$|V|=p^n$. An ordered basis is chosen successively in $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$ ways, which also counts bijective endomorphisms. Counting ordered independent $k$-tuples and dividing by the number of bases of ${\mathbb{F}}_p^k$ gives the Gaussian binomial ${\prod }_{i=0}^{k-1}(p^n-p^i)/(p^k-p^i)$.

An extreme point is not a nontrivial convex combination of two distinct points of the set. Write $x=x^{+}-x^{-}$ with nonnegative parts and set $y=(x^{+},x^{-})$, $c=(1,1)$, $M=(A,-A)$, $a=b$. An extreme feasible point of this standard-form LP has at most $m$ positive coordinates (otherwise the corresponding columns are dependent and permit a two-sided feasible perturbation); an optimum may be chosen extreme, and cancelling simultaneous positive pairs gives an $x$ with at most $m$ nonzeros. For the final problem, fix $b=A{x}^{\ast }$ for any optimum and replace $x^{\ast }$ by such a sparse minimum-${\ell }^1$ solution of $Ax=b$; $f(Ax)$ is unchanged and the penalty cannot increase.

The MLE is $\hat{\theta }=n/S$, $S=\sum X_i$. Since $S\sim \Gamma (n,\theta )$, $\theta S\sim \Gamma (n,1)$ (in general $\gamma Y\sim \Gamma (\beta ,\lambda /\gamma )$). If $q_r$ is the $r$-quantile of $\Gamma (n,1)$, a central confidence interval is $[(q_{\alpha /2}/n)\hat{\theta },(q_{1-\alpha /2}/n)\hat{\theta }]$. A $\Gamma (\beta ,\lambda )$ prior yields posterior $\Gamma (\beta +n,\lambda +S)$ and mean $\tilde{\theta }=(\beta +n)/(\lambda +S)$. With $q_r\prime$ the quantiles of $\Gamma (\beta +n,1)$, take $l\prime =q_{\alpha /2}\prime /(\beta +n)$ and $u\prime =q_{1-\alpha /2}\prime /(\beta +n)$.

Vertical hydrostatic balance gives $p=\rho g(h-z)$; depth-independent horizontal momentum and depth-integrated incompressibility yield the two shallow-water equations. Curling momentum and combining with mass conservation gives $D[(\zeta +f)/h]/Dt=0$. Dot momentum with $\rho hu$ and use continuity to obtain $E_t+\nabla \cdot J=0$ with $E=\rho h|u{|}^2/2+\rho g{h}^2/2$ and $J=u(E+\rho g{h}^2/2)$. It is kinetic plus gravitational potential energy. A no-normal-flow boundary kills the flux and conserves its integral; a localized nonuniform height has excess positive energy relative to the uniform rest state with the same mass, so it cannot decay to that state without dissipation.

$\rho =|\psi {|}^2$ and $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }_x)$. Schrödinger’s equation and its conjugate give ${\rho }_t+j_x=0$; a stationary state has $j_x=0$. Continuity of $\psi ,\psi \prime$ at $0,a$ and matching plane waves giveso $F=1/3$ and $G=a\sqrt{3m}/\hslash$. Transmission is minimised when $G\sqrt{U_0}=\pi /2+n\pi$, i.e. $U_0=(\pi /2+n\pi {)}^2/G^2$.

Characteristics satisfy $\dot{x}=a$, $\dot{y}=b$, $\dot{u}=c$, with data prescribed on a transverse curve. If $(a,b)=(H_y,-H_x)$ then $dH/dt=H_x{H}_y-H_y{H}_x=0$. For the displayed field the divergence is zero and one may take $H=x{y}^2-x^3/3$. Thus characteristics are its level curves (a cubic foliation), and because $c=0$ the general solution is $u(x,y)=F(x{y}^2-x^3/3)$ for arbitrary differentiable $F$.

Twice integrating by parts, with both endpoint variations fixed, gives $g_x-d(g_{\dot{x}})/dt+d^2(g_{\ddot{x}})/d{t}^2=0$. In the given integral the mixed $\dot{x}\ddot{x}$ term is a boundary term, and the equation is $(D^2-4{)}^{2}x=0$. Decay removes the growing solutions, so $x=(A+Bt)e^{-2t}$. The initial data give $A=1,B=3$, hence $x=(1+3t)e^{-2t}$.

$Y(s)={\int }_0^{\infty }{e}^{-st}y(t)dt$, with transforms $sY-y(0)$ and $s^{2}Y-sy(0)-\dot{y}(0)$; $e^{\lambda t}$ transforms to $(s-\lambda {)}^{-1}$ and $\delta$ to $1$. Algebra gives $Y=GH$ with $H=F+(s+2\gamma )y(0)+\dot{y}(0)$, so $y=g\ast h$. Writing $a=\sqrt{|{\gamma }^2-{\omega }^2|}$ gives $g=e^{-\gamma t}\sinh (at)/a$ for $\gamma >\omega$, $e^{-\gamma t}\sin (at)/a$ for $\gamma <\omega$, and $t{e}^{-\gamma t}$ at equality. If $y(0)=0$, $h=f+\dot{y}(0)\delta$, so the initial velocity contributes $\dot{y}(0)g(t)$.

$P$ is a square with two open square holes; its boundary is the three square boundary curves. Using their twelve corners gives a triangulation with $V=12,E=27,F=14$. Doubling two compact connected copies along their common boundary is compact and connected, and boundary half-discs glue to discs, so the double is locally Euclidean and is a closed surface. Under the quotient the doubled triangulation has $V=12,E=42,F=28$, hence $\chi =-2$. The classification theorem says a connected compact orientable surface is a sphere or a connected sum of $k$ tori with $\chi =2-2k$; thus $k=2$.

Bounded means finite diameter; sequential compactness means every sequence has a convergent subsequence. A failure of uniform continuity supplies two close sequences whose image distances stay apart; a convergent subsequence contradicts continuity. Into a discrete metric, continuity is local constancy, and compactness upgrades the radii uniformly. The Cantor set is closed in compact $[0,1]$, hence sequentially compact. Uniform local constancy gives a positive separation scale, so finitely many initial Cantor-code (binary) digits determine $f$. Removing zero destroys compactness: the hinted function reading the digit after the first $1$ is continuous at every remaining sequence but depends on arbitrarily late digits.

$V^{\ast }$ is the space of linear functionals and the dual basis satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Nondegeneracy and equal finite dimensions make $u\mapsto ⟨u,\cdot ⟩$ an isomorphism; evaluation similarly identifies $V$ with $V^{\ast \ast }$. Riesz representation applied to $v\mapsto ⟨Tv,u⟩$ gives the unique adjoint. If $T$ is diagonalizable, declare an eigenbasis orthonormal to make it self-adjoint. Conversely, for self-adjoint $T$ and invariant $W$, $⟨Tv,w⟩=⟨v,Tw⟩=0$ shows $W^{\perp }$ invariant.

The communicating classes are $\{1,2,3,6\}$ and $\{4,5\}$; only the latter is closed. Let $h_i=P_i(T_2<\infty )$, with $h_2=1$ and $h_4=h_5=0$. The first-step equations give $h_3=1/2$, $h_6=(h_1+h_3)/2$, and $h_1=1/3+h_3/3+h_6/6$, hence $h_1=13/22$.

Unsteady Bernoulli gives the surface pressure from $p+\rho {\partial }_t\varphi +\rho |\nabla \varphi {|}^2/2=C(t)$. The velocity-squared contribution integrates to zero net force, while the unsteady term gives $F=-m_a\dot{V}$ with added mass $m_a=\frac{1}{2}(4\pi \rho a^3/3)=2\pi \rho a^3/3$. Hence $(m+m_a)a=G$, so $m^{\ast }=m+2\pi \rho a^3/3$: accelerating the bubble also accelerates surrounding fluid.

$L_a={\epsilon }_{abc}{x}_b{p}_c$, with $[x_a,p_b]=i\hslash {\delta }_{ab}$ and position-position and momentum-momentum commutators zero. Expansion gives $[L_a,L_b]=i\hslash {\epsilon }_{abc}{L}_c$ and $[L_a,p_d]=i\hslash {\epsilon }_{adc}{p}_c$. Thus $[L_1,p_2]=i\hslash p_3\ne 0$, so they are not simultaneously diagonalisable.

Integration by parts gives ${\tilde{g}}_{\sigma }\prime (k)=-{\sigma }^{2}k{\tilde{g}}_{\sigma }(k)$ and ${\tilde{g}}_{\sigma }(0)=1$, so ${\tilde{g}}_{\sigma }=e^{-{\sigma }^2{k}^2/2}$. The convolution theorem says $\widetilde{f\ast g}=\tilde{f}\tilde{g}$; multiplying the two Gaussian transforms and inverting yields $g_{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}$.

For fixed boundary values, integration by parts gives $L_u-{\partial }_x{L}_{u_x}-{\partial }_y{L}_{u_y}=0$. For $L=u^2+u_x{u}_y$, this is $2u-2u_{xy}=0$, or $u_{xy}=u$, with the independently prescribed boundary data $u=xy$ on the circle.

Integrate $(1+z^p{)}^{-1}$ around the wedge $0\le \arg z\le 2\pi /p$. The radial integrals differ by the factor $e^{2\pi i/p}$, the arc vanishes, and the wedge contains the simple pole $e^{i\pi /p}$ with residue $-[p{e}^{-i\pi /p}{]}^{-1}$. Solving the resulting identity gives $I=(\pi /p)\csc (\pi /p)$.

The subspace opens are $A\cap U$; quotient opens are those whose inverse image is open; a homeomorphism is a continuous bijection with continuous inverse. Here $q(x,y)=(|x|,y)$ identifies exactly the required pairs and induces a homeomorphism $Y\to [0,1]\times [-1,1]$. An affine rescaling maps that rectangle homeomorphically onto $S$.

$A={\mathbb{Z}}^2/⟨(6,9)⟩$. The class of $(2,3)$ has order $3$, so $A$ has torsion and is not free. A unimodular change of basis (Smith normal form, using $\gcd (6,9)=3$) gives $A\cong \mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$.

Gradient descent is $x_{t+1}=x_t-\eta \nabla f(x_t)$. Integrating the Hessian along a segment gives the descent lemma. Applying it to the hinted $z$ and convexity yields cocoercivity: $(\nabla f(x)-\nabla f(y){)}^T(x-y)\ge \Vert \nabla f(x)-\nabla f(y){\Vert }^2/\beta$. Apply this to $\varphi =f-\alpha \Vert x{\Vert }^2/2$, whose Hessian lies between $0$ and $(\beta -\alpha )I$, and rearrange to obtain the displayed strengthened inequality. With $y=x^{\ast }$, $\nabla f(x^{\ast })=0$, and $\eta =2/(\alpha +\beta )$, expansion of the squared update gives contraction factor $((\beta -\alpha )/(\beta +\alpha ){)}^2$ per step and the stated bound.