Neumann boundary data are prescribed values of the outward normal derivative of a function on the boundary of a boundary value problem. They enter a Neumann boundary condition as ${\partial }_{n}u=g_N$. Directional derivatives defined using a fixed coordinate direction must be converted to the outward-normal convention: for example, on the bottom of an upper domain, ${\partial }_{n}u=-u_y$.

In a directed network with no negative-cost directed cycle, add a source having zero-cost arcs to all vertices and compute shortest-path distances $d_i$. The inequalities $d_j\le d_i+c_{ij}$ make ${\pi }_i=-d_i$ dual feasible for uncapacitated minimum-cost flow. Shortest-path tree arcs are tight. An arbitrary spanning tree also fixes tight candidate potentials up to a constant, but it does not guarantee the non-tree inequalities; these must be checked.

The view factor $F_{i\to j}$ is the fraction of uniformly distributed, diffuse outgoing radiant flux from surface $i$ that directly reaches surface $j$. It depends on geometry, including obstruction, rather than on temperature or emissivity. All destinations in a closed enclosure have fractions summing to one. For a convex body fully enclosed by walls, $F_{\mathrm{body}\to \mathrm{walls}}=1$; the reverse view factor is generally smaller. Reciprocity gives $A_i{F}_{i\to j}=A_j{F}_{j\to i}$. Two unobstructed spheres with outer radius ten times the inner radius have $F_{\mathrm{inner}\to \mathrm{outer}}=1$ and $F_{\mathrm{outer}\to \mathrm{inner}}=1/100$. Do not multiply both directions by an assumed identical view factor.^[ref]

For the volume-preserving family $F=\mathrm{diag}(\lambda ,{\lambda }^{-1/2},{\lambda }^{-1/2})$, equal dead nominal tension $T$ gives potentialIts derivative factors as ${\Pi }^{\prime }=(1-{\lambda }^{-3/2})[\mu (\lambda +{\lambda }^{-1/2})-T]$. The nontrivial branch has a fold at $\lambda =2^{-2/3}$, $T/\mu =32^{-2/3}$, and meets the trivial branch at $T/\mu =2$. Stability in this one-parameter family is determined by ${\Pi }^{\prime \prime }$, not by branch count alone. At $T/\mu =2$ there is only one nontrivial positive solution, $\lambda =(3-\sqrt{5})/2$; the other branch value is $\lambda =1$.

The dielectric response differs along and perpendicular to the nematic director. In Gaussian units the orientational electric contribution at prescribed field is $-\Delta ϵ(\mathbf{E}\cdot \mathbf{n}{)}^2/(8\pi )$; positive anisotropy favors alignment with the electric field. The electric term does not receive the additional one-half convention used in writing the Frank elastic terms.

The Fréedericksz transition is a field-driven orientational instability of a confined nematic liquid crystal. Elasticity and surface anchoring resist the electric field or magnetic torque; a lowest allowed distortion mode becomes unstable at a finite threshold. Positive dielectric anisotropy of a nematic permits a normal electric field to tilt a planar anchored director.

The twisted nematic field effect switches a nematic liquid crystal between a twisted optically rotating state and a largely field-aligned state. With adiabatic optical following in a twisted nematic and crossed polarizers aligned to the two planar surface directors, a quarter-turn cell passes light at low voltage and suppresses it at high voltage. A rear reflector permits reflective operation with little power expended on light generation.

Let $k$ be a positive integer, $0<p<1$, and $\lambda >0$. For independent rate-$\lambda$ exponential distribution claims and a zero-based negative binomial distribution count of shape $k$ and parameter $p$, the random-sum transform identity gives $M_S(t)=[p(\lambda -t)/(p\lambda -t){]}^k$. This equals the transform of a compound binomial distribution with count $\mathrm{Bin}(k,1-p)$ and rate-$p\lambda$ claim sizes. Conditional on its positive count $n$, the total has Erlang distribution, whose survival function is the inner exponential-polynomial sum. Conditioning yields the displayed finite mixture distribution formula for $x>0$. It replaces an infinite mixture and includes the zero-total atom $p^k$ through the omitted $n=0$ term.

For $k=1$, the formula reduces to $(1-p)e^{-p\lambda x}$ for $x>0$: there is an atom $p$ at zero and a rate-$p\lambda$ exponential distribution positive component. This checks both the omitted zero-count term and the parameter scaling.

A uniform algebra is natural exactly when functions with no common zero generate the unit ideal. If a generated ideal is proper, it is contained in a maximal ideal. That maximal ideal is closed: the closure of a proper ideal stays proper, since an element within distance less than one of the identity is invertible by the Neumann series. The Gelfand-Mazur theorem identifies the maximal-ideal quotient with the complex numbers, yielding a character of an algebra. Naturality would make that character a common-zero evaluation, a contradiction. Conversely, the asserted unit-ideal property makes the zero sets of every finite collection in a character's kernel intersect. Compactness gives a point common to all these zero sets; applying this to $f-\phi (f)1$ proves $\phi (f)=f(x)$. For $C(K)$, one can directly take $g_j=\overline{f_j}/{\sum }_k|f_k{|}^2$.

For $F,G:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$, a dinatural transformation has components ${\alpha }_C:F(C,C)\to G(C,C)$ satisfying, for $f:C\to D$,Both composites have domain $F(D,C)$ and codomain $G(C,D)$. It is the compatibility used for categorical ends and categorical coends, rather than ordinary naturality on just one variance.

The interval dependency problem is the loss of relationships between repeated occurrences of the same input. For $X=[0,1]$, interval subtraction gives $X-X=[-1,1]$: it permits $u-v$ with independent choices $u,v\in X$. But the function $x\mapsto x-x$ has actual image of a function $\{0\}=[0,0]$. Giving both operands the name $X$ does not tell an interval arithmetic calculation to enforce $u=v$.

For the polynomial $f(x)=x(1-x)$ on $X=[0,1]$, the natural interval extension isThis allows the impossible combination $x=1$ in the first factor and $x=0$ in the second, producing the spurious upper bound $1$. In the actual function, both occurrences use the same $x$. Completing the square gives $x(1-x)=1/4-(x-1/2{)}^2$, which lies between $0$ and $1/4$ on $[0,1]$; every value between these bounds occurs, by continuity and the intermediate value theorem. The maximum $1/4$ occurs at $x=1/2$, while either endpoint gives the minimum $0$.

A natural exponential family is full when its parameter set is the entire natural parameter space, with no restrictions on the allowable canonical parameters beyond finiteness of its normalizing integral. Fullness must not be confused with minimality or regularity.

A natural exponential family is regular when its parameter space is open. In a full regular family, differentiation under the normalizing integral is justified locally by exponential moments in a neighbourhood. Regularity does not guarantee that every possible data set has a finite maximum-likelihood estimator.

The convex support is the closed convex hull of the canonical sufficient-statistic support. For a minimal exponential family it has nonempty interior. Means at finite interior natural parameters lie in that interior, because a supporting hyperplane at a boundary mean would force the sufficient statistic itself onto that hyperplane.

For each player in a finite bimatrix game, add the positive pure-strategy gains to its current mixed strategy and normalize. This gives a continuous self-map of the product of probability simplices. At a Brouwer fixed-point theorem fixed point, the gain coordinates equal the current strategy times their total gain. If that total were positive, every supported action would beat the current average payoff, contradicting that average. Thus every pure deviation has nonpositive gain for both players, proving a Nash equilibrium.

Define $M_2=K_2$ and $M_{k+1}=\mu (M_k)$ for $k\ge 2$. By the Mycielskian chromatic increase and Mycielskian triangle preservation, each $M_k$ is a finite triangle-free graph with chromatic number $k$. The first steps are $K_2,C_5$, the Grötzsch graph, and then a 23-vertex graph needing five colours. The vertex recurrence $n_{k+1}=2n_k+1$ gives $n_k=3\cdot 2^{k-2}-1$. These explicit examples show how arbitrarily strong graph coloring constraints can accumulate without any three-vertex clique.

A finite simple graph $G$ is a triangle-free graph if and only if its Mycielskian is a triangle-free graph. No triangle in a graph can contain the Mycielski apex, whose neighbours are an independent set, or two Mycielski shadow vertices. A triangle in a graph containing one shadow $u_i$ and two originals $v_j,v_{\ell }$ requires all three original edges $v_i{v}_j,v_i{v}_{\ell },v_j{v}_{\ell }$, hence an original triangle in a graph. Conversely every original triangle in a graph remains. In fact each original triangle in a graph gives four in $\mu (G)$: itself and the three choices of replacing one of its vertices by its shadow. Thus triangle-freeness is preserved, but a positive triangle count is multiplied by four.

For a nonempty finite simple graph, the upper bound copies an optimal proper vertex colouring to its Mycielski shadow vertices and gives the Mycielski apex a fresh colour. For the lower bound, suppose $\mu (G)$ uses $k=\chi (G)$ colours, with $w$ coloured $k$. Shadows avoid $k$. Replace colour $k$ at each original $v_i$ by $u_i$'s colour. Changed originals form an independent set; unchanged neighbours differ because $u_i{v}_j$ is an edge. This proper vertex colouring of $G$ uses $k-1$ colours, contradicting its chromatic number.^[ref]

The nonempty condition means $V(G)\ne \varnothing$, not $E(G)\ne \varnothing$. An edgeless $n$-vertex graph with $n\ge 1$ has chromatic number one; its Mycielskian is a star graph on $w,u_1,\dots ,u_n$ together with $n$ isolated vertices, and needs two colours. If $G$ has no vertices, $\mu (G)$ is a single vertex, so the same formula also holds with the convention $\chi (G)=0$.

The Mycielski apex is the added vertex adjacent to every Mycielski shadow vertex and to no original vertex. Its graph neighbourhood is precisely $\{u_1,\dots ,u_n\}$.

A Mycielski shadow vertex $u_i$ has graph neighbourhood $N_G(v_i)\cup \{w\}$ in $\mu (G)$. Its neighbours other than the Mycielski apex are original vertices. All Mycielski shadow vertices form an independent set.