Nondecreasing functions $\mathbb{N}\to \mathbb{N}$ form an uncountable set: every subset $A$ is encoded by the strictly increasing function $n+{\sum }_{k=1}^n{1}_A(k)$. In contrast, nonincreasing functions form a countable set because each is determined by a finite initial segment and its eventual constant value.

// Target: classical-mechanics.bigb

A finite subset $A$ of the natural numbers, numbered from one, has the binary code ${\sum }_{j\in A}{2}^{j-1}$. Distinct finite subsets give distinct codes, because the largest differing exponent exceeds the sum of all smaller powers of two. Thus the finite subsets form a countable set, although the entire power set is an uncountable set by the Cantor diagonal argument. An enumeration transfers this coding to any countable set.

A family of pairwise disjoint real intervals of positive length is countable. Each interval contains a rational number in its interior by the density of the rational numbers. Fix an enumeration of the rational numbers and select the first member lying inside each interval. Disjointness makes this selection an injection into a countable set. The argument also applies to disjoint nonempty open subsets of the real line.

For $\mathrm{Im}z>0$, writing cotangent as $i(e^{2\pi iz}+1)/(e^{2\pi iz}-1)$ and expanding a geometric series gives this Fourier series. Differentiating the cotangent partial-fraction expansion and this Fourier series $k-1$ times yields

The incremental cost-effectiveness ratio divides an intervention's additional mean cost by its additional mean effect relative to a comparator. Its units are cost per outcome unit. When both increments are positive, comparison with a threshold $\lambda$ is equivalent to checking whether the incremental net monetary benefit $\lambda \Delta E-\Delta C$ is positive. When $\Delta E<0$ the inequality reverses on division, and when $\Delta E=0$ the ratio is undefined. A ratio alone cannot identify the quadrant of the cost-effectiveness plane.

The cost-effectiveness plane plots an effect increment horizontally and a cost increment vertically. The southeast quadrant represents greater effectiveness at lower cost, while the northwest quadrant represents lower effectiveness at greater cost. A threshold $\lambda$ corresponds to the line $\Delta C=\lambda \Delta E$; points below the line have positive incremental net monetary benefit, in every quadrant. A joint confidence region or paired bootstrap cloud reveals uncertainty and preserves the dependence between cost and effect estimates.

At a specified willingness-to-pay threshold $\lambda$, the incremental net monetary benefit is the value of the effect increment minus the cost increment. It avoids dividing by a potentially near-zero or negative effect estimate. For estimates with variances $v_E,v_C$ and covariance $v_{CE}$, its estimated variance is ${\lambda }^2{v}_E+v_C-2\lambda v_{CE}$. A normal approximation therefore gives a confidence interval directly on the net-benefit scale. The economic sign criterion does not require a positive effect increment.

A light signal sent from a comoving observer to a comoving reflector at separation $\chi$, and immediately returned, consumes conformal time $2\chi$. The supremum of the initial physical reflector distances allowing a return is therefore half the cosmological event horizon radius. Equality generally requires infinite return time; finite-time reception requires a strictly smaller distance. For a cosmological domain wall network with $a\propto t^2$, the supremum is $t/2=1/H$.

For separately conserved pressureless matter, ${\rho }_m\propto a^{-3}$. The deceleration parameter identity $\dot{H}/H^2=-(1+q)$ and ${\Omega }_m\propto {\rho }_m/H^2$ give ${\dot{\Omega }}_m=H{\Omega }_m(2q-1)$. A dust-plus-vacuum universe has $q={\Omega }_m/2-{\Omega }_{\Lambda }$; conversion to conformal time yields the displayed flow. The familiar matter-only equation requires negligible cosmological constant. In a spatially flat dust-plus-vacuum universe, ${\Omega }_m+{\Omega }_{\Lambda }=1$ and ${\Omega }_m^{\prime }=-3\mathcal{H}{\Omega }_m{\Omega }_{\Lambda }$, so flatness does not require ${\Omega }_m=1$.

For a scale-invariant potential spectrum with ${\mathcal{A}}_{\Phi }=k^3{P}_{\Phi }/(2{\pi }^2)$ constant, the matter-era Sachs-Wolfe effect gives $C_{\ell }=(4\pi {\mathcal{A}}_{\Phi }/9)\int j_{\ell }(x{)}^{2}dx/x$. The logarithmic spherical Bessel square integral then gives $\ell (\ell +1)C_{\ell }=2\pi {\mathcal{A}}_{\Phi }/9$. This plateau applies to large-angle temperature anisotropies, neglecting acoustic evolution and late evolving potentials; it is not a constant $C_{\ell }$ spectrum.

For an outward sightline, Fourier velocity $v_b^i=i{k}^i{\theta }_b$, and sudden last scattering, intrinsic photon temperature projects with a Spherical Bessel function, while the Doppler term projects with its derivative. Density and velocity acoustic sources are approximately a quarter cycle out of phase. Their amplitude is suppressed by photon diffusion, giving a power damping envelope $e^{-2k^2/k_D^2}$. Projection maps the sound scale to peak spacing approximately $\pi D_{\ast }/r_s$ and the damping scale to angular multipoles of order $k_D{D}_{\ast }$.

If a measured cluster baryon fraction represents the cosmic baryon fraction ${\Omega }_b/{\Omega }_m$, combining it with the cosmological baryon density inferred from Big Bang nucleosynthesis gives the displayed estimate. Gas and stellar baryons must both be counted, and all distance or Hubble constant dependence must be consistent. Baryon depletion or feedback modifies the identification: with $f_{b,\mathrm{cluster}}=b{\Omega }_b/{\Omega }_m$, the corrected answer is ${\Omega }_m=b{\Omega }_b/f_{b,\mathrm{cluster}}$. The method measures total matter rather than vacuum or radiation density.

At colatitude $\theta$, ignore terms quadratic in Earth angular speed and use uniform vertical gravity. A vertically launched particle has zeroth-order height $z=vt-g{t}^2/2$. Its eastward Coriolis acceleration is $-2\omega \sin \theta \dot{z}$, so its horizontal velocity is $-2\omega \sin \theta z$. Integrating over the flight time $2v/g$ gives the westward displacement displayed above. Descent reverses the horizontal acceleration but does not cancel the accumulated displacement.

Core collapse can fail to produce an ordinary successful ejecting supernova, leaving a black hole and retaining much of the processed core. Some envelope material can still escape, depending on the event; failure is not equivalent to a universal zero-yield prescription.

A list of integers $a_1,\dots ,a_r$, with $r\ge 2$, has joint coprimality when its greatest common divisor is one; define that greatest common divisor by repeated application of the two-input operation. For positive integers, no prime factor may divide every entry. Thus $6,10,15$ have joint coprimality but fail pairwise coprimality. For two entries, the conditions coincide.

A list of integers $a_1,\dots ,a_r$, with $r\ge 2$, has pairwise coprimality when every two entries in distinct positions are coprime. This implies joint coprimality. For positive integers, it means no prime factor occurs in two different entries. For example, $4,9,25$ satisfy this condition.

Two players independently choose venues $A$ or $B$. Meeting at $A$ gives the ordered payoff pair $(3,2)$; meeting at $B$ gives $(2,3)$; choosing different venues gives $(0,0)$. As throughout this article, the row player's payoff comes first:

Against a column player choosing $A$, the row player gets $3$ from $A$ and $0$ from $B$, so $A$ is the unique best response. Against $B$, the row player gets $0$ from $A$ and $2$ from $B$, so $B$ is the unique best response. The column player likewise prefers to match the row choice. Thus $(A,A)$ and $(B,B)$ are pure strategy Nash equilibria, while each mismatched strategy profile has a profitable unilateral deviation.

For mixed strategies, let $p$ and $q$ be the row and column probabilities of choosing $A$. The pure-choice expected payoffs areThe row player is indifferent when $3q=2(1-q)$, giving $q=2/5$. The column player is indifferent when $2p=3(1-p)$, giving $p=3/5$. The best responses are thereforeThese give exactly three Nash equilibria: $(p,q)=(0,0),(1,1),(3/5,2/5)$. If either player uses both choices, their indifference condition places the other player's probability strictly between zero and one, so both conditions must hold; otherwise both choices are pure and must match. Notice whose equation determines whose probability: the row player's indifference fixes the column player's $q$, and the column player's indifference fixes the row player's $p$.

At the mixed Nash equilibrium, independence givesThus they meet with probability $12/25$, and each receives expected payoff $3(6/25)+2(6/25)=6/5$. Holding $q=2/5$ fixed, every row deviation $p^{\prime }$ gives $p^{\prime }(6/5)+(1-p^{\prime })(6/5)=6/5$; holding $p=3/5$ fixed, every column deviation $q^{\prime }$ gives $q^{\prime }(6/5)+(1-q^{\prime })(6/5)=6/5$. This directly verifies the Nash equilibrium inequalities.

Both coordinated pure strategy outcomes strictly improve both players' expected payoffs over $(6/5,6/5)$: $(3,2)$ and $(2,3)$ each do so. The mixed Nash equilibrium therefore fails Pareto efficiency, even though no unilateral deviation helps. Switching together can help when switching alone cannot. Neither coordinated outcome improves both players over the other, and both are Pareto efficient, even when a lottery over joint action profiles is allowed: any positive weight on another outcome reduces the payoff of the player receiving $3$.

For $a,b>0$, the convolution of $e^{-a|x|}$ with the indicator of $[-b,b]$ is $2e^{-a|x|}\sinh (ab)/a$ outside the interval and $2[1-e^{-ab}\cosh (ax)]/a$ inside it. With Fourier convention $\tilde{f}(k)=\int f(x)e^{-ikx}dx$, their transforms are $2a/(a^2+k^2)$ and $2\sin (bk)/k$. Fourier inversion therefore evaluates ${\int }_{\mathbb{R}}\sin (bk)e^{ikx}/[k(a^2+k^2)]dk$ as $\pi h(x)/(2a)$.

If a convex roof extension is non-increasing on average for every pure input to a quantum instrument, then it is non-increasing on average for mixed inputs too. For an input ensemble $\{p_j,{\pi }_j\}$, let $q_{k|j}$ and ${\sigma }_{jk}$ be its conditional outcome probabilities and states. Linearity of each unnormalized outcome map gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. Convexity gives ${\sum }_k{P}_{k}F({\rho }_k)\le {\sum }_{j,k}{p}_j{q}_{k|j}F({\sigma }_{jk})\le {\sum }_j{p}_j\mu ({\pi }_j)$. Taking the infimum over input ensembles proves the result, even when conditional outputs of pure inputs are mixed, provided the averages are well defined.

A convex chain between two fixed points inside a triangle is a finite point set whose union with the endpoints forms the vertices of a convex polygon, with the endpoint segment as one boundary edge. In the triangle $0\le y\le x\le 1$, with endpoints $(0,0)$ and $(1,1)$, its ordered lower boundary has strictly increasing segment slopes. Every subset of its intermediate vertices remains a convex chain. This hereditary property lets selected point coordinates certify the existence of a long chain.