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Write $R=\mathbb{Z}[\sqrt{-13}]$ and $I=(2,1+\sqrt{-13})$. The mapis a surjective ring homomorphism: in ${\mathbb{F}}_2$, the image of $\sqrt{-13}$ is $1$ and $1^2=-13=1$. Its kernel is exactly $I$. Indeed, both generators lie in the kernel; conversely, if $a+b$ is even, thenand $a-b$ is even. Hence $R/I\cong {\mathbb{F}}_2$ and the index of a subgroup is $[R:I]=2$.

If $I=(\alpha )$ were a principal ideal, multiplication by $\alpha =a+b\sqrt{-13}$ would identify its additive lattice with a sublattice of indexThus principality would require $a^2+13b^2=2$. This Diophantine equation has no integer solution: $b=0$ would give $a^2=2$, while $b\ne 0$ makes the left side at least $13$. Therefore

Let the perturbed interface be $z=\eta (x,y,t)$. For inviscid disturbances about rest, the initially irrotational velocities remain potential flows,in the upper layer $0<z<h$ and lower layer $-h<z<0$. The rigid walls impose no-penetration boundary conditionand zero normal derivative at $x,y=0,2h$. Linearizing the kinematic and pressure-continuity conditions at $z=0$ gives

The horizontal Neumann eigenfunctions arewhere $m,n$ are nonnegative integers, not both zero. For time dependence $e^{-i\omega t}$, the vertical factors satisfying the rigid lids areThe kinematic condition gives their interface valuesSubstitution into the dynamic condition yields the interfacial gravity-wave dispersion relationFor ${\rho }_2>{\rho }_1$, the lowest nontrivial wavenumber is $k=\pi /(2h)$, with the two degenerate modes $(m,n)=(1,0)$ and $(0,1)$. If ${\rho }_1\ll {\rho }_2$, the relation reduces to the finite-depth free-surface result ${\omega }^2\simeq gk\tanh (kh)$. If ${\rho }_1>{\rho }_2$, then ${\omega }^2<0$ and disturbances grow exponentially: this is the Rayleigh-Taylor instability.

For a boost of speed $v$ in the $x$ direction, withthe Lorentz transformation of electromagnetic fields isExpanding $E_y^{\prime }{B}_y^{\prime }+E_z^{\prime }{B}_z^{\prime }$, the mixed terms cancel and the remaining transverse terms acquire the factor ${\gamma }^2(1-v^2/c^2)=1$. Together with the unchanged longitudinal product, this proves the Lorentz invariantAn independent invariant isIf only $E_y=E$ and $B_y=B$ are nonzero, thenand direct subtraction gives

Now put $cB=\lambda E$ and $\beta =v/c$. Invariance of the dot product gives ${\mathbf{E}}^{\prime }\cdot {\mathbf{B}}^{\prime }=\lambda E^2/c$, while the displayed components giveHenceAt $\beta =0$ this equals $\mathrm{sgn}\lambda$, and as $|\beta |\uparrow 1$ it tends continuously to zero with the same sign. Thus boosts realize every $0\le \theta <\pi /2$ when $\lambda >0$, and every $\pi /2<\theta \le \pi$ when $\lambda <0$.

Inside the infinite square well, the stationary Schrödinger equation for unit mass isIts normalized eigenstates and energies areThe general normalized time-dependent solution is

Immediately after removing the barrier, the normalized wavefunction isIts Fourier sine series coefficients in the full-well basis areThey areThe Born rule therefore gives

In the sense of distribution theory, $H^{\prime }(t)=\delta (t)$. Since $\sin (\alpha t)$ vanishes at zero,and therefore

Let $G_{\mathrm{ret}}(\mathbf{x},t;\mathbf{y})$ solveThe spatial Fourier transform turns this into the preceding oscillator equation with $\alpha =ck$, soInverting with the supplied radial integral gives the three-dimensional retarded Green function

For zero initial displacement and initial velocity $f$, convolution with this kernel gives the Kirchhoff formulaThus the value at $(\mathbf{x},t)$ depends only on the initial data on the sphere of radius $ct$, not on its interior. This is the Huygens principle and describes propagation at the finite wave speed $c$.

For a closed piecewise smooth path $\gamma$ avoiding $w$, its winding number isA meromorphic function has a zero of order $m>0$ at $z_0$ whenwith $g$ holomorphic; it has a pole of order $m$ at $w_0$ when $(z-w_0{)}^{m}f(z)$ extends holomorphically and nonvanishingly there.

The argument principle states that, if $f$ has no zero or pole on $\gamma$,For a positively oriented simple contour this is the number of enclosed zeros minus poles, counted with multiplicity. It follows from the residue theorem because the residue of $f^{\prime }/f$ at a zero of order $m$ is $m$, while at a pole of order $m$ it is $-m$.

LetOn the imaginary axis,Thus $p(iy)$ always lies in the open right half-plane, has no zero there, and its continuous argument has net change zero as the imaginary-axis part of a large right-half-disc contour is traversed. On the right semicircle, $p(z)/z^4\to 1$ uniformly, so the argument change tends to that of $z^4$, namely $4\pi$. The argument principle therefore givesroots in the open right half-plane, counted with multiplicity.