For smooth positive $f$, let $L=f^{-1}\nabla \cdot (f\nabla )+k^2$. Its unweighted bilinear transpose is the displayed operator. If $v$ satisfies $\alpha v+\beta {\partial }_{n}v=0$, its transpose boundary condition is $\alpha u+\beta ({\partial }_{n}u-u{\partial }_n\log f)=0$. The resulting Green kernels satisfy $G_L({\mathbf{r}}_2,{\mathbf{r}}_1)=G_{L^{\mathrm{T}}}({\mathbf{r}}_1,{\mathbf{r}}_2)$. Equivalently $G_{L^{\mathrm{T}}}(\mathbf{r},{\mathbf{r}}_2)=f(\mathbf{r})G_L(\mathbf{r},{\mathbf{r}}_2)/f({\mathbf{r}}_2)$, exhibiting weighted acoustic Green-function reciprocity.

The formal adjoint of an infinite matrix does not specify the domain of the true adjoint. For the weighted difference generator $(Qf{)}_n=4^n(f_{n-1}/2-f_n)$, its formal positive-eigenvalue solution is $h_n=2^{-n}{h}_0{\prod }_{k<n}(1+\alpha 4^{-k})$. This is square summable but does not belong to the true adjoint domain of the maximal operator: testing against $f_n=2^{-n}$ contradicts the adjoint identity. The nonzero limiting boundary term is the obstruction.

// Target: analysis.bigb

With positive stable buoyancy destruction $B$ and positive dissipation $ϵ$, this normalized sink fraction need not equal the Flux Richardson number $B/P$ away from equilibrium. If $R=(B+ϵ)/P$, then $J={\mathrm{Ri}}_f/R$. In the fixed-correlation equilibrium model for stratified turbulence, a minimum of $R=a/q+bq$ has $B=ϵ$, hence $J=1/2$. The actual flux Richardson number there is $R_{\min }/2$ and equals $1/2$ only when that minimum also satisfies the equilibrium condition $R=1$.

At zero frequency the graded-layer transfer equation integrates to $\exp [L\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)]$, where $L=\frac{1}{2}\log (Z_b/Z_a)$. Eliminating the outgoing wave amplitude at the left endpoint gives the displayed incoming-to-outgoing scattering matrix. It is real orthogonal, so it preserves total incoming and outgoing flux. It coincides with continuity of velocity and traction at a sharp seismic impedance interface; the profile between the endpoints enters at nonzero frequency.

If the rate functions are bounded and Lipschitz on an invariant compact state space, the Poisson time-change representation of a Markov chain has drift $F={\sum }_j{v}_j{\beta }_j$. Each centered Poisson term is uniformly small after division by $n$. The Gronwall inequality then bounds the distance to $\dot{z}=F(z)$ by the initial error and the centered counting error, multiplied by $e^{LT}$. A Poisson maximal concentration bound gives exponentially decreasing error probabilities for a fixed horizon and fixed error tolerance.

For a density-dependent Markov jump process with finitely many jump vectors, bounded nonnegative rates and Lipschitz drift $F(x)={\sum }_{\ell }\ell {\beta }_{\ell }(x)$, an initial state converging in probability to $x_0$ gives uniform convergence in probability on every finite horizon to the ODE $\dot{x}=F(x)$, $x(0)=x_0$. The compensated martingale has expected maximal squared size $O(N^{-1})$ by the Doob L2 maximal inequality; Gronwall's inequality transfers that estimate to the solution error.

// Target: probability-and-statistics.bigb

For bending stiffness $D$ and mass per area $m$, a plane wave of surface displacement $\eta$ satisfies $m{\eta }_{tt}+D{\nabla }_H^4\eta =p$. An inviscid deep-water velocity potential has vertical dependence $e^{kz}$; the kinematic boundary condition and linear Bernoulli equation give $p=({\rho }_w{\omega }^2/k-{\rho }_{w}g)\eta$. Eliminating $p$ gives the displayed dispersion relation. With water depth $d$, replace it by ${\omega }^2=(gk+D{k}^5/{\rho }_w)\tanh (kd)/[1+(m/{\rho }_w)k\tanh (kd)]$.

Write $\beta =D/{\rho }_w$ and $\alpha =m/{\rho }_w$ in the dispersion relation of a floating elastic plate. A stationary point of squared phase velocity satisfies $\beta k^4(3+2\alpha k)=g(1+2\alpha k)$. If plate inertia is neglected, its solution is $k=(g/(3\beta ){)}^{1/4}$. A steadily translating load can resonate only when its speed matches a phase velocity; this threshold differs from the group-velocity minimum of a flexural-gravity wave.

The average of the two flat translation connections on a Lie group is a torsion-free connection. For left-invariant vector fields, the Jacobi identity gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$ with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. Contracting gives $\mathrm{Ric}(X,Y)=-\frac{1}{4}K(X,Y)$ for the Killing form $K$. It is flat exactly when all double Lie brackets vanish, including the case of a two-step nilpotent group. For a semisimple Lie algebra, $K$ is nondegenerate and supplies a bi-invariant indefinite metric whose Levi-Civita connection is ${\nabla }^0$.

Declare a global frame of left-invariant vector fields parallel on a semisimple Lie group. This affine connection is flat and preserves a left-invariant metric tensor because its frame components are constant. If $[e_a,e_c]=C_a{}^b{}_c{e}_b$, its torsion tensor is $T_a{}^b{}_c=-C_a{}^b{}_c$. For fixed $c$, the matrix with entries $T_a{}^b{}_c$ is $\mathrm{ad}{e}_c$. Its trace vanishes because a semisimple Lie algebra is perfect and adjoints of brackets are commutators. The trace of the product for $c,d$ is the Killing form $K_{cd}$, proving both contractions. This flat metric connection differs from the curved torsion-free Levi-Civita connection of the Killing metric.

Let a bounded probability density function have Fourier transform supported in the plateau of a flat-top kernel for density estimation. At fixed smoothing bandwidth one, the convolution theorem gives $K\ast f=f$, so the kernel density estimator is pointwise unbiased. Its variance is at most $\Vert f{\Vert }_{\infty }\Vert K{\Vert }_2^2/n$, and the Cauchy-Schwarz inequality proves the displayed rate. An explicit example is $f(x)=(1-\cos x)/(\pi x^2)$, with value $1/(2\pi )$ at zero and transform $(1-|t|{)}_{+}$. Direct inverse Fourier integration gives the formula. The function is nonnegative, integrable and has integral one because its transform at zero is one. Compact frequency support also gives bounded derivatives of every order.

Every map from a flat sublocale $A_j$ of $A$ to a compact locale that is also a regular locale extends uniquely to $A$. On frames the extension iswhere the terms are first regarded as fixed elements in the ambient frame. Regularity proves agreement after applying $j$; compactness reduces arbitrary covers to finite subcovers on a well-inside open, and flatness identifies those finite joins with ambient joins. These facts prove that $g^{\ast }$ preserves arbitrary joins and finite meets.

For a Noetherian ring $A$, an injective module $I$ gives a flasque sheaf on $\mathrm{Spec}A$. If $U=\mathrm{Spec}A\setminus V(\mathfrak{a})$, clearing denominators on finitely many principal opens identifies $\Gamma (U,\tilde{I})$ with ${\underrightarrow{\lim }}_n{\mathrm{Hom}}_A({\mathfrak{a}}^n,I)$. Each map extends to $A$ by injectivity. Thus every section on $U$ extends globally, and all restriction maps are surjective. This concerns sheafification of an injective module, rather than the different assertion that an injective sheaf of modules is flasque.

With known size $\theta >0$, the mean-dependent log mass of a negative binomial distribution is $y\log \mu -(y+\theta )\log (\mu +\theta )$. Its derivative is $\theta (y-\mu )/[\mu (\mu +\theta )]$, so the saturated mean is $y$, including its boundary value zero. Subtracting the fitted log mass from that maximum gives the displayed deviance. This formula needs no fitted-intercept cancellation. As $\theta \to \infty$, its second term tends to $-2(y_i-e_i)$ and it becomes the full Poisson deviance.

With $A_f=\lambda x.f(xx)$, the combinator $Y=\lambda f.A_f{A}_f$ satisfies $Yf{\equiv }_{\beta }f(Yf)$. Every finite reduct of $Y$ is, up to alpha equivalence, $\lambda f.f^n(A_f{A}_f)$: the remaining copy of $A_f{A}_f$ is its only beta-redex.

Put $D=\lambda xf.f(xxf)$ and $\Theta =DD$. Then $\Theta f{\to }_{\beta }^{\ast }f(\Theta f)$. Unlike the Curry fixed-point combinator, every finite reduct retains a closed subterm $DD$. Contracting this subterm recreates one; every other generated redex substitutes a variable and preserves it. The Church-Rosser theorem therefore proves $Y/{\equiv }_{\beta }\Theta$.

Take the original sphere on the tube axis, with diagonal positive hydrodynamic resistance matrix $R_0$ and axial force $W>0$. Add a freely moving neutral sphere and extend its rigid velocity through its interior. The resulting field is admissible in the original domain, so its dissipation obeys $D>q^T{R}_{0}q\ge R_{zz}{V}_z^2$ for any finite nonoverlapping placement of the extra sphere. Since only the first sphere does external work, $D=W{V}_z$, giving the displayed bound without assuming zero transverse velocity or rotation. Strictness follows because equality would require the original nontrivial analytic flow to have zero strain throughout the added ball.

For $x>1$, moving the Mellin contour past the pole at one, the Nontrivial zeros of the Riemann zeta function, and the remaining poles givesZeros are included with multiplicity. The Riemann–von Mangoldt formula makes the zero sum absolutely convergent. If every nontrivial zero has real part below one, division by $x^2$ and dominated convergence give ${\Psi }_1(x)\sim x^2/2$. Monotonicity of $\psi$ then recovers $\psi (x)\sim x$ by upper and lower difference quotients.

Equip the vector space $c_{00}$ of finitely supported sequences with the norm inherited from the absolutely summable sequence space. Consider the truncationsFor $M>N$, the geometric series givesTherefore $(x^{(N)})$ is a Cauchy sequence. Any possible norm limit must have coordinate $k$ equal to $2^{-k}$, since $|x_k-y_k|\le \Vert x-y{\Vert }_1$. That forces infinitely many nonzero coordinates, so no limit lies in $c_{00}$. There is nothing wrong with the approximations: the space is missing their limit.

Inside the absolutely summable sequence space, the limit is $x=(2^{-k}{)}_{k\ge 1}$ and its exact truncation error is $2^{-N}$. Four retained coordinates leave error $1/16$; ten leave $1/1024$. More generally, truncating any element of the absolutely summable sequence space leaves a tail whose norm tends to zero. Together with completeness, this identifies that Banach space as the completion of a normed space $c_{00}$ with this norm. Finite storage and an infinite mathematical object fit together through a controlled error.

Normalization of the reverse law gives an exact finite identity:The sum notation describes discrete paths; the corresponding path integral has the same change-of-variable argument. Jensen's inequality for the exponential function then gives $e^{-⟨{\Sigma }_F{⟩}_F}\le 1$, hence $⟨{\Sigma }_F{⟩}_F\ge 0$. Equivalently this mean is the Kullback-Leibler divergence between the forward path law and the reverse path law pulled back by $\Theta$. Equality requires those two laws to agree. ^[ref]

Rare negative entropy production contributes disproportionately to this expectation through $e^{-{\Sigma }_F}$. Missing such events in a finite sample can spoil an estimate of the identity. Matching probability support is also essential: reverse paths absent from the forward law cannot be recovered by cancelling forward weights. This Second law of thermodynamics statement concerns an ensemble average; an individual negative value is compatible with it.