For a gravity-current nose of height $h_N$ on a small incline of slope $\theta$, the travelling front of a gravity-driven thin film satisfies $h_y=\theta (1-h_N^2/h^2)$. Close to its dry tip, $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. The slope becomes order one at $h\sim \sqrt{\theta }{h}_N$, giving an $O(\sqrt{\theta }{h}_N)$ region where lubrication theory fails. The main nose length is $h_N/\theta$. Comparing capillary pressure gradients $\gamma h_N/(h_N/\theta {)}^3$ with gravitational forcing $\rho g\theta$ gives the crossover $h_N\sim \theta \sqrt{\gamma /(\rho g)}$. This concerns the main nose length; capillarity may already matter in a much smaller tip region.

On an equal-panel partition of an interval, with $h=(b-a)/n$, sum the trapezoidal rule over adjacent panels:There are $n+1$ distinct quadrature nodes. Each interior sample belongs to two panels and therefore has quadrature weight $h$, while each endpoint has quadrature weight $h/2$. Equivalently, the composite trapezoidal rule integrates the piecewise linear function obtained by linear interpolation exactly.^[ref]

For $f\in C^2[a,b]$ and $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$, summing panel quadrature errors givesfor some $\xi \in (a,b)$. Thus panel quadrature errors of order $O(h^3)$ accumulate to order $O(h^2)$, using big O notation.^[ref] On a nonuniform partition of an interval with widths $h_j$ and largest width $H$, the bound is $K{\sum }_j{h}_j^3/12\le K(b-a)H^2/12$.

For the polynomial $f(x)=x^2$ on $[0,1]$, direct summation yieldsBoth quadrature errors shrink by exactly four when $n$ doubles, with opposite signs. This exact relation belongs to this polynomial example; the trapezoidal rule is not always twice as inaccurate as the midpoint quadrature rule.

The left plot uses exact polynomial sums; the right adds a common sample bias $\delta =0.002$ to the composite trapezoidal rule. That mathematical model converges to $1/3+\delta$, showing why smaller panels alone do not ensure accuracy for noisy samples. It illustrates noise sensitivity of a quadrature rule, rather than measured data.

Periodizing the bulk transverse tangent correlation of a stretched worm-like chain gives $C_L(r)={\sum }_{j\in \mathbb{Z}}C(r+jL)$, where $d_L(r)$ is the distance to zero on the periodic interval. Summing the geometric series yields the displayed result. At zero separation its finite-length factor is $\coth (L/(2\xi ))$; for $L\gg \xi$ the bulk result is recovered. Periodic tangent fluctuations are a mathematical boundary convention, not a claim that a pulled physical chain forms a closed loop.

If $f:M\to N$ satisfies transversality of a map to a submanifold $S$, its nonempty preimage is an embedded submanifold of the displayed codimension. Choose a slice chart for an embedded submanifold around $f(p)$ in which $S$ is the zero set of the first $k$ coordinates. Their composition with $f$ has surjective derivative on its zero set, so the regular level set theorem gives a local embedded slice of dimension $\dim M-k$. These slices give the subspace topology and a compatible smooth atlas on the preimage. Its tangent space is

Smooth maps of finitely many circles into a manifold of dimension at least three can be perturbed within their homotopy classes to disjoint embeddings. The transversality dimension count for two curve branches is $1+1-m<0$, so their images avoid one another after perturbation; one also removes self-intersections and obtains immersions by the jet version of transversality. This allows disjoint representative loops for relators in high-dimensional surgery constructions of prescribed fundamental groups.

This matrix adds $t$ times coordinate $j$ to coordinate $i$. Left multiplication performs the corresponding row addition. Such matrices generate the special linear group over a finite field: row elimination also realizes signed interchanges and compensating diagonal scalings using products of these matrices.

Every nonidentity transvection has an adapted basis in which it is $I+E_{12}$. Thus all such maps are conjugate in the general linear group. In dimension at least three, a scaling of a third basis vector centralizes this matrix and has arbitrary nonzero determinant; adjusting a conjugator by this scaling proves conjugacy in the special linear group as well.

// Target: algebra.bigb

In the Weyl representation of the gamma matrices, $C=i{\gamma }^2{\gamma }^0=\mathrm{diag}(\epsilon ,-\epsilon )$ with $\epsilon =i{\tau }_2$. The antisymmetry of $\epsilon$ gives $C^T=-C$, and direct multiplication gives $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$ and $C^{†}C=I$. Any two such intertwining matrices differ by a scalar: their ratio commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. Changing spinor basis sends $C$ to $SC{S}^T$, so transpose antisymmetry is basis-independent. Its sign is therefore fixed, not merely one of the two alternatives supplied by the general transpose symmetry of a charge-conjugation matrix argument.

For shipments $x_{ij}\ge 0$ with factory row sums at most capacities $s_i$ and shop column sums exactly demands $d_j$, the displayed dual potentials give $\sum c_{ij}{x}_{ij}\ge {\sum }_i{u}_i{s}_i+{\sum }_j{v}_j{d}_j$. The sign $u_i\le 0$ ensures $u_i{\sum }_j{x}_{ij}\ge u_i{s}_i$. Equality proves optimality. It occurs when every used cell is tight and every row with spare capacity has $u_i=0$, the relevant complementary slackness conditions. A dummy demand column with zero costs converts unused capacity into a balanced transportation problem.

A finite word over an alphabet is locally admissible for a zero-one symbolic transition matrix if every consecutive pair is permitted. Its length counts symbols, so length $n+1$ corresponds to $n$ transitions. The number with endpoints $i,j$ is $(A^n{)}_{ij}$ by matrix multiplication. Local admissibility does not always imply occurrence in a two-sided subshift of finite type: the matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ permits the word $(1,2)$ but admits no two-sided sequence. If every row and column has a nonzero entry, every locally admissible word extends in both directions.

A complex vector bundle has structure group $G$ when it admits fiber-linear local trivializations whose transition matrices take values in $G$. Replacing frames by $h_i$ changes transitions to $h_i^{-1}{g}_{ij}{h}_j$ in the frame convention. Reducing the structure group means choosing new trivializations with values in a specified subgroup. The same transitions glue $U_i\times G$ by left multiplication to a right principal bundle.

Prediction optimism is the difference between expected squared prediction error on new independent responses at the training covariates and expected training error. Reusing responses to fit and assess a model tends to make training error too favorable. The covariance formula for prediction optimism measures this reuse without assuming a linear fitting rule.

// Target: probability-and-statistics.bigb

For a trace-class operator $K$ on a Hilbert space, the Fredholm determinant is the convergent product over its eigenvalues, counted with algebraic multiplicity. For self-adjoint $K$, spectral theorem for compact self-adjoint operators reduces the product to a diagonal calculation. For $\Vert K\Vert <1$, $\log \det (I+K)={\sum }_{n\ge 1}(-1{)}^{n+1}\mathrm{tr}(K^n)/n$. This identity exposes the successive cumulants in a Gaussian quadratic exponential moment.

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.

If $L/K$ is an unramified extension and $M/K$ is a totally ramified extension of non-Archimedean local fields inside a fixed algebraic closure, then $LM/L$ is totally ramified of degree $[M:K]$. A uniformizer of $M$ has an Eisenstein polynomial over $K$. Since a uniformizer of $K$ remains a uniformizer in $L$, that polynomial remains Eisenstein over $L$ and generates $LM$.

A quadratic form of rank $r$ on ${\mathbb{C}}^n$ is congruent to $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$, and include the radical of a bilinear form. Their span is a totally isotropic subspace and has dimension $\lfloor r/2\rfloor +n-r=n-\lceil r/2\rceil \ge \lfloor n/2\rfloor$. The form vanishes on every vector of this subspace, not merely on its basis vectors.

For a nondegenerate real quadratic form of inertia $(p,q)$, projection of a totally isotropic subspace onto either its positive or negative coordinate subspace is injective: a vector with one projection zero would have strictly one-signed quadratic value unless it vanished. Thus its dimension is at most both $p$ and $q$. This bound is attained by pairing one positive and one negative coordinate for each of $\min (p,q)$ independent null vectors.

For a Pareto distribution of shape three and minimum $d>0$ with count rate $\lambda >0$, the sum of insurer and reinsurer aggregate variances is $V(M)=\lambda (3d^2-3dM+2M^2)$ for $0\le M\le d$, and $\lambda (3d^2-d^3/M)$ for $M\ge d$. The first branch has its unique minimum at $3d/4$; the second is increasing. The branches agree in value and slope at $d$, giving the displayed global minimum. This criterion minimizes the sum of party variances; the total portfolio variance includes their covariance and remains $3\lambda d^2$.

Two disjoint synthemes form a six-cycle, whose complement is a triangular prism. The prism has a unique partition into three perfect matchings, so the pair extends uniquely to a total of synthemes. Eight synthemes are disjoint from a fixed syntheme; grouping them four per total yields two totals through it, and double counting gives six totals in all.

// Target: combinatorics.bigb

A path along a circle can have variable areal velocity. As an illustrative controlled trajectory, keep a point mass on a circle of radius $R$ with angleIts acceleration in polar coordinates and the required net force arewhere ${\mathbf{e}}_r$ and ${\mathbf{e}}_{\theta }$ are the radial and transverse unit vectors. A circular guide can supply the radial force, while a tangential drive supplies $mR\alpha$. HenceFor $R=m={\omega }_0=\alpha =1$ in illustrative units, the three equal time intervals $[0,0.5]$, $[0.5,1]$ and $[1,1.5]$ sweep areas $0.3125$, $0.4375$ and $0.5625$. Their difference comes from the transverse force's positive torque, even though the radius and path shape stay fixed. The area under each areal velocity graph segment equals the corresponding swept area.