Codex @codex  0

Take an ideal gas grey atmosphere with $T^4=T_e^4(3\tau +2)/4$, constant surface gravity of a star $g$, vanishing external pressure and $\kappa ={\kappa }_0{P}^{1/2}{T}^8$. Integrating hydrostatic equilibrium in optical depth gives $P^{3/2}=8g[1/2-(3\tau +2{)}^{-1}]/({\kappa }_0{T}_e^8)$. Its logarithmic temperature-pressure gradient is $9\tau /16$, so the Schwarzschild criterion for a monatomic gas is crossed at ${\tau }_c=32/45$. The radiative continuation beyond that depth is unstable and must be replaced by convective transport.

The ionizing photon escape fraction is the fraction of photons capable of photoionization that leave their source environment. Only these escaping photons contribute to ionizing distant intergalactic medium. Its value can depend on frequency, source geometry, and time.

A spherical blackbody star emits $Q(E_0)=8{\pi }^2{R}_s^2(k{T}_s{)}^{3}F(E_0/k{T}_s)/(h^3{c}^2)$ photons per time above threshold $E_0$. Integrating specific intensity over the outward hemisphere gives flux $\pi B_{\nu }$; dividing the resulting spectral luminosity by photon energy and substituting $x=h\nu /k{T}_s$ proves the formula. The convergent series $F(y)={\sum }_{m\ge 1}{e}^{-my}(y^2/m+2y/m^2+2/m^3)$ follows by integrating the geometric-series expansion termwise.

For two same-height throats carrying the same light-layer volume flux and reduced gravity, the narrower throat has the larger critical minimum energy. A narrower downstream throat can back up and submerge an upstream throat. If the first throat is narrower, it can control discharge and feed a supercritical current through the second; a hydraulic jump followed by a second control is also possible when the available energy drop and intermediate geometry permit it. Equal throat widths are a marginal lossless case. Relative width alone does not fix jump position or mixing in the intervening reservoir.

For independent real sequences, put the faster sequence at the slower speed by slower-speed degeneration of a large deviation principle, then use the product large-deviation principle and contraction principle for large deviations for addition. Unequal speeds give a translate of the slower sequence's rate. Equal speeds give the infimal convolution of the two rates. Its unique zero is the sum of their zero points, and a translated positive finite point has positive finite cost, since compactness attains each finite fiber infimum. Thus the defining nondegeneracy is preserved.

// Target: queueing-theory.bigb

If $X$ is simply connected and $h_{\ast }:{\pi }_{\ast }(X)\otimes \mathbb{Q}\to {\tilde{H}}_{\ast }(X;\mathbb{Q})$ is onto, choose a homogeneous homology basis and representing sphere maps into $X_{\mathbb{Q}}$. Their wedge sum gives a rational homology equivalence, hence a rational homotopy equivalence by the rational Whitehead theorem. All positive cup products therefore vanish. In a minimal model, the induced map from cohomology to the indecomposable quotient is injective; replacing a set of generators by cocycle representatives produces a quasi-isomorphism to the square-zero cohomology algebra, proving formality of a topological space.

Let $f_0(t)$ be the initial instantaneous forward rate curve and choose $r_0=f_0(0)$. In the Hull-White model, take $\beta (t)=f_0^{\prime }(t)+a{f}_0(t)+{\eta }^2(1-e^{-2at})/(2a)$. To verify the fit, write $r_t=x_t+\varphi (t)$ with $d{x}_t=-a{x}_{t}dt+\eta d{W}_t^Q$, $x_0=0$, and $\varphi (t)=f_0(t)+{\eta }^2(1-e^{-at}{)}^2/(2a^2)$. The Gaussian exponential formula gives $-{\partial }_T\log P(0,T)=\varphi (T)-{\eta }^2(1-e^{-aT}{)}^2/(2a^2)=f_0(T)$.

At equally spaced distinct times, a mean profile is affine exactly when its consecutive second differences vanish. These form $p-2$ independent contrasts. At unequal times, compare adjacent slopes instead: $({\mu }_{i+1}-{\mu }_i)/(t_{i+1}-t_i)-({\mu }_i-{\mu }_{i-1})/(t_i-t_{i-1})$. A Hotelling test of linear hypotheses tests these restrictions jointly with unknown covariance.

In a local trivialization, a vector-bundle connection with matrix $A$ defines the displayed complement of the vertical tangent space. The frame-change law makes these complements independent of the trivialization. Their dependence on the fiber coordinate is linear; an arbitrary horizontal complement need not satisfy this condition and need not define a connection on a vector bundle. Horizontal lifts satisfy the ordinary differential equation defining parallel transport.

The holonomy of a connection along a closed curve is the group element relating the endpoints of its horizontal lift. A flat connection can have nontrivial holonomy around a noncontractible loop.

For $M$ binary units and desired patterns ${\xi }^{\mu }$, use these symmetric weights and set $w_{ii}=0$. For one pattern, the field at that pattern is $(M-1){\xi }_i/M$, so it is stable. Several patterns add cross-talk terms; storage by this rule alone does not guarantee stability of every arbitrary collection of patterns.

For symmetric zero-diagonal weights and zero thresholds, update one binary unit to the sign of its current field, retaining it at a zero field. The displayed energy change is negative on every actual flip. There are finitely many states, so a fair asynchronous update schedule eventually reaches a fixed point that is a minimum against single-unit flips. Synchronous updates can instead cycle, and local minima need not be global minima.

The leg consists of the cells strictly below $(i,j)$ in its column of the Young diagram; its length is $l_{\lambda }(i,j)={\lambda }_j^{\prime }-i$, where ${\lambda }^{\prime }$ is the conjugate partition. It excludes the anchor cell. At $(1,2)$ in $(3,2)$ the leg contains only $(2,2)$, so its length is one. Transposing the Young diagram exchanges the leg and the arm of a Young-diagram cell.

The arm consists of the cells strictly to the right of $(i,j)$ in its row of the Young diagram; its length is $a_{\lambda }(i,j)={\lambda }_i-j$. It excludes the anchor cell. At $(1,2)$ in $(3,2)$ the arm contains only $(1,3)$, so its length is one. Together with the leg of a Young-diagram cell and the anchor, it forms the hook of a Young diagram.

Take independent ensembles for two quantum channels and their Cartesian product ensemble. The Von Neumann entropy of a tensor product is the sum of the two entropies, both for its average output and for each labelled output. Thus its Holevo quantity is the sum of the two individual quantities. Taking ensembles arbitrarily close to each supremum proves the displayed inequality without requiring maximizing ensembles to exist. It does not assert additivity for arbitrary channels.

For $dX=dB+a{X}^{-1}dt$, put $\gamma =1-2a\in (0,1)$ and $v_R(x)=(R^{2-\gamma }{x}^{\gamma }-x^2)/(1+2a)$. It is nonnegative on $[0,R]$, vanishes at both endpoints and solves $\frac{1}{2}{v}_R^{\prime \prime }+(a/x)v_R^{\prime }=-1$ on $(0,R)$. Optional stopping at $T_{\epsilon }\wedge T_R$ gives $\mathbb{E}(T_{\epsilon }\wedge T_R)\le v_R(x)$ uniformly as $\epsilon \downarrow 0$. Thus the limiting exit to zero or $R$ occurs in finite time, not merely as an asymptotic approach to zero. Scale probabilities give ${\mathbb{P}}_x(T_0<T_R)=1-(x/R{)}^{\gamma }$; letting $R\to \infty$ gives almost sure finite hitting of zero.

Choose $\chi (z)=\sqrt{z^2-a^2}\sim z$ at infinity, cut on the finite crack. Its upper crack value is $i\sqrt{a^2-t^2}$, so $H=\chi G$ has jump $-2i\sqrt{a^2-t^2}p(t)$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral solution. It decays as $z^{-2}$ for integrable weighted traction. A possible homogeneous term $C/\chi$ would give a logarithmic displacement period; it is removed for a dislocation-free finite crack with no extra remote loading.

For two semi-infinite cracks separated by a ligament, choose $\chi (z)=\sqrt{a^2-z^2}>0$ on the ligament. Its upper crack values are $-i\sqrt{t^2-a^2}$ on the right and $+i\sqrt{t^2-a^2}$ on the left. Multiplication by $\chi$ converts the Hilbert problem for an antiplane crack to an additive jump; the Sokhotski–Plemelj theorem gives the displayed solution, assuming convergence. A homogeneous term $C/\chi$ transfers an extra resultant $\pi C$ through the ligament and has logarithmic displacement at infinity. The usual correction with no additional remote resultant excludes it. The far-field normalization is essential to uniqueness.

For the moving semi-infinite crack, take the branch cut of $\sqrt{z}$ on the negative real axis and $\sqrt{x}>0$ ahead of the tip. With steadily moving antiplane shear, face tractions ${\sigma }_{23}^{\pm }=-f(x)$ imply $\mathrm{Im}{F}^{\prime \pm }=f/\mu$. The antisymmetric displacement response gives $F^{\prime -}=-\overline{F^{\prime +}}$. Hence $G=\sqrt{z}{F}^{\prime }$ has jump $G^{+}-G^{-}=-2\sqrt{-x}f(x)/\mu$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral, assuming convergence and no extra remote homogeneous loading. For $x>0$ it is purely imaginary, so ${\sigma }_{23}={\int }_{-\infty }^0\sqrt{-s}f(s)/(x-s)ds/(\pi \sqrt{x})$. At fixed co-moving loading this ahead-of-tip stress is independent of speed; the displacement still has the factor $1/\beta$.

Under $\mathrm{diag}(t,t^{-1})$, acting by inverse substitution, $X^{d-j}{Y}^j$ has weight $2j-d$. If $r$ is the smallest active exponent of $Y$, the multiplicity at $[1:0]$ is $r$ and the Hilbert-Mumford value is $d-2r$. Conjugating replaces this point by any projective point. Thus strict positivity is exactly the displayed condition; replacing the strict inequality by $m_p\le d/2$ gives semistability. A binary quartic is stable precisely when it has no repeated projective root.