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There is no compact boundaryless minimal surface in ${\mathbb{R}}^3$. For an immersion $X$ of a minimal surface, the coordinate functions are harmonic: ${\Delta }_{S}X=0$. Consequently ${\Delta }_S|X{|}^2=2|{\nabla }_{S}X{|}^2=4$, because the immersion has a two-dimensional orthonormal tangent frame. Integrating over a compact surface without boundary gives zero on the left by the divergence theorem and $4\mathrm{Area}(S)>0$ on the right, a contradiction. If surfaces with boundary are allowed, a closed planar disk is a compact minimal example; the boundaryless convention matters.

A minimal surface has $H=0$, so $k_2=-k_1$ and $\boxed{K=-k_1^2\le 0}$. For a nonflat example take the catenoid parametrization $X(u,v)=(\cosh v\cos u,\cosh v\sin u,v)$. Its first fundamental form has $E=G={\cosh }^{2}v$ and $F=0$. Choosing the normal appropriately, its second fundamental form has coefficients $e=-1$, $f=0$, $g=1$. Thus its principal curvatures are $\pm {\mathrm{sech}}^{2}v$, proving $H=0$ and $\boxed{K=-{\mathrm{sech}}^{4}v<0}$.

For an oriented surface with unit normal $N$, the Gauss map sends $p$ to $N(p)\in S^2$. The shape operator is $-dN$ on the tangent plane; its eigenvalues are the principal curvatures $k_1,k_2$. The Gaussian curvature is $K=k_1{k}_2$ and the mean curvature is $H=(k_1+k_2)/2$. Reversing orientation reverses both principal curvatures and $H$, but preserves $K$. The Theorema Egregium says that Gaussian curvature is determined by the first fundamental form, so is invariant under local isometries.

A separable algebraic element has a minimal polynomial with distinct roots in an algebraic closure. A separable field extension is an algebraic extension all of whose elements are separable. In characteristic $p$, the extension ${\mathbb{F}}_p(t)/{\mathbb{F}}_p(t^p)$ is inseparable: the minimal polynomial of $t$ is $T^p-t^p$, irreducible over the smaller field but having the single root $t$ with multiplicity $p$.

If $L$ is a finite field of order $q$, every element of $L$ is a root of $T^q-T\in K[T]$, because $K$ contains its prime field. This polynomial has derivative $-1$, so it has no repeated roots. Every element's minimal polynomial divides it and is therefore separable. Hence $L/K$ is separable.

An algebraic element $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its minimal polynomial, equivalently $[K(\alpha ):K]$. Since $\alpha$ satisfies $T^2-{\alpha }^2$ over $K({\alpha }^2)$, the tower law gives $[K(\alpha ):K({\alpha }^2)]\le 2$ and makes that degree divide the odd integer $[K(\alpha ):K]$. It must therefore be one, proving $\boxed{K(\alpha )=K({\alpha }^2)}$.

For the addition assertion, let polynomial relations of degrees $r,s$ express ${\alpha }^r$ and ${\beta }^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by ${\alpha }^i{\beta }^j$ for $0\le i<r$, $0\le j<s$ contains one and is closed under multiplication by $\alpha +\beta$. Thus $1,\alpha +\beta ,\dots ,(\alpha +\beta {)}^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so $\alpha +\beta$ is algebraic.

Normalize $a(t_0)=1$. In the matter era, horizon crossing $c{t}_H=2\pi a(t_H)/k$ with $a=(t/t_0{)}^{2/3}$ gives $t_H/t_0=(k_0/k{)}^3$. In the radiation era, match $a=(t_{\mathrm{eq}}/t_0{)}^{2/3}(t/t_{\mathrm{eq}}{)}^{1/2}$ to obtainThe boundary mode therefore has $k_{\mathrm{eq}}=k_0(1+z_{\mathrm{eq}}{)}^{1/2}$. For $k<k_{\mathrm{eq}}$, the growing mode multiplies the primordial amplitude by $(t_0/t_H{)}^{2/3}=(k/k_0{)}^2$. For $k>k_{\mathrm{eq}}$, growth is approximately frozen until equality, then multiplied by $(t_0/t_{\mathrm{eq}}{)}^{2/3}=1+z_{\mathrm{eq}}=(k_{\mathrm{eq}}/k_0{)}^2$. Squaring these factors gives the matter power spectrumThe large-$k$ estimate omits the logarithmic radiation-era transfer correction, consistently with the stipulated approximation.

In a flat matter-dominated universe, $a\propto t^{2/3}$, $H=2/(3t)$ and $4\pi G{\bar{\rho }}_c=2/(3t^2)$. Substitution of ${\delta }_k\propto t^{\beta }$ gives $\beta (\beta -1)+(4/3)\beta -2/3=0$, whose roots are $2/3$ and $-1$. HenceDuring radiation domination, $a\propto t^{1/2}$ while ${\bar{\rho }}_c/{\bar{\rho }}_{\mathrm{total}}\propto a\ll 1$. The matter self-gravity term is suppressed relative to $H^2$, leaving approximately ${\ddot{\delta }}_k+t^{-1}{\dot{\delta }}_k=0$, with solutions constant and $\log t$. Thus there is no growing power-law mode comparable to the later $t^{2/3}$ growth. The elementary estimate in part (ii) neglects this slow logarithmic growth of matter perturbations during radiation domination.

The second-order equation is equivalent to the planar system after setting $y=\dot{x}-ax+b{x}^3$. Any nonconstant periodic solution has a maximum $M$ and minimum $m$. At an extremum $\ddot{x}=x^3-x$, so $M>1$ and $m<-1$ are impossible. Equality at either endpoint would give one of the equilibrium points, and uniqueness would force a constant solution. Therefore a nonconstant periodic orbit lies in $-1<x<1$.

If $b/a<1$, throughout this striphas the sign of $a$ whenever $x\ne 0$. A nonconstant orbit spends a positive time away from $x=0$, so its integral cannot vanish, contradicting part (b). Thus there are no nonconstant periodic solutions. The constant equilibrium solutions of course remain; as usual the question's periodic-orbit assertion excludes equilibria.

Take $H(x,y)=y^2/2+x^2/2-x^4/4$. Along a trajectory,Integrating over a periodic orbit returns $H$ to its starting value, and henceThis is the energy obstruction for a cubic planar oscillator.

The equilibrium points are $(0,0)$ and $(\pm 1,\pm (b-a))$, with correlated signs. The Jacobian matrix is $\left(\begin{array}{cc}a-3b{x}^2& 1\\ 3x^2-1& 0\end{array}\right)$. At the origin its eigenvalues satisfy ${\lambda }^2-a\lambda +1=0$. It is a focus for $0<|a|<2$, a repeated-eigenvalue node for $|a|=2$, and a node for $|a|>2$; stability is determined by the sign of $a$, with stability for $a<0$. Its Poincaré index is $+1$. At either other equilibrium the determinant is $-2$, so it is a saddle point with Poincaré index $-1$.

The matrix recurrence gives $p_{n+1}{q}_n-p_n{q}_{n+1}=(-1{)}^n$ and $q_{n+1}=a_{n+1}{q}_n+q_{n-1}$. Successive continued fraction convergents bracket $x$, henceThe infinite continued fraction is irrational: if $x=p/q$ were rational, a distinct convergent would be at distance at least $1/(q{q}_n)$, contradicting the first bound once $q_{n+1}>q$. If $x^2$ were rational, this irrational $x$ would have degree two. The Liouville approximation theorem would then give $|x-p_n/q_n|\ge C/q_n^2$. Comparing with the displayed bound would force $a_{n+1}\le 1/C$ for every $n$, contrary to the unbounded coefficients. Thus $x^2$ is irrational, by the unbounded continued-fraction coefficients exclude quadratic irrationality argument.

The Liouville approximation theorem says that for a real algebraic number $\xi$ of degree $d\ge 2$, there is a constant $C(\xi )>0$ such thatfor all integers $p$ and $q\ge 1$ with $p/q\ne \xi$. In particular a quadratic irrational cannot have arbitrarily accurate rational approximations of order $o(q^{-2})$.

Insert the Fourier mode $u_m^n=g^n{e}^{im\theta }$ into the leapfrog advection scheme. The amplification roots satisfyFor $0<\mu <1$, both roots have modulus one and remain uniformly separated by at least $2\sqrt{1-{\mu }^2}$. The two-level Fourier amplification matrices are therefore uniformly diagonalizable, and their powers stay bounded. The Parseval identity gives stability in the discrete two-norm. For $\mu >1$, some frequencies have $|\mu \sin \theta |>1$ and one root has modulus greater than one, giving instability.

For two-level stability at the leapfrog Courant boundary, at $\mu =1$ and $\theta =\pi /2$, the repeated root is $i$ and the recurrence has solutions $a_n=(A+Bn)i^n$. For example $a_0=0,a_1=1$ gives $a_n=n{i}^{n-1}$. Thus arbitrary two-level perturbations grow without a uniform power bound. The precise answer for unrestricted starting data isThe often-used root-modulus CFL condition is $\mu \le 1$, but by itself it misses this boundary Jordan growth. A startup that selects the exact translating branch at $\mu =1$ removes that growth for the selected data; it does not make the unrestricted two-level update power-bounded.

Substitute the exact smooth solution into the leapfrog advection scheme and expand the centered differences. With $\tau =\Delta t$ and $h=\Delta x$, the normalized local truncation error isBecause $u_t=u_x$, repeated differentiation gives $u_{ttt}=u_{xxx}$. HenceEquivalently the unnormalized one-update residual is $\tau ({\tau }^2-h^2)u_{xxx}/3+O({\tau }^5+\tau h^4)$. Thus the scheme is second order in space and time. When $\tau =h$, the exact translating solution $u(x,t)=F(x+t)$ satisfies the grid recurrence exactly; initialization still matters for the independent computational branch.

Apply the volume mean value property to the two balls and use nonnegativity together with inclusion:For any $w,z\in B(x,r)$, the triangle inequality gives $B(z,r)\subset B(w,3r)\subset B(x,4r)\subset \Omega$. Taking $R=3r$ in the previous bound therefore gives $u(w)\ge u(z)/27$. Take the infimum over $w$ and supremum over $z$ to obtainNo positivity stronger than nonnegativity is needed; if a value vanishes, this argument forces every value in the smaller ball to vanish too.

The mean value property for harmonic functions states that if $u$ is harmonic on a neighborhood of a closed ball, then its average on the ball or its boundary sphere equals its value at the center. To prove it in three dimensions, define $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(x+r\omega )d\omega$. Differentiation and the divergence theorem giveSince $M(r)\to u(x)$ as $r\to 0$, the spherical average equals $u(x)$. Integrating $4\pi r^{2}M(r)$ from zero to the ball radius gives the volume version,For balls merely contained in the open harmonicity domain, apply the result first to smaller radii and pass to the radius of interest by continuity.

A fundamental solution of a linear differential operator $P(D)$ is a distribution $E$ satisfying $P(D)E={\delta }_0$. For the three-dimensional negative Laplacian,Away from zero the function is harmonic. For a test function $\varphi$, integrate $E(-\Delta \varphi )$ outside a ball of radius $\epsilon$ and use Green's identity. The small inner sphere contributes $\varphi (0)+o(1)$ because the normal derivative of $E$ has total outward-from-the-ball flux $-1$; the other boundary term is $O(\epsilon )$. Letting $\epsilon \to 0$ verifies the distributional equation and its normalization.

Define the continuous map $T(z)=-(z^6-z^5+2z^2+1)/6$. For $|z|\le 1$, the triangle inequality gives $|T(z)|\le (1+1+2+1)/6=5/6$. Thus $T$ maps the closed unit disk into itself. By the Brouwer fixed-point theorem, some $z$ has $T(z)=z$, which rearranges toThis supplies the requested existence and even places a root strictly inside the unit disk.

The planar Brouwer fixed-point theorem says that every continuous map from the closed disk $D$ to itself has a fixed point. If a retraction $r:D\to \partial D$ existed, $x\mapsto -r(x)$ would be a continuous self-map of $D$. At a fixed point $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.

Conversely, suppose a continuous self-map $f:D\to D$ has no fixed point. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,The ray exits beyond or at $x$, so $\lambda \ge 1$, and when $x\in \partial D$ its exit point is $x$. Hence $r$ is a retraction. Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.

For equal rates, the change of numeraire from the Euro bank account to the Dollar account has density process $Z_t=Y_t/Y_0$, a strictly positive mean-one martingale under $Q_E$ by part (b). Conditional change of measure givesThus the reciprocal is a genuine martingale under $Q_D$, directly from the Euro-measure martingale property, without recalculating its stochastic differential equation.