The curvature-minus-one hyperbolic metric on the punctured disk assigns distance $2\log n$ along the radial geodesic from $e^{-1}$ to $e^{-n^2}$. Its exponential distances therefore sum to a finite value; the triangle inequality gives the same conclusion for any fixed basepoint. Nevertheless no nonzero bounded holomorphic function has these zeros: the Riemann removable singularity theorem extends it at zero, and the identity theorem then makes it identically zero. Simple connectivity is necessary in the hyperbolic characterization of bounded analytic zero sets.

For points $ib$ and $a+ib$ with $a,b>0$, the hyperbolic distance is $2\mathrm{arsinh}(a/(2b))$. It equals their Euclidean distance $a$ exactly when the displayed relation holds. The function of $a$ decreases continuously from one to zero: its derivative has the sign of $\sinh (a/2)-(a/2)\cosh (a/2)<0$. Hence every height $b\in (0,1)$ has exactly one positive matching horizontal separation.

For a proper arc $I$ of the boundary of the unit disc, $Q(I)$ is the side of the hyperbolic geodesic between its endpoints adjacent to $I$. For short arcs this is a curved boundary cap. For arcs longer than a semicircle it contains the center. With endpoints $e^{\pm i\alpha }$ and $0<\alpha <\pi /2$, the geodesic is the circle with center $\sec \alpha$ and radius $\tan \alpha$, orthogonal to the unit circle. It is a diameter at $\alpha =\pi /2$.

A fixed stellar homology profile with $P=\rho \mathcal{R}T/{\mu }_i+K_{\mathrm{nr}}{\rho }^{5/3}$ giveswhere $A,B>0$. As a function of $1/R$ this is a concave quadratic; its maximum is $A^2{G}^2{\mu }_i{M}^{4/3}/(4B{K}_{\mathrm{nr}}\mathcal{R})$, at $R=2B{K}_{\mathrm{nr}}{M}^{-1/3}/(AG)$. At the maximum, the central electron degeneracy pressure and thermal pressure are equal. If this temperature is insufficient for sustained hydrogen burning, the object becomes a brown dwarf. Only the positive-temperature segment represents a physical sequence.

Consider two widely separated rigid spheres of equal radius $a$, with separation $R\mathbf{n}$, in Stokes flow. Apply a force $6\pi \mu a{\mathbf{U}}_0$ to the first sphere and let the second be force-free and torque-free. The second sphere samples the rate-of-strain tensor of the first sphere's Stokeslet, $\mathbf{E}=3a({\mathbf{U}}_0\cdot \mathbf{n})(\mathbf{I}-3\mathbf{nn})/(4R^2)$. Its leading disturbance is the stresslet of a sphere in a uniform straining Stokes flow. Evaluating that disturbance back at the first sphere givesThe hydrodynamic mobility matrix decreases at fixed force, consistently with the increase of resistance caused by an additional rigid inclusion.

For the nonzero spectral decomposition $\rho ={\sum }_{a=1}^r{\lambda }_a|e_a⟩⟨e_a|$, every pure-state quantum state ensemble of $\rho$ is specified by the displayed column linear isometry. To prove this, write $w_j=\sqrt{p_j}{\varphi }_j$. Positivity forces each $w_j$ into the support of $\rho$, and the coefficients $U_{ja}=⟨e_a|w_j⟩/\sqrt{{\lambda }_a}$ have orthonormal columns because ${\sum }_j|w_j⟩⟨w_j|=\rho$. Conversely this column identity reconstructs $\rho$; the squared vector norms are the probabilities and sum to one. Complete the columns to unitary bases after adding zero ensemble vectors to obtain the usual unitary-mixing formulation.

Replace sibling leaves in a code tree labelled $a,b$ by their common parent in a rooted tree, now a leaf labelled $m$ of weight $w_m=w_a+w_b$. All other codewords stay fixed; each of $a,b$ loses one bit. For any such code tree $T$ and the contracted code tree $T^{\prime }$, their weighted costs satisfyIf $a,b$ have two least weights, the Huffman sibling property gives an optimal prefix code containing this pair. Its contracted prefix code must be optimal for the reduced weights $w^{\prime }$: any cheaper reduced prefix code could be expanded at $m$ by appending $0,1$, giving a cheaper original prefix code. Conversely, expanding any optimal reduced prefix code adds precisely $w_a+w_b$ and attains the original minimum. This is optimal substructure, the justification for repeating the merge in Huffman coding. ^[ref]

In the exchange diagram, $C,D$ contract to weight $0.3$ at leaf depth in a code tree $2$. The reduced expected codeword length is $0.4\cdot 1+0.3\cdot 2+0.3\cdot 2=1.6$ bits; expanding restores $1.6+0.3=1.9$ bits. For $n=2$, the reduction leaves one symbol: allow its temporary codeword to be empty, of cost zero, so expansion gives the two one-bit codewords. This convention is only the terminal reduced problem; the original source has at least two symbols.

Two leaves of a code tree with the same parent in a rooted tree. Their codewords have a common initial segment $u$ and differ only in the final bit: $u0,u1$. Their leaf depth in a code tree is equal. The Huffman sibling property places two chosen least weights at such leaves in some optimal prefix code.

For $m\ge 2$ equiprobable symbols, the maximum codeword length of a binary Huffman code is $s=\lceil {\log }_{2}m\rceil$. Exactly $2^s-m$ codewords have length $s-1$ and $2m-2^s$ have length $s$. The balancing exchange for a uniform optimal prefix code restricts the depths to these two levels; counting leaves and using equality in the Kraft inequality determine the numbers.

// Target: probability-and-statistics.bigb

An orthogonal similarity preserves skew symmetry, so upper-Hessenberg reduction of a real skew-symmetric matrix is automatically tridiagonal with zero diagonal. For a Householder reflection, the scalar $v^{T}Av$ vanishes, eliminating the fourth term of the general similarity update. Storing only one triangle and using the displayed skew rank-two update costs $2n^3/3+O(n^2)$ scalar products, the same leading product count as optimized symmetric tridiagonalization; the simplification saves lower-order work.

The displayed pointwise control limits follow from the normal approximation to an observed binomial proportion. Clip plotted limits to the probability range. Their nominal coverage is approximate and can be poor for small expected counts; exact binomial tails give a discrete alternative. Pointwise coverage does not provide simultaneous coverage of all hospitals; use Bonferroni correction or another familywise error rate procedure when that is the required guarantee.

If $K$ obeys the Hörmander integral kernel condition with constant $C$, and an integrable $b$ is supported in a cube $Q$ with $\int b=0$, then ${\int }_{(LQ{)}^c}|K\ast b|\le C\Vert b{\Vert }_1$ for $L=4\sqrt{d}$. Here $LQ$ is the concentric cube with multiplied side length. Subtract the kernel value at the cube center inside the convolution; outside $LQ$, every source displacement is smaller than half the observation displacement. Tonelli theorem and the integral kernel condition give the bound.

Suppose a convolution operator has $L^2$ operator norm at most $A$, and its kernel satisfies the Hörmander integral kernel condition with constant $C$. Under hypotheses making its integrals well-defined, the Calderón–Zygmund decomposition gives weak type (1,1). Indeed, the bounded part has squared $L^2$ norm at most $2^d\lambda \Vert f{\Vert }_1$. The bad cubes occupy at most $\Vert f{\Vert }_1/\lambda$, while outside their $4\sqrt{d}$ dilations the cancellation estimate outside a dilated cube bounds the bad image in $L^1$. Applying distributional estimates to these three regions gives $|\{|Tf|>\lambda \}|\le ((4\sqrt{d}{)}^d+2^{d+2}{A}^2+4C)\Vert f{\Vert }_1/\lambda$.

A measurable convolution kernel satisfying the displayed size and cancellation properties together with the Hörmander integral kernel condition defines uniformly bounded Fourier multipliers after deletion of a ball at the origin. The deleted-ball kernels belong to $L^2$, although they need not belong to $L^1$ at infinity. Cancellation controls low-frequency integration near the deleted ball; translation by half an oscillation controls the far tail through the oscillatory tail estimate for a singular kernel.

For downward depth $z=ϵy$, take $(u,w)=({\psi }_z,-{\psi }_x)$, surface temperature $T=x$, a flat impermeable stress-free upper boundary and a no-slip insulated bottom. In the steady infinite-Prandtl number thin-layer interior, ${\psi }_{zzzz}=-\mathrm{Ra}$ with $\psi ={\psi }_{zz}=0$ at the surface and $\psi ={\psi }_z=0$ at the bottom. The displayed profile follows. Sidewall end layers close the circulation; the interior profile must not be imposed unchanged at those walls.

In a right Artinian ring, the powers of its Jacobson radical stabilize at an ideal $I$ satisfying $I^2=I$. If $I\ne 0$, choose a minimal right ideal $K$ with $KI\ne 0$. Since $(KI)I=KI$, minimality gives $KI=K$. Choosing $k\in K$ with $kI\ne 0$ gives $K=kR$ by the same minimality. Consequently $k=kj$ for some $j\in J(R)$. The unit criterion for the Jacobson radical makes $1-j$ invertible, forcing $k=0$, a contradiction. This proof avoids incorrectly applying the finitely generated form of Nakayama lemma to an arbitrary Artinian module.

The Hopf link exterior is $T^2\times I$. Each component's Seifert longitude is the other's meridian. Filling the two boundary tori along these longitudes therefore attaches solid tori with meridians intersecting once on a common torus, recovering the standard genus-one splitting of the three-sphere. Merely specifying two linked unknots without specifying their link type does not force this answer.

Let $a,b\ge 2$ with $a+b=2n+1$. Collapse the lower cells of $S^a\times S^b$ to obtain $q:S^a\times S^b\to S^{2n+1}$ and compose with the Hopf fibration $p:S^{2n+1}\to {\mathbb{CP}}^n$. Both factor inclusions are killed by q, so every induced homotopy group map is zero; the induced reduced homology maps also vanish. If the composite were nullhomotopic, homotopy lifting property would deform q into the circle fibre. Every map from the simply connected product to that circle is nullhomotopic, contradicting the degree-one top homology map of q.

A linear spring stores elastic energy $k{x}^2/2$ when displaced by $x$ from its relaxed position. The external force maintaining that displacement is $kx$; its restoring force on the moving attachment has the opposite sign. Its spring constant $k$ has units force per length.

The martini lattice is obtained from the honeycomb lattice by replacing every vertex in one of its two vertex classes by a triangle, attaching its three former incident bonds to the triangle's three vertices. It is a periodic degree-three planar graph, with triangular faces and nine-sided faces. Each decorated vertex and its incident bonds is a three-terminal cell with three internal triangle bonds and three terminal spokes. Its independent bond percolation threshold is $1/\sqrt{2}$, as follows from three-terminal cell partition duality.