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Let $C$ be the middle-third Cantor set and choose a positive integer $d$ with $d\log 2/\log 3\ge k$; this is possible for every real $k$, including negative $k$. Set $X=C^d\subset {\mathbb{R}}^d$. It is closed and bounded since each factor is compact. If a subset of $X$ is connected, each coordinate projection is a connected subset of $C$ and hence a singleton; therefore the subset itself is a singleton. Thus $X$ is totally disconnected.

For completeness, the dimension of these Cartesian powers of the Cantor set can be checked directly. At stage $n$ there are $2^{dn}$ cubes of side $3^{-n}$, giving vanishing $s$-dimensional cover cost whenever $s>d\log 2/\log 3$. Conversely put independent equiprobable zero-or-two ternary digits on every coordinate. Each stage-$n$ cube has probability $2^{-dn}$. If $3^{-n}\le r<3^{-n+1}$, a Euclidean ball of radius $r$ meets at most a constant depending on $d$ such cubes. Hence its probability is at most $C_d{r}^{d\log 2/\log 3}$. For any sufficiently fine cover, summing this bound over covering sets shows that its cost at exponent $d\log 2/\log 3$ is bounded below by a positive constant, since total probability is one. These upper and lower Hausdorff measure bounds proveThe construction does not claim arbitrarily large Hausdorff dimension in one fixed ambient Euclidean dimension.

In the shock rest frame the upstream and downstream speed magnitudes are $U$ and $U-V$. Denote the downstream density by ${\rho }_2$, and write $v_i=1/{\rho }_i$ for specific volume. The Rankine-Hugoniot conditions for a perfect gas express conservation of mass, momentum and total energy:These follow by balancing the respective fluxes across a stationary infinitesimal control volume around the shock wave. The mass relation also gives $V=j(v_1-v_2)$, and momentum gives $j^2(v_1-v_2)=p_2-p_1$.

Eliminate the kinetic-energy difference in the energy condition using momentum:Substitute the given perfect-gas internal energy and collect the coefficients of $v_1,v_2$:HenceFinally $V^2=j^2(v_1-v_2{)}^2=(p_2-p_1)(v_1-v_2)$ givesFor the advancing piston choose the compressive branch $p_2>p_1$, $U>V>0$; the algebraic squared relation alone does not specify that branch.

Write the common density as $\rho$ and take positive bending stiffness $\beta$. With $q=|k|>0$, use decaying velocity potentialsadded to the upper base potential $Ux$ and the lower zero base potential. They solve the Laplace equation, as required by incompressibility and irrotational flow. The two linear kinematic boundary conditions on the membrane areThe linearized Bernoulli equation gives perturbation pressures $p_{+}^{\prime }=-\rho (\sigma +ikU)A_{+}$ and $p_{-}^{\prime }=-\rho \sigma A_{-}$. Substituting the kinematic conditions and imposing the stipulated pressure difference givesCompleting the square yields the bending-membrane Kelvin-Helmholtz dispersion relationThe imaginary shift convects the disturbance at mean fluid speed $U/2$. For $U\ne 0$ there is a growing root exactly whenAt $k=0$ the uniform displacement is neutral; at $|k|=k_{\max }$ the square-root growth vanishes. Above this cutoff both roots are imaginary. To maximize the growth, maximize its square $U^2{q}^2/4-\beta q^5/(2\rho )$ over the unstable interval. Its derivative is $U^{2}q/2-5\beta q^4/(2\rho )$, so its unique positive maximum isFor $U=0$ there is no Kelvin-Helmholtz instability. Positivity of $\beta$ is the physical stiffness assumption behind the finite cutoff: $\beta =0$ gives no upper unstable cutoff, and $\beta <0$ destabilizes arbitrarily short waves.

Set $f(r)=1-2M/r$ and use proper time along a timelike geodesic. In the equatorial plane, the radial Christoffel symbols needed for a circular orbit areThe radial geodesic equation with $\dot{r}=0$ gives $(fM/r^2){\dot{t}}^2-fr{\dot{\phi }}^2=0$. Outside the horizon this implies

For stability use the conserved energy $E=f\dot{t}$ and angular momentum $L=r^2\dot{\phi }$. The normalization of the timelike tangent in the Schwarzschild metric gives ${\dot{r}}^2+V(r)=E^2$, with the effective potential for timelike Schwarzschild geodesicsAt a circular orbit $V^{\prime }=0$, so $L^2=M{r}^2/(r-3M)$, requiring $r>3M$. Differentiate $V$ holding this conserved $L$ fixed and then substitute the circular value:The linear radial perturbation equation is $\delta \ddot{r}=-V^{\prime \prime }(r)\delta r/2$. Thus circular timelike orbits are stable for $r>6M$, unstable for $3M<r<6M$, with the marginal case at $6M$.

For the orbiting astronaut at radius $R$, use $d\phi /dt=\sqrt{M/R^3}$ to obtainThe static Earth twin has $d{\tau }_0/dt=\sqrt{1-2M/R_0}$. During equal coordinate-time intervals of the long circular-orbit phase, the astronaut ages less precisely whenThe outward and return transfers have finite additional proper times; the stated long-orbit comparison is the limit in which the circular phase dominates them. At equality the circular phase contributes no age difference, so transfer details can decide the finite-trip result.

Use SI units and normalize the condensate wavefunction so $|\psi {|}^2=n$. With mechanical momentum operator $-i\hslash \nabla -qA$, the charge current density isFor uniform $n$ and $\psi =\sqrt{n}{e}^{i\vartheta }$, this becomes $j=(nq/m)(\hslash \nabla \vartheta -qA)$.

Under the gauge transformation $A\mapsto A+\nabla \chi$, the wavefunction transforms as $\psi \mapsto e^{iq\chi /\hslash }\psi$, so $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two added gradients cancel in $\hslash \nabla \vartheta -qA$, while $|\psi {|}^2$ stays fixed. Thus the superconducting current is gauge invariant. The same cancellation in the full displayed current proves it without requiring a constant amplitude.

For uniform carrier density in a region with smooth single-valued phase, taking a curl gives the London equation $\nabla \times j=-n{q}^{2}B/m$. In the time-independent setting Ampère's circuital law gives $\nabla \times B={\mu }_{0}j$, and Gauss's law for magnetism gives $\nabla \cdot B=0$. Thereforeso the Helmholtz equation and London penetration depth areThe equation has the modified, exponentially decaying Helmholtz sign. For a planar surface at $x=0$ with material in $x>0$, the bounded solution is $B(x)=B(0)e^{-x/\ell }$; the exponentially growing solution is excluded in the bulk. Hence a macroscopic region far from the surface has negligible field: magnetic flux is expelled from the bulk, the Meissner effect. The literal assertion of zero flux everywhere is the bulk limit: finite penetration remains within distance $\ell$ of a surface. Vortices or flux trapped through a hole require different phase or topology assumptions and are not ruled out by this smooth-phase derivation.

For potential period $a$, a Bloch state has the form ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$, or equivalently ${\psi }_k(x+a)=e^{ika}{\psi }_k(x)$. Quasimomentum is defined modulo the reciprocal-lattice vector $2\pi /a$. A Brillouin zone is one fundamental interval for that identification, usually $[-\pi /a,\pi /a)$. Here $a=2\pi$, so it is $[-1/2,1/2)$.

For small $V_0>0$, the first free-particle degeneracy occurs at the zone edge $k=1/2$, between plane waves $e^{ix/2}$ and $e^{-ix/2}$ of energy $1/4$. In this two-dimensional subspace, the diagonal matrix elements of the potential vanish. Its Fourier coefficient at wave number one is $-V_0/2$; the $\cos 2x$ term cannot connect these two waves at first order. Thus degenerate perturbation theory gives the matrixwith energies $1/4\pm V_0/2+O(V_0^2)$. Near the edge the two uncoupled energies have opposite slopes, and diagonalization gives the avoided crossing, so the lowest band gap is

For large $V_0$, expand about each deepest minimum $x=2\pi j$:The local quantum harmonic oscillator $-d^2/d{x}^2+(5V_0/2)x^2$ has levels $(2n+1)\sqrt{5V_0/2}$. Its localization width is $O(V_0^{-1/4})$, so the quartic term changes these low levels by $O(1)$. Tunnelling between successive deep wells gives exponentially narrow energy bands. The secondary minima at $x=(2j+1)\pi$ have potential zero, and their levels lie far above the first two deep-well levels, whose energies are $-2V_0+O(\sqrt{V_0})$. Therefore the gap between the first and second bands is the first oscillator spacing to leading order:The $O(1)$ anharmonic shifts and exponentially small bandwidths do not change this leading estimate.

Let ${\mathrm{ad}}_A(B)=[A,B]$. Differentiating both exponentials and using that $A$ commutes with $e^{\lambda A}$ givesApplying the same calculation repeatedly gives $F^{(n)}(\lambda )=e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$. Its Taylor series at zero, evaluated at one, proves the commutator expansion for exponential conjugationFor bounded operators the series converges in operator norm by the bound $(2\Vert A\Vert {)}^n\Vert B\Vert$. The quantum computations below hold on a common invariant domain, such as Schwartz functions, with the commutator series either terminating or closing on a finite-dimensional span.

Use the canonical commutation relations $[\hat{x},\hat{p}]=i\hslash$ and the Heisenberg picture $\hat{B}(t)=e^{itH/\hslash }\hat{B}{e}^{-itH/\hslash }$. For the first Hamiltonian,All further relevant commutators vanish, soThe explicit energy check isThe scalar shifts commute with all operators, so there is no ordering ambiguity.

For the two-dimensional system put $L={\hat{x}}_1{\hat{p}}_2-{\hat{x}}_2{\hat{p}}_1$ and $V=-\beta L$. Direct commutation gives $[L,{\hat{p}}_1]=i\hslash {\hat{p}}_2$, $[L,{\hat{p}}_2]=-i\hslash {\hat{p}}_1$. Thus the Heisenberg equations of motion are ${\dot{\hat{p}}}_1=\beta {\hat{p}}_2$, ${\dot{\hat{p}}}_2=-\beta {\hat{p}}_1$, whose solution, equivalently the resummed commutator series, isExpanding both squares cancels their cross terms and gives ${\hat{p}}_1(t{)}^2+{\hat{p}}_2(t{)}^2={\hat{p}}_1^2+{\hat{p}}_2^2$.

In the interaction picture generated by $H_0$, $[H_0,{\hat{x}}_i]=-i\hslash {\hat{p}}_i$ and $[H_0,{\hat{p}}_i]=0$. The terminating expansion therefore gives ${\overline{x}}_i(t)={\hat{x}}_i+t{\hat{p}}_i$. Also $[H_0,L]=0$, either by the momentum-square calculation or by direct commutation. Conjugating ${\overline{x}}_i$ by $e^{itV/\hslash }$ rotates position and momentum by the same matrix, since $[L,{\hat{x}}_1]=i\hslash {\hat{x}}_2$ and $[L,{\hat{x}}_2]=-i\hslash {\hat{x}}_1$. Hence

The lowest Liouville–Green approximation comes from $f=e^S$, for which $S^{\prime \prime }+(S^{\prime }{)}^2=Q$. First take $S^{\prime }=\pm \sqrt{Q}$; the next transport term is $-Q^{\prime }/(4Q)$. Integrating givesMultiplicative constants and the lower limit are arbitrary. The negative sign selects the recessive approximation in the usual WKB approximation regime, where the exponent dominates the amplitude and the derivative corrections are small. Positivity of $Q$ by itself does not prove that $f_{-}$ tends to zero or that the approximation is valid: for example $Q=e^{-x}$ makes this expression grow like $e^{x/4}$ with a bounded integral in its exponent. For $Q=x$, all the required large-$x$ conditions hold and $f_{-}=x^{-1/4}{e}^{-2x^{3/2}/3}\to 0$.

Put $a=f_{-}$. Direct differentiation givesSubstituting $\mathrm{Ai}=wa$ into the Airy ordinary differential equation ${\mathrm{Ai}}^{\prime \prime }=x\mathrm{Ai}$ and dividing by $a$ gives $w^{\prime \prime }+2(a^{\prime }/a)w^{\prime }+5w/(16x^2)=0$. Multiplication by $x^2$ provesThe recessive branch has a constant leading amplitude $c$. To find its first correction write $w=c(1+b{x}^{-\delta }+\cdots )$. Balancing $-2x^{5/2}{w}^{\prime }$ against the constant term $5c/16$ forces $3/2-\delta =0$, so $\delta =3/2$. The constant coefficients then give $3b+5/16=0$. The remaining terms are of order $x^{-3/2}$ and are cancelled by a next term of order $x^{-3}$. ThusThe multiplicative constant depends on the chosen normalization of the Airy function and of $f_{-}$. The calculation selects the recessive solution; a growing Airy solution would not have this constant-amplitude expansion.

Write the Fourier coefficients of the two boundary functions as $a_{j,n}=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_j(\phi )e^{-in\phi }d\phi$. The Laplace equation in polar coordinates is $u_{rr}+r^{-1}{u}_r+r^{-2}{u}_{\phi \phi }=0$. A separated mode $R(r)e^{in\phi }$ satisfies $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. The ansatz $R=r^{\alpha }$ gives ${\alpha }^2=n^2$, so the solutions are $r^{|n|},r^{-|n|}$ for $n\ne 0$ and $1,\log r$ for $n=0$.

Set $s=\log (r/R_1)$ and $L=\log (R_2/R_1)$. Solving the two boundary equations for every mode yieldsFor real boundary data, the conjugate coefficients ensure this is real. On any compact subannulus $0<\delta \le s\le L-\delta$, the multipliers and all their differentiated versions are bounded by a polynomial in $|n|$ times $e^{-\delta |n|}$. The boundary coefficients are square summable by Parseval identity, so Cauchy-Schwarz inequality proves locally uniform convergence of the series and of all its derivatives. Thus it is smooth and satisfies the Laplace equation term by term.

Because the data are only $L^2$, the Dirichlet boundary condition is interpreted in $L^2$ on circles. As $s\downarrow 0$, the first multiplier tends to one and the second to zero; both lie between zero and one. Dominated convergence of the square-summable Fourier coefficients gives $u(r,\cdot )\to u_1$ in $L^2$, and similarly $u\to u_2$ as $s\uparrow L$. Finally, any harmonic solution with these $L^2$ traces has Fourier coefficients satisfying the radial equation above with the same two endpoint values. Uniqueness of those scalar boundary-value problems makes every coefficient equal to the displayed one. Hence this is the unique harmonic solution with the prescribed $L^2$ traces; pointwise values at every boundary angle are not required by the given data.

The Liouville approximation theorem says that if $\alpha$ is an irrational algebraic number of degree $d$, there is $c_{\alpha }>0$ such that $|\alpha -p/q|\ge c_{\alpha }{q}^{-d}$ for all integers $p$ and $q\ge 1$. To see the source of the bound, a degree-$d$ integer minimal polynomial $P$ gives $q^{d}P(p/q)$ a nonzero integer. On a fixed neighbourhood of $\alpha$, the mean value theorem bounds $|P(p/q)|$ by a constant times $|p/q-\alpha |$; outside that neighbourhood a smaller uniform constant suffices.

Let $\alpha$ denote the factorial-denominator sum and let $p_N/q_N$ be its partial sum through $N$, with $q_N=10^{N!}$ for $N\ge 2$. This is an integer numerator even though $0!=1!=1$ makes the first two terms equal. Positivity and the rapid increase of factorial exponents implyFor example the first omitted term is $10^{-(N+1)!}$ and every succeeding factorial exponent increases by at least one, giving a geometric tail bound less than $10/9$ times that first term.

First $\alpha$ is irrational: if $\alpha =a/b$, the positive difference from a partial sum is at least $1/(b{q}_N)$, contradicting this bound for sufficiently large $N$. If it were algebraic of any fixed degree $d$, Liouville approximation theorem would require $c_{\alpha }{q}_N^{-d}<2q_N^{-(N+1)}$, again impossible for large $N$. Hence $\boxed{\alpha \text{ is transcendental}.}$

The terminal wealth of portfolio $\psi$ is the scalar normal distribution $\psi \cdot X$, with mean $\psi \cdot \mu$ and variance $\psi \cdot V\psi$. Its Gaussian exponential moment givesSince $-e^{-\gamma q}$ is strictly increasing in $q$, strict preference is exactly $q(\psi )>q(\theta )$. The nonsingular covariance matrix is positive definite. Completing the square givesso the unique optimum is $\boxed{{\psi }^{\ast }={\gamma }^{-1}{V}^{-1}\mu }$.

Paying the deterministic transaction cost reduces the certainty equivalent of the new portfolio by $ϵ\Vert z{\Vert }_1$. Put $g=\mu -\gamma V\theta$. Its gain over the current portfolio is exactlyIf every $|g_i|\le ϵ$, then $g\cdot z\le ϵ\Vert z{\Vert }_1$, so every nonzero trade has strictly negative gain, by positive definiteness. Conversely, if $|g_i|>ϵ$, choose $z=t\mathrm{sign}(g_i)e_i$. Its gain isfor $0<t<2(|g_i|-ϵ)/(\gamma V_{ii})$. Therefore