For a point electric charge $q$ in a homogeneous medium of permittivity ${ϵ}_1$ beside a plane interface with permittivity ${ϵ}_2$, its potential on the source side is reproduced by a mirrored image charge $q^{\prime }=q({ϵ}_1-{ϵ}_2)/({ϵ}_1+{ϵ}_2)$. The interface has no free surface charge; continuity of tangential electric field and normal electric displacement field fixes this coefficient.

For conducting sheets $y=\pm x$ at common zero potential and a charge in $y>|x|$, successive reflections give two negative images and one positive double-reflection image. On the bisector, the negative images lie at $(\pm y,0)$ and the positive image at $(0,-y)$. Summing their fields gives the displayed downward force. Image forces exclude the real charge's self-field; replacing the induced charge by images does not make the images physical particles.

An idempotent residue modulo a positive integer $m$ is a class $e\in \mathbb{Z}/m\mathbb{Z}$ satisfying $e^2=e$, equivalently $m\mid e(e-1)$. Since consecutive integers are coprime, an idempotent modulo a prime power must be either zero or one.

If $m$ has $r$ distinct prime factors, the Chinese remainder theorem therefore gives $2^r$ idempotent residues: independently choose zero or one at each prime-power factor. Modulo six the four possibilities are $0,1,3,4$. They expose the decomposition of the quotient ring into its prime-power components.

The epicenter is the projection of an earthquake's hypocenter onto the surface directly above it.

For nonzero real $a,b$, the identity ${\cosh }^{2}t-{\sinh }^{2}t=1$ eliminates the parameter. The additional condition $x/a=\cosh t\ge 1$ is essential: the parametrization traces only the branch with the sign of $a$. Every point of that branch is obtained by $t=\mathrm{arsinh}(y/b)$. Its vertex is $(a,0)$ and its asymptotes are $y=\pm (b/a)x$. An algebraic equation alone would also include the opposite branch, which this real parametrization never reaches.

A hyperbola whose two asymptotes are perpendicular. After a suitable choice of coordinates it has equation $xy=c$ for a nonzero constant $c$. In the form $x^2/a^2-y^2/b^2=1$, it is rectangular exactly when $a=b$.

In atomic units, use the radial trial wavefunction $\psi (r)=\max (1-r/\beta ,0)$ with $\beta >0$. Its radial norm integral is $\int r^2{\psi }^{2}dr={\beta }^3/30$, its gradient contribution is $\int r^2{\psi }^{\prime 2}dr=\beta /3$, and its Coulomb contribution is $-2\int r{\psi }^{2}dr=-{\beta }^2/6$. In the hydrogen radial Rayleigh quotient, this gives the displayed energy, minimized at $\beta =4$. It is above the exact hydrogen ground-state energy $-1/2$, as required by the quantum variational principle. The derivative jumps at $r=\beta$, but the function has a square-integrable weak first derivative and is an admissible trial for the energy quadratic form. One must use that form rather than assume that a classical second derivative exists at the join.

For pressureless matter, a positive cosmological constant, flat spatial geometry and negligible radiation, the continuity equation gives ${\rho }_m\propto a^{-3}$. Substitution in the Friedmann equation yields the displayed Hubble parameter, with ${\Omega }_{m0}+{\Omega }_{\Lambda 0}=1$. radiation must be restored at early epochs; the formula does not require matter to be the largest present-day density component.

If a tight-binding Hamiltonian is written with coefficient $-t_{ij}$ multiplying $c_i^{†}{c}_j$, projecting the one-electron Hamiltonian gives the displayed relation. An alternative convention defines the coefficient itself as the hopping integral and has no leading minus. Orthogonality removes constant scalar potentials from off-diagonal matrix elements, but does not remove a spatially varying potential. Orbital rephasing changes both matrix elements together and cannot repair an algebraic mismatch between these two conventions in a fixed basis.

For $x>0$ on a domain where the tangent is finite, put $v=y/x$. The homogeneous first-order differential equation becomes $x{v}^{\prime }=-\tan v$. Direct differentiation gives $(x\sin v{)}^{\prime }=0$, hence $x\sin (y/x)=C$, without dividing by $\sin v$ and losing constant-ratio solutions. If $C=x_0>0$ and $y/x\to \pi /2$ as $x\downarrow x_0$, two branches are $y=x\arcsin (x_0/x)$ and $y=x[\pi -\arcsin (x_0/x)]$, defined for $x>x_0$. They approach the same finite endpoint but have unbounded derivatives. The original tangent is undefined at the endpoint, so neither is a classical solution through that point. An implicit relation extending continuously to an initial point does not by itself solve an initial value problem there.

On $0\le n\le 2S$, the raising/lowering amplitudes are $\sqrt{n(2S-n+1)}$ and $\sqrt{(n+1)(2S-n)}$. Their squared difference gives $[S^{+},S^{-}]=2(S-n)$. The order of the number-dependent square root is essential.

For $-\gamma \ddot{\psi }=-{\psi }^{\prime \prime }+\beta (|\psi {|}^2-1)\psi$ with positive $\beta ,\gamma$, expand about $\psi =1$ as $1+\rho /2+i\theta$. The real and imaginary equations become $\gamma \ddot{\rho }={\rho }^{\prime \prime }-2\beta \rho$ and $\gamma \ddot{\theta }={\theta }^{\prime \prime }$. Plane waves give a gapped radial amplitude branch and a gapless tangential phase branch. This second-order model has two independent real mode coordinates, unlike density-phase conjugacy in first-order dynamics.

A profile hidden Markov model represents the position-dependent structure of a sequence alignment using an ordered series of match states and optional insertion states, commonly with silent deletion states as well. An emitting state chooses one character from its probability mass function; transitions describe which profile position comes next. If deletion states are forbidden, an alignment column containing gaps cannot be a compulsory match state.

A match state in a profile hidden Markov model emits one aligned character at a particular profile position whenever it is visited. “Match” names its alignment role, not a requirement to emit one deterministic character or to match another sequence exactly.

An insertion state in a profile hidden Markov model emits characters between successive match states. A bypass allows zero inserted characters and a self-transition allows several. Under a fixed loop probability $r<1$, the length of an insertion run conditional on entering has a geometric distribution on positive integers with exit probability $1-r$.

The forward algorithm computes sequence likelihoods by summing over all hidden paths through a Hidden Markov model. For emitting states, ${\alpha }_t(i)=P(o_1,\dots ,o_t,Z_t=i)$ obeys ${\alpha }_{t+1}(j)=b_j(o_{t+1}){\sum }_i{\alpha }_t(i)a_{ij}$. It reuses partial sums rather than enumerating exponentially many state sequences. Silent states require additional transitions without consuming an observation. Scaling or log-sum-exp prevents floating-point underflow.

The backward algorithm computes suffix likelihoods in a Hidden Markov model. With ${\beta }_t(i)=P(o_{t+1},\dots ,o_T\mid Z_t=i)$, its recurrence is ${\beta }_t(i)={\sum }_j{a}_{ij}{b}_j(o_{t+1}){\beta }_{t+1}(j)$. Together with the forward algorithm, it gives posterior state probabilities proportional to ${\alpha }_t(i){\beta }_t(i)$. The terminal condition must include any explicit End-state transition.

An emission probability in a Hidden Markov model is the probability of an observation conditional on its hidden state. For a finite observation alphabet, each state's emission probabilities sum to one. A match state can therefore emit several different characters, each with its own probability.

A regular hexagon is a hexagon with all side lengths equal and all interior angles equal to $2\pi /3$. Joining its centre to the six vertices divides it into six equilateral triangles. Consequently its circumradius equals its side length $a$, its inradius is $\sqrt{3}a/2$, and its area is $3\sqrt{3}{a}^2/2$. Three regular hexagons meet around a vertex because $3(2\pi /3)=2\pi$, producing the familiar edge-to-edge tessellation of the Euclidean plane.