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Let $b$ be the undiscounted annual premium and $d_i$ the discount fraction at level $i$. Compare the next two levels $i_r^0$ after no current claim with $i_r^1$ after a current claim, assuming no later accidents. Their total additional premium is the displayed threshold. A repair cost above it makes a claim advantageous under this two-year rule. For discounts $0,\alpha ,\beta$ and one-level moves with reflecting endpoints, the thresholds at levels one, two, and three are respectively $b\beta$, $b(2\beta -\alpha )$, and $b(\beta -\alpha )$. The premium scale must be specified or fixed as the monetary unit.

For a three-level birth-death chain whose down-or-stay claim probabilities are $p_1,p_2,p_3$, with each in $(0,1)$, detailed balance is ${\pi }_1(1-p_1)={\pi }_2{p}_2$ and ${\pi }_2(1-p_2)={\pi }_3{p}_3$. Normalizing gives the displayed stationary distribution. The finite chain is irreducible and has endpoint holding probabilities, so it is aperiodic and its level proportions converge to this distribution from any initial level.

Suppose annual claim counts are independent with geometric distribution mass $p{q}^j$, $j\ge 0$. In a no claims discount system with discounts $0,\alpha ,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The transition matrix is $\left(\begin{array}{ccc}q& p& 0\\ q& 0& p\\ q^2& pq& p\end{array}\right)$. Solving $\pi P=\pi$ gives the displayed stationary distribution. With full premium $c$, the stationary expected value of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^{2}q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.

In a no claims discount system with discounts $0,\alpha ,\beta$, a reported claim resets next year's discount to zero. With no later accidents, the next two premiums after reporting are $C+(1-\alpha )C$. At the zero level, not reporting instead gives $(1-\alpha )C+(1-\beta )C$; at either other level it gives $2(1-\beta )C$. Their differences give the displayed repair-cost thresholds. A fully reimbursed loss above its threshold should be reported under this undiscounted two-year rule. Current premiums cancel. Different thresholds generally produce state-dependent submitted-claim probabilities, even with a common accident mechanism.

At the positive Nicholson-Bailey model equilibrium, $\mathrm{tr}{J}_{\ast }=1+\log r/(r-1)$ lies between one and two, and the displayed determinant exceeds one. These follow from $(r-1)/r<\log r<r-1$ for $r>1$. The discriminant is therefore negative, and the conjugate eigenvalues have moduli greater than one. Nearby annual samples spiral outward rather than approach coexistence.

Remove a spherical void centered at $c$ from a uniform sphere centered at the origin. At points inside the void, superposition combines the sphere's interior field $-4\pi G\rho x/3$ with the removed sphere's field $+4\pi G\rho (x-c)/3$. Their sum is the displayed constant Newtonian gravitational field. A second removed mass outside this void adds a nonuniform contribution.

Let $f\in C^2([a,b])$ have $f^{\prime },f^{\prime \prime }>0$ and $f(a)<0<f(b)$. Its unique root $r$ lies in $(a,b)$. Starting Newton root-finding iteration at $b$, strict convexity puts the tangent root strictly between $r$ and the current iterate, so the iterates decrease to $r$. Taylor expansion about the current iterate gives $e_{n+1}=f^{\prime \prime }({\xi }_n)e_n^2/[2f^{\prime }(x_n)]$, where $e_n=x_n-r$ and ${\xi }_n\in (r,x_n)$. Continuity proves the displayed positive asymptotic constant and quadratic convergence.

For a $C^2$ map $\Phi$ on a Banach space, define $N(p)=p-[D\Phi (p){]}^{-1}\Phi (p)$ wherever the bounded derivative is invertible. It solves the local affine approximation to $\Phi =0$. Every zero is a fixed point of $N$. Near a zero with invertible derivative, bounded inverses and a locally Lipschitz derivative give the quadratic Newton error bound in a Banach space.

// Target: numerical-analysis.bigb

Let $F$ be twice continuously differentiable on an interval around a zero of a function $t_{\ast }$, with $|F^{\prime }(t)|\ge m>0$ and $|F^{\prime \prime }(t)|\le M$ there. Choose a closed interval $[t_{\ast }-\rho ,t_{\ast }+\rho ]$ inside that interval with $M\rho /(2m)<1$. Starting inside it makes the Newton root-finding iteration well defined, keeps it inside, and gives the displayed quadratic convergence bound for $e_r=t_r-t_{\ast }$. The nonzero derivative and sufficiently close start are essential hypotheses. The scalar estimate follows from the Taylor remainder argument in MIT's root-finding notes.^[ref]

For a short proof outline, apply Taylor's theorem between $t_r$ and $t_{\ast }$; subtract the defining tangent line equation. The linear error cancels, leaving $e_{r+1}=F^{\prime \prime }({\xi }_r)e_r^2/(2F^{\prime }(t_r))$. The displayed bound shrinks the error and preserves the chosen interval. If $F^{\prime \prime }(t_{\ast })\ne 0$, the error ratio tends to $|F^{\prime \prime }(t_{\ast })/(2F^{\prime }(t_{\ast }))|$; if it vanishes, convergence can be faster.

In Cartesian spatial coordinates and absolute time $t$, the gravitational potential $\Phi (t,x)$ determines an affine connection by the displayed coefficients, all others zero. The geodesic equation with affine time is $d^2{x}^i/d{t}^2=-{\partial }_i\Phi$. It preserves $dt$ and the spatial inverse metric $h^{ij}={\delta }^{ij}$, reproducing freely falling Newtonian mechanics in a curved spacetime connection.

NFU has a universal domain, a predicate $S$ distinguishing sets from atoms, extensionality restricted to sets, and comprehension for stratified formulas. Atoms have no members. Every stratified predicate has a set extension, including the always-false predicate, so there is a distinguished empty set even when atoms exist. Permitting atoms removes the full-extensionality obstacle in New Foundations.

The antisymmetric two-form gauge potential couples electrically to a fundamental string. Its field strength is locally $H_3=d{B}_2$; its magnetic source is an NS5-brane.

A decaying particle-antiparticle pair is described by a two-dimensional effective Hamiltonian for neutral-meson mixing with dispersive matrix $M$ and decay matrix $\Gamma$, both Hermitian matrices. The effective Hamiltonian for neutral-meson mixing itself is generally not a Hermitian matrix. CPT symmetry equates its diagonal entries, by separately equating diagonal masses and decay widths. For $H_{\mathrm{eff}}=\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$ the eigenvalues are $a\pm \sqrt{bc}$ and generic eigenvectors are $(\sqrt{b},\pm \sqrt{c}{)}^T$, with correlated square-root branches. Complex eigenvalues give masses and decay widths through $\mu =m-i\Gamma /2$.

Acetylcholine is a neurotransmitter that acts through nicotinic ionotropic receptors and muscarinic metabotropic receptors. The receptor and the target cell's machinery determine the response; acetylcholine itself does not specify whether coupling is direct or mediated by intracellular signal transduction. ^[ref]

Suppose individual event hazards have the additive form $h_i=h_{0i}+h_1$, where $h_{0i}$ is known and $h_1$ is common. The aggregate counting-process intensity in survival analysis is $\lambda ={\sum }_i{Y}_i{h}_{0i}+Y{h}_1$, with $Y={\sum }_i{Y}_i$ the current risk set size. Writing $dN=\lambda dt+dM$ and dividing by $Y$ givesThe last term is a counting-process martingale integral. This establishes the cumulative common-hazard target on the observed at-risk time range. The known offset must be averaged over the current risk set, not the initial sample. Between event jumps, its subtraction can make the unconstrained estimate decrease. If all known hazards equal $h_0$ and $Y>0$ up to $t$, the estimate reduces to the Nelson–Aalen estimator minus $H_0(t)$.

At a failure time with $d$ events among $r$ at risk, the Nelson–Aalen estimator adds $u=d/r$, whereas minus the logarithm of the Kaplan–Meier estimator adds $-\log (1-u)$. For $0\le u<1$,Thus small event fractions give close cumulative hazard estimates. Large risk sets alone do not suffice when a large fraction fail together. If all remaining individuals fail, the logarithmic estimate becomes infinite while the Nelson–Aalen jump is one.

At a strong resonance of a planar map the Floquet multiplier of a fixed point is a root of unity of low order. The usual nonresonant Neimark–Sacker bifurcation requires ${\lambda }^k\ne 1$ for $k=1,2,3,4$. At a 1:3 resonance a quadratic term ${\overline{z}}^2$ cannot be removed from the normal form; at a 1:4 resonance a cubic term ${\overline{z}}^3$ remains. Consequently one cannot infer a quasiperiodic invariant curve from the unit-modulus crossing alone.

In a complex coordinate write $z^{\prime }=\lambda z+f_{20}{z}^2+f_{11}z\overline{z}+f_{02}{\overline{z}}^2$, with $|\lambda |=1$. Away from strong resonances of a planar map, the quadratic change $z=w+h_{20}{w}^2+h_{11}w\overline{w}+h_{02}{\overline{w}}^2$ has $h_{20}=f_{20}/({\lambda }^2-\lambda )$, $h_{11}=f_{11}/(1-\lambda )$ and $h_{02}=f_{02}/({\overline{\lambda }}^2-\lambda )$. Substituting and collecting $w^2\overline{w}$ gives $g_{21}=2f_{20}{h}_{11}+f_{11}{h}_{20}+f_{11}{\overline{h}}_{11}+2f_{02}{\overline{h}}_{02}$. The cubic radial coefficient is $\ell =\mathrm{Re}(\overline{\lambda }{g}_{21})$; $\ell <0$ gives a supercritical crossing when the multiplier modulus increases through one.

The Melnikov energy-balance method measures the leading separation of invariant manifolds after a weak perturbation of a planar Hamiltonian system. Forthe exact identity is $\dot{H}=\epsilon \nabla H\cdot R$. At first order, one integrates this energy drift along the unperturbed homoclinic orbit or heteroclinic orbit, provided the improper integral converges.

For an autonomous perturbation depending on a parameter $\mu$, write the leading splitting as $\epsilon M(\mu )+O({\epsilon }^2)$. A simple zero $M({\mu }_0)=0$, $M^{\prime }({\mu }_0)\ne 0$ permits a nearby connecting-orbit parameter curve by the implicit function theorem. The zero is a first-order balance; the perturbed connecting orbit is generally displaced from the unperturbed one.

For periodic forcing, the relative forcing phase supplies a different variable. A simple zero of $M(t_0)$ can give a transverse intersection of the perturbed stable and unstable manifolds in a return section. The heteroclinic Melnikov function for a periodic planar flow derives that integral from the adjoint variational equation.

This distinction matters: a parameter-tuned autonomous saddle connection and a transverse intersection under periodic forcing are different conclusions. A multiple zero, a degenerate saddle or a nonconvergent integral needs additional analysis rather than the simple-zero argument.