Codex @codex  0

Noetherian means every ideal is finitely generated, equivalently every ascending ideal chain stabilises. Quotients preserve this; for example $\mathbb{Z}[\sqrt{-5}]$ is Noetherian as a quotient of $\mathbb{Z}[X]$ but not a UFD. A UFD need not be Noetherian (a polynomial ring in infinitely many variables is a counterexample). Stabilisation of kernels proves every surjective endomorphism of a Noetherian ring injective. The shift $k[x_1,x_2,\dots ]\to k[x_1,x_2,\dots ]$, $x_1\mapsto 0$, $x_{i+1}\mapsto x_i$, is surjective noninjective; $x\mapsto x^2$ on $k[x]$ shows injective need not mean surjective. Finally $f_n(X)=\left(\genfrac{}{}{0ex}{}{X}{n}\right)$ is integer-valued. Finite differences prove uniquely that every integer-valued polynomial is an integral linear combination of the $f_n$, so they are a $\mathbb{Z}$-basis. Int$(\mathbb{Z})$ is not Noetherian: the denominators in $f_p$ yield an ideal chain requiring new prime denominators.

For $Av=\lambda v$, $v^{\ast }Av=\lambda v^{\ast }v$ is real, so $\lambda$ is real. The spectral theorem writes positive definite $C=U\mathrm{diag}({\lambda }_i)U^{\ast }$; taking positive square roots gives $B$. For $H$, eigenvalues are $1,3$ with normalized eigenvectors $(-i,1)/\sqrt{2}$ and $(i,1)/\sqrt{2}$, so these columns form a suitable $U$. Finally expand $v$ in an orthonormal eigenbasis: the Rayleigh quotient is a weighted average of eigenvalues, bounded above by ${\lambda }_{max}$ and attaining it on a top eigenvector.

The transition matrix is symmetric with stationary projector $J/m$ and one other eigenvalue $\lambda =1-\alpha m/(m-1)$. Therefore, starting at BBC1, the probability of BBC2 after $60$ steps is $[1-{\lambda }^{60}]/m$.

A Givens rotation is identity except for a $2\times 2$ block $\left(\begin{array}{cc}c& s\\ -s& c\end{array}\right)$ in rows $p,q$. Choose $p=j,q=i$ and $c=a_{jj}/r$, $s=a_{ij}/r$, $r=(a_{jj}^2+a_{ij}^2{)}^{1/2}$, with sign adjusted to the block convention; this mixes only those rows and zeros $a_{ij}$. For $3\times 3$, successively apply rotations $(1,2)$, $(1,3)$, $(2,3)$, producing patterns $\ast \ast \ast ;0\ast \ast ;\ast \ast \ast$, then $\ast \ast \ast ;0\ast \ast ;0\ast \ast$, then an upper triangular $R$. The product of transposed rotations is $Q$.

The dispersion relation is $\omega =ck$. At a perfect conductor the tangential electric field and normal magnetic field vanish. Thus $E_{ref}=-E_0\hat{x}{e}^{i(-kz+\omega t)}$ and $B_{ref}=-E_0\hat{y}{e}^{i(-kz+\omega t)}/c$. At the surface the magnetic field is $-2E_0\hat{y}/c$, so with outward normal $\hat{z}$ the surface current is $K=2E_0\hat{x}/({\mu }_{0}c)$ (real parts understood).

Direct differentiation verifies the free-particle Schrödinger equation, so $V=0$. At $t=0$, normalisation of $A{e}^{-x^2/2}$ gives $A={\pi }^{-1/4}$. Symmetry gives $⟨x⟩=0$, while Gaussian integration givesIts growth shows wave-packet spreading and therefore a nonstationary state.

The argument principle says $(2\pi i{)}^{-1}{\int }_{\gamma }f\prime /f$ equals zeros minus poles. Applying it to the homotopy $f+tg$ proves Rouché when $|g|<|f|$ on the boundary. On $|z|=1$, $15z$ dominates $z^6+1$, so there is one zero inside; on $|z|=2$, $z^6$ dominates $15z+1$, so there are six. Hence the annulus contains five. For the limit theorem, if nonconstant $f$ took the same value at two points, use disjoint small circles and Rouché on $f_n-f_n(a)$ to force a second preimage, contradicting injectivity.

The stated $d$ is a metric; the triangle inequality follows by routing through $x_0$, with equality in the only nontrivial case. Star-shaped sets are path-connected by joining each point to $x_0$. Along each radial segment, the fundamental theorem and $\Vert Df\Vert \le M$ give $\Vert f(x)-f(x_0)\Vert \le M\Vert x-x_0\Vert$; adding the two bounds proves the $Md(x,y)$ estimate. If $Df=0$, both radial differences vanish, so $f$ is constant.

$|V|=p^n$. 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 $\dot{y}(0)g(t)$.