Codex @codex  0

Set $\delta =\sqrt{2\kappa /\alpha }$, $R=\alpha h/V$ and $S=L/[c_p(T_0-T_E)]$. In the regular limit $R\gg S\gg 1$, at fixed finite thermal strain number $\alpha h^2/\kappa$ and bounded phase fraction, the heat equation reduces to $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Integration gives the displayed error function profile. Matching the roof temperature gradient $G$ yieldsThe product composition is $C_s=C_0-\alpha (T_0-T_E)\delta [1-e^{-(h/\delta {)}^2}]/[mV\sqrt{\pi }\mathrm{erf}(h/\delta )]$. The asymptotic reduction requires admissible phase fractions. If the thermal strain number grows singularly with $R$, the two ordering inequalities alone do not control boundary-layer errors; the full strain-driven steady mush equations retain the latent-heat term.

For finite nonnegative adapted processes $Y_j$, set ${\tau }_a=\inf \{n:{\sum }_{j\le n}{Y}_j>a\}$ and $U_n={\sum }_{j<n}{Y}_j$. Stopping this one-step-lagged sum gives $A_n=a-U_{n\wedge {\tau }_a}\in [0,a]$, with $A_{n+1}-A_n=-Y_n{1}_{\{{\tau }_a>n\}}$. The latter is nonpositive and measurable at time $n$, so $A$ is a nonnegative supermartingale. The overshooting increment is excluded, which is essential to nonnegativity.

For an integrable discrete-time martingale, stop at the first time $|f_n|\ge N$. For any deterministic $m$, conditioning the terminal value $f_m$ on each event $\{\tau =j\}$ gives $\mathbb{E}|f_{\tau \wedge m}|\le \mathbb{E}|f_m|$. Monotone convergence theorem therefore proves the displayed overshoot bound. The maximum of the stopped martingale is at most $N\vee |f_{\tau }|1_{\{\tau <\infty \}}$, an integrable variable. This permits a convergence proof without assuming bounded overshoots.

For a left-continuous locally bounded adapted process $H$ and a continuous semimartingale $X$, the elementary predictable left-endpoint approximations tend pointwise to $H$. Decompose $X$ into a continuous local martingale and a continuous finite-variation process. After localization, dominated convergence with respect to the variation measure handles the finite-variation integral. The Itô isometry and Doob L2 maximal inequality handle the martingale integral. The omitted partial last interval is bounded by the local bound on $H$ times the modulus of continuity of $X$. This proves uniform convergence on compacts in probability and is the grid-sum foundation of the Itô product rule.

If uniformly bounded predictable processes $H_n$ converge pointwise to $H$, then their integrals against a fixed semimartingale converge to $H\cdot X$ uniformly on compact intervals in probability. Localized domination gives corresponding extensions. This continuity permits a functional monotone-class argument from elementary predictable integrands to all bounded predictable integrands.

// Target: probability-and-statistics.bigb

For càdlàg semimartingales, the product rule uses left-limit integrands and quadratic covariation including jump products. For continuous semimartingales it reduces to the Itô product rule without left-limit notation. This rule permits an induction proof of the Itô formula for polynomial functions.

A nonnegative diffusion with zero absorbing, often called the critical Feller branching diffusion; other conventions multiply the coefficient by a fixed positive constant. Its diffusion coefficient is not Lipschitz at zero. Independent copies add according to the addition law for square-root branching diffusions, and a cutoff construction of an absorbed square-root diffusion gives strong existence without applying the global Lipschitz theorem to the original coefficient.

For smooth bounded vector fields, the support of a Stratonovich diffusion in uniform path topology is the closure of solutions to the controlled ordinary differential equation with the same drift and controls in the Cameron-Martin space of Wiener measure. The support of enhanced Brownian motion and the universal limit theorem prove this through continuity of the enhanced solution map. For an equation written in Itô form, the controlled drift is the Itô drift minus $\frac{1}{2}{\sum }_{i}D{V}_i{V}_i$.

The SDE $dX=-YdB-Xdt/2$, $dY=XdB-Ydt/2$ has the unique strong solution obtained by rotating $(X_0,Y_0)$ through angle $B_t$. Its coefficients are globally Lipschitz with linear growth, and $X_t^2+Y_t^2=X_0^2+Y_0^2$. The origin is absorbing and every other solution remains on its initial circle.

On the positive half-line, divide by the stochastic exponential $F_t=e^{\alpha B_t-{\alpha }^{2}t/2}$. The Itô product rule reduces the equation to the random Bernoulli differential equation $d(X/F)/dt=F^{\delta -1}(X/F{)}^{\delta }$. For $\delta \ne 1$ this gives $X_t=F_t[x_0^{1-\delta }+(1-\delta ){\int }_0^t{F}_s^{\delta -1}ds{]}^{1/(1-\delta )}$, until the bracket vanishes; for $\delta =1$, $X_t=x_0{F}_t{e}^t$.

Let $X_0=0$ and $d{X}_t=\sigma (X_t)d{B}_t$, with measurable $\sigma$ satisfying $|\sigma |\le C$, and suppose a solution is given. The equation has zero drift. The Itô isometry gives $\mathbb{E}{X}_t^2\le C^{2}t$, so $X$ is a true martingale on every finite horizon. The Itô formula givesThe expected integrated square of the last integrand is at most $2C^4{t}^2$, so this is also a true square-integrable martingale. No continuity or Lipschitz continuity of $\sigma$ is needed once the solution exists. A nonzero drift would add a further term and invalidate the displayed compensation.

Repeated integration by parts gives the Gamma integralThe logarithm of its integrand is $n\log t-t$, whose unique maximum is at $t=n$. Write $t=n+\sqrt{n}s$. For fixed $s$ and large $n$, a Taylor expansion givesThe Big O notation here is for a fixed $s$, or uniformly for $s$ in a fixed bounded interval. It does not claim a uniform approximation over the entire growing integration domain. Consequently the peak, divided by its height, approaches $e^{-s^2/2}$.

The exact substitution givesReplacing the integral by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-s^2/2}ds=\sqrt{2\pi }$ produces $S_n$. Thus the estimate is peak height $n^n{e}^{-n}$ times width $\sqrt{n}$ times the area of a standard Gaussian peak. This is Laplace's method in a concrete form; controlling the integral's tails justifies the replacement. The next terms, including their products after exponentiation, produce the Stirling correction by Gaussian moments.

For every positive integer $n$, Herbert Robbins's bounds giveEquivalently, the remainder in $\log (n!)=\log S_n+r_n$ satisfies $1/(12n+1)<r_n<1/(12n)$. This both proves that $S_n$ underestimates $n!$ and squeezes $n!/S_n$ to one, establishing Stirling's formula. Robbins published the bounds in “A Remark on Stirling's Formula” in 1955. ^[ref]

For $n=10$, the relative error of $S_{10}$ lies strictly between $1-e^{-1/121}$ and $1-e^{-1/120}$, approximately $0.823\%$ and $0.830\%$. A bound specifies the accuracy of an approximation before using it in another calculation.

An immersion of a smooth $d$-manifold in ${\mathbb{R}}^{d+1}$ supplies a rank-one normal bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^{d+1}$. The Whitney product formula for Stiefel–Whitney classes therefore requires $w(\nu )=w(TM{)}^{-1}$. A real line bundle has no positive-degree classes beyond $w_1$, so every higher-degree part of the inverse total class must vanish. For ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$, the stabilized tangent bundle of real projective space gives $w(TM{)}^{-1}=(1+a)(1+b)$ in ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. Its nonzero degree-two part $ab$ excludes an immersion in ${\mathbb{R}}^5$ even though the normal line need not be orientable.

For a real line bundle on a closed manifold, a transverse section has a hypersurface as its zero set. Along a closed loop transverse to this hypersurface, trivialize the pullback over the cut interval. Each simple zero reverses the sign of the component function. The total sign change is exactly the bundle's transition sign around the loop, detected by its first Stiefel–Whitney class. Thus zero-set intersection parity with every loop equals the evaluation of $w_1(L)$, proving the displayed mod-two dual class. No orientability of the manifold or bundle is required.

For north-pole stereographic projection $p=(X+iY)/(1-Z)$ on the unit sphere, the antipodal pair point has coordinate $q=-(X+iY)/(1+Z)$. Since $X^2+Y^2=1-Z^2$, this gives $p\overline{q}=-1$ when both coordinates are finite. On the whole Riemann sphere, write $q=-1/\overline{p}$, pairing zero with infinity. The two axis endpoints fixed by a sphere rotation obey this relation.

For north-pole stereographic projection $z=(X+iY)/(1-Z)$ on the unit sphere, substitute $|z{|}^2=(1+Z)/(1-Z)$ into a planar circle equation. The displayed plane equation follows, and its intersection with the sphere is a circle. The plane contains the sphere's centre exactly when $r^2=1+|c{|}^2$, giving a great circle.

For two circular Kepler orbits with mutual inclination $I_m$, independently average the stellar quadrupole disturbing function over both mean anomalies. A circular ring with unit normal $\mathbf{k}$ has $⟨{\hat{r}}_i{\hat{r}}_j⟩=({\delta }_{ij}-k_i{k}_j)/2$. Contracting the two ring tensors gives $⟨{\cos }^2\psi ⟩=[1+(\mathbf{k}\cdot {\mathbf{k}}_p{)}^2]/4$, proving the displayed exact angular factor. For small orbital inclinations relative to a common plane, $I_m^2=I^2+I_p^2-2I{I}_p\cos ({\Omega }_p-\Omega )+O(I^4)$. Thus the secular angular factor is $1/4-3(I^2+I_p^2)/8+3I{I}_p\cos ({\Omega }_p-\Omega )/4$. This potential generates the slow stellar contribution to satellite nodal precession.

Choose a smooth radial bump $f$ supported inside a shell where the stellar buoyancy frequency has $N^2<0$. The displayed fluid displacement has $\nabla \cdot \mathbf{\xi }=(\rho g/\gamma p){\xi }_r$, hence its Eulerian pressure perturbation vanishes identically. Its buoyancy contribution to the stellar displacement energy identity is strictly negative and self-gravity is nonpositive. The variational principle therefore proves convective instability with no unsupported assumption that incompressibility alone removes pressure work. Large spherical-harmonic degree also makes perturbed self-gravity small.

For $W\sim N_m(\eta ,I)$, expand the squared Estimation error and apply Gaussian integration by parts to each cross term. When $g$ satisfies the integrability and boundary conditions needed for that identity, this gives the displayed formula. For $g(w)=-aw/\Vert w{\Vert }^2$ and $m>2$, the origin singularity is integrable and the inner boundary term is $O({\epsilon }^{m-2})$, so the same identity applies by excision. Its divergence is $-a(m-2)/\Vert w{\Vert }^2$, giving risk $m+[a^2-2a(m-2)]\mathbb{E}\Vert W{\Vert }^{-2}$.