <(1+t_{n-1,1-\alpha /2}^2/(n-1){)}^{-n/2}$. No large-sample chi-squared approximation is needed for this normal-mean likelihood ratio with unknown variance.

For a state in the domain of the operator, its expectation value isFor a normalized state the denominator is one. Use an orthonormal energy eigenstate basis, choosing orthogonal vectors within degenerate eigenspaces. Inserting an expansion into the time-dependent Schrödinger equation gives $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$. ConsequentlyThe phases preserve normalization. Projection onto a specified basis eigenstate gives probability $\boxed{|a_p{|}^2}$, independent of time. If the measurement distinguishes only the energy and $E_p$ is degenerate, sum $|a_n{|}^2$ over all $n$ with $E_n=E_p$.

Using orthonormality to eliminate all cross terms,This is time independent whenever the expectation is defined. The time-independent Hamiltonian changes only relative phases, not the energy probabilities.

At a regular point of the constraint curve, $\nabla g\ne 0$. A constrained stationary point has $Df$ zero on every tangent vector annihilated by $Dg$, so its gradient is normal to the curve. The Lagrange multiplier equations are therefore $\nabla f=\lambda \nabla g$ together with $g=0$. Singular constraint points require separate examination.

Assume $a,b>0$. For the ellipse, the equations areNeither coordinate can vanish at a stationary point, because the first two equations would then force both to vanish. Multiplication gives $4{\lambda }^2=a^2{b}^2$. Therefore $y=\pm (b/a)x$, and the constraint gives $x^2=a^2/2$, $y^2=b^2/2$. The stationary values areThe compact ellipse has global extrema, all its points are regular, and these are all stationary points. Hence the first value is the global maximum and the second the global minimum.

A real harmonic function on a planar open set is a twice continuously differentiable function satisfying $f_{xx}+f_{yy}=0$. The ordered pair $(f,g)$ consists of harmonic conjugates when $f+ig$ is holomorphic; the corresponding Cauchy-Riemann equations are $f_x=g_y$ and $f_y=-g_x$.

To prove composition of harmonic conjugate pairs, denote $P=p(u,v)$ and $Q=q(u,v)$, and assume the inner pair's image lies in the outer pair's domain. Their equations are $u_x=v_y$, $u_y=-v_x$, $p_u=q_v$ and $p_v=-q_u$. The chain rule givesThus the composite satisfies the Cauchy-Riemann equations. Its components are $C^2$, and differentiating gives $P_{xx}+P_{yy}=Q_{yx}-Q_{xy}=0$ and similarly $Q_{xx}+Q_{yy}=0$. The composite pair is harmonic conjugate, including when the inner map has a critical point.

Include the empty set, and take all unions of intervals $[a,b)$ with $a<b$. These intervals cover the real line, and the intersection of two is either empty or $[\max (a,c),\min (b,d))$. Distributing intersections over unions therefore makes finite intersections of open sets open; arbitrary unions are open by construction. The whole line is open, so the axioms of a topology hold. This is the lower limit topology, whose space is the Sorgenfrey line.

If $x<y$, the neighborhoods $[x,y)$ and $[y,y+1)$ are disjoint. Hence $\boxed{(\mathbb{R},{\tau }_1)\text{ is Hausdorff}}$.

For the identity into the cofinite topology, an inverse image is the same subset of $\mathbb{R}$. Let $F$ be finite and $x\notin F$. If $F$ has a point greater than $x$, choose $b$ to be its least such point; otherwise take $b=x+1$. Then $[x,b)$ avoids every point of $F$, including $b$ itself. Taking these intervals over all $x\notin F$ writes $\mathbb{R}\setminus F$ as a ${\tau }_1$-open set. The inverse image of every cofinite open set, and of the empty set, is open. The identity map is continuous, because ${\tau }_2\subseteq {\tau }_1$.

For curvature $-1$, the Poincare disc model has metric and area elementThe given minimizing-diameter property makes the distance from zero to Euclidean radius $R$ equal toRadial symmetry therefore identifies the whole hyperbolic circle in the Poincare disc. Integrating its enclosed hyperbolic area gives

For a geodesic hyperbolic triangle with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem states $A=\pi -(\alpha +\beta +\gamma )$, in particular $A\le \pi$, with strict inequality for ordinary finite vertices. Let $d$ be the distance of the interior point $P$ to the nearest side. The open disc of radius $d$ about $P$ is contained in the triangle: a path from $P$ to an exterior point must first cross the boundary at distance at least $d$. A hyperbolic isometry carries its center to zero, so its area is the area already calculated. ConsequentlyThis uses hyperbolic area bounds an inscribed disc and establishes the requested bound without assuming that $P$ is an incenter. A sharper universal triangle bound is possible, but is not needed here.