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A calcium salt of bilirubin, important among the components of pigment gallstones.

A clear aqueous chemical solution of calcium hydroxide. Carbon dioxide can form a cloudy calcium carbonate precipitate in it.

Calcium hydroxide prepared by hydrating calcium oxide.

A program-directed hint to fetch a cache line early. A useful nonblocking hint produces no destination processor register value and allows execution to continue before the transfer completes. Fault and dropping behavior depend on the specified instruction contract.

The deprotonated counterpart of butyric acid.

Suppose a finite group acts on two finite position sets $A$ and $B$. Color the first set using $r$ colors and the second using $s$ colors, with both colorings moved by the same group element. If $c_A(g)$ and $c_B(g)$ are the cycle counts of its two induced permutations, the number of simultaneous colorings fixed by $g$ is $r^{c_A(g)}{s}^{c_B(g)}$: each cycle must be monochromatic, and different cycles can be colored independently. Burnside's lemma therefore givesfor the number of orbits. When the two position actions are equivariantly identified, this simplifies to the usual coloring count with $rs$ combined colors. For example, a tetrahedron's face is paired with its opposite vertex, and this pairing commutes with every symmetry.

Pair a tetrahedron vertex with its opposite face. This is an equivariant correspondence under all tetrahedral symmetries, so three face colors and two vertex colors become six combined symbols at each of four positions. Burnside's lemma on the alternating group $A_4$ gives $(6^4+3\cdot 6^2+8\cdot 6^2)/12=141$. Including reflections gives $S_4$ and reduces the count to multisets of four symbols from six, namely $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$.

The twelve rotational symmetries of a tetrahedron act on its six edges. Their edge cycle patterns are six fixed edges for the identity, two three-cycles for each of eight rotational symmetries of order three, and two fixed edges plus two transpositions for each of three half-turns. Burnside's lemma therefore counts edge colourings with $q$ labelled colours by $(q^6+8q^2+3q^4)/12$.

The Burgers' equation with constant linear damping has the displayed form. For zero viscosity and $\gamma >0$, characteristics are $u=u_0(a)e^{-\gamma t}$ and $x=a+u_0(a)(1-e^{-\gamma t})/\gamma$. A negative initial slope can make this coordinate map cease to be invertible, producing shock formation by characteristic intersection. The transformation $u=e^{-\gamma t}v$, $\tau =(1-e^{-\gamma t})/\gamma$ converts the inviscid equation to the ordinary inviscid Burgers' equation. Across a thin viscous layer the damping contributes only at lower order to the leading Rankine-Hugoniot condition.

For ${\alpha }_t+[V_s\alpha (1-\alpha ){]}_x=0$ in a finite upward pipe, bubbles move upward at $V_s(1-\alpha )$, but dense-mixture phase-fraction characteristics move downward. Hence an open upper end may be an incoming boundary for the scalar conservation law even though the bubbles themselves leave through it. Atmospheric pressure alone does not prescribe that incoming phase-fraction trace.

The bottom-clearing shock from zero to a uniform initial ${\alpha }_0$ has speed $V_s(1-{\alpha }_0)$. Continuing that constant right state to the outlet gives clearance time $L/[V_s(1-{\alpha }_0)]$, but at ${\alpha }_0>1/2$ it requires a compatible maintained incoming trace. For ${\alpha }_0=3/4$, the shock is $x=V_{s}t/4$, and a right-state characteristic traced backward from it has initial position $3V_{s}t/4$. It reaches the upper boundary at time $4L/(3V_s)$, before the formal clearance time $4L/V_s$. Later evolution therefore depends on the top closure. This boundary issue is distinct from the validity of the local slip and conservation derivation.

A continuous vector field on a closed Euclidean ball whose boundary values have nonnegative outward component has a zero in the ball. Otherwise $T(x)=-Rg(x)/|g(x)|$ is a continuous self-map of the ball. Its fixed point, supplied by the Brouwer fixed-point theorem, lies on the boundary and satisfies $g(x)\cdot x=-R|g(x)|<0$, a contradiction. The argument turns a zero-finding problem into a fixed point problem without any differentiability assumption.

A continuous Euclidean self-map with this growth has a fixed point. Choose $R\ge (2A{)}^{1/(1-\alpha )}$ and $R\ge 2B$. Then $|f(x)|\le R$ whenever $|x|\le R$, so the closed radius-$R$ ball is invariant and Brouwer applies. Existence does not imply convergence of iterating the original map: $x\mapsto 1-x$ on the unit interval has a fixed point but a two-cycle from either endpoint.

Return of breast cancer after treatment, distinguished as local, regional or distant disease.

For offspring $\mathrm{Bin}(k,p)$ with $0<p<1$ and $kp>1$, let $\eta \in (0,1)$ be the branching extinction probability. Conditioning on extinction changes the offspring law to $\mathrm{Bin}(k,\tilde{p})$ with the displayed parameter. Indeed, the probability of $j$ children is multiplied by ${\eta }^{j-1}$, so the new probability generating function is $f(\eta s)/\eta$. Since $f(s)=(1-p+ps{)}^k$ and $f(\eta )=\eta$, this equals $((1-p+p\eta s)/(1-p+p\eta ){)}^k$. Its mean is $k\tilde{p}=f^{\prime }(\eta )<1$, because strict convexity of $f(s)-s$ and its zeros at $\eta$ and one force a negative derivative at the smaller zero. The conditioned process therefore has an exponentially decaying total-progeny tail. Unconditionally that tail instead converges to the survival mass $1-\eta$, with an exponentially small correction. When extinction has probability zero, conditioning on it does not define this law.

If a minimum one-tree is not a tour, some vertex has degree greater than two, since there are equally many edges and vertices. Every tour must omit at least one of its incident selected edges. Partition tours by the first omitted edge in an ordered list, forcing all preceding listed edges and forbidding that edge. These disjoint children cover all tours. Recompute the constrained one-tree lower bound in each child and compare it with the incumbent solution to prune.