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.

Taking the divergence of the constant-density Navier-Stokes equation and using incompressibility gives the displayed elliptic Poisson equation for pressure. Its Green function is nonlocal. For an isolated eddy, the far pressure has a quadrupolar contribution proportional to $r^{-3}$, obtained by differentiating the kernel $1/r$ twice. Boundary conditions can contribute additional pressure fields.

The Moon is Earth's natural satellite. Its synchronous rotation keeps approximately the same near side of the Moon facing Earth, while both the near side of the Moon and the far side of the Moon experience sunlight and darkness. See illumination of a synchronously rotating moon for the geometry. ^[ref].

A feasible payoff vector $u^{\prime }$ jointly dominates $u$ if every coordinate is at least as large and at least one is strictly larger. A Pareto optimal vector has no feasible joint dominator. Strict increase of every coordinate is not required. This compares each player's own utilities, not the numerical size of utilities across different players.

For a compact convex set of feasible payoff vectors $F$ and a specified disagreement point $d$, the negotiation set is its individually rational Pareto frontier. It is the set of efficient agreements each player prefers at least weakly to disagreement. In a maximin bargaining solution, $d$ is the vector of the two security level payoffs. This usage is distinct from objection-and-counterobjection bargaining sets in coalitional-game theory.

At an equilibrium $x_{\ast }=f(y_{\ast })$, $y_{\ast }=g(x_{\ast })$, the Jacobian matrix has eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its product of repression gains is also $(\partial \log f/\partial \log y)(\partial \log g/\partial \log x)$. Product below one gives a stable node, above one a saddle, and equality the threshold for a zero eigenvalue. Robust multiple stable equilibria require an intervening unstable equilibrium; a gain exceeding one alone is not a sufficient global criterion for a particular pair of repression functions.

// Target: mathematical-biology.bigb

For positive $m,n$, this mutual repression system can display bistability for some $\lambda$ exactly when both $m>1$ and $n>1$. If either exponent is at most one, it has a unique stable equilibrium for every $\lambda$. To prove the less obvious case $m\le 1<n$, put $a=x/(1+x)$ and $b=y/(1+y)$ at equilibrium. Equality of the two expressions for $\lambda$ gives $a\le b(1-b{)}^{n-1}$. The loop gain is then bounded by $mn{b}^2(1-b{)}^{n-1}$, whose maximum is $4mn(n-1{)}^{n-1}/(n+1{)}^{n+1}<1$. Swap the two components for the other case. If both exponents exceed one, sufficiently large $\lambda$ yields two stable equilibria with $(x,y)\sim (\lambda ,{\lambda }^{1-m})$ and $(x,y)\sim ({\lambda }^{1-n},\lambda )$. The weaker condition $mn\le 1$ is sufficient to exclude multistability, but does not exhaust all exclusions when the same production prefactor is imposed.

// Target: mathematical-biology.bigb

For $\dot{x}=\lambda /(1+y^m)-x$, $\dot{y}=\lambda /(1+x^n)-y$, the equilibrium slope product is $K=mn{x}^n{y}^m/[(1+x^n)(1+y^m)]$. A value $K>1$ gives a saddle separating attracting alternatives. Both exponents greater than one allow this at sufficiently large common $\lambda$. If both are at most one it is impossible. If $0<m\le 1<n$, define $Y(x)$ as the unique positive solution of $Y^{-1}+Y^{m-1}=x^{-1}+x^{n-1}$ and set $K_{m,n}(x)=mn{x}^{n}Y(x{)}^m/[(1+x^n)(1+Y(x{)}^m)]$. All equilibria are parameterized by $\lambda =x(1+Y^m)$, whose derivative is $(1+Y^m)(1-K)/(1-m{Y}^m/(1+Y^m))$. Thus multistability for some common strength occurs exactly when ${\max }_{x>0}{K}_{m,n}(x)>1$. For $m=1$ the maximum is $(n-1{)}^2[(n+1)/(n-1){]}^{(n+1)/n}/(4n)$. Consequently $mn>1$ alone is not sufficient when the two synthesis strengths are required to coincide.

For finite discrete random variables, suppose $X\to Y\to Z$ is a Markov chain: the final channel uses only $Y$, so $X$ and $Z$ have conditional independence given $Y$. In terms of the joint probability mass function,and hence $I(X;Z\mid Y)=0$. Then the two data processing inequalities areAll mutual informations here use base-two logarithms. Thus deterministic or randomized processing cannot increase the information shared with an unprocessed variable.^[ref]

For a concrete cascade, take a fair bit $X$ and Bernoulli random variables $N_q,N_r$ with probabilities $q=0.1$, $r=0.2$ of being one. Assume $X,N_q,N_r$ are jointly independent random variables, and pass through two binary symmetric channels:The second noise is independent of both $X$ and $Y$, so this is a Markov chain. Both $Y$ and $Z$ are fair bits. The net exclusive or error is one exactly when one noise flips the bit and the other does not:Using the binary entropy $h_2(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$, the three mutual informations are fully determined:All numbers are in bits. The two losses are $I(X;Y\mid Z)\approx 0.3577507789$ and $I(Y;Z\mid X)\approx 0.1048182776$. Thus the cascade retains less information about the original source than either individual channel shares between its own input and output.

With $q=0.1$ fixed and $0\le r\le 1/2$, the effective error is $s=0.1+0.8r$. At $r=0$, $Z=Y$ and $I(X;Z)=I(X;Y)$; at $r=1/2$, $Z$ is independent of the previous variables and $I(X;Z)=I(Y;Z)=0$. The diagram plots these endpoints and the computed $r=0.2$ example.

The data processing inequality is useful for testing what a proposed decoder or data summary could possibly reveal: if its only input is $Y$, even randomized processing cannot provide more mutual information about $X$ than $Y$ already contains. The Markov chain hypothesis matters. In the exclusive or example from conditional mutual information, allowing a processor to use both $Y$ and $Z$ reveals $X$ exactly; that processor has additional input, so the conclusion for processing $Y$ alone does not apply.

The chain rule for mutual information proves both inequalities and identifies their exact gaps. First expand the information about $X$ in the two observation orders:Therefore $I(X;Y)-I(X;Z)=I(X;Y\mid Z)\ge 0$. Next use $Z$ as the variable being learned:Symmetry of mutual information gives $I(Y;Z)-I(X;Z)=I(Y;Z\mid X)\ge 0$. Both signs follow from the nonnegativity of conditional mutual information; the first gap also gives equality in data processing inequality.

The monomial $x^i{y}^j$ is an eigenvector of the differential linear operator $x^2{\partial }_x^2+y^2{\partial }_y^2$, with eigenvalue $i(i-1)+j(j-1)$. Over the real or complex numbers this vanishes exactly for $i,j\in \{0,1\}$. On polynomials of bounded total degree, the kernel is therefore spanned by the permitted members of $1,x,y,xy$, while the image of a linear map is spanned by all remaining monomials. The trace is the sum of those diagonal eigenvalues.