A vertex with at least one child in a rooted tree. Internal vertices and leaves of a rooted tree partition the vertex set. The ancestors of an internal vertex are also internal; consequently the internal vertices of a nontrivial rooted tree induce a connected subgraph containing its root.

Two distinct vertices are siblings when they have the same parent in a rooted tree.

A vertex reachable from a specified vertex by following child edges in a rooted tree. Allowing the length-zero graph path includes the starting vertex; a proper descendant excludes it. The subtree at a vertex consists of that vertex and its descendants. State the properness convention when using strict discovery-time inequalities.

A rose tree is an ordered rooted tree in which each vertex can have any finite number of children in a rooted tree. A labelled Haskell representation is data RTree a = Node a [RTree a]. Its tree fold has type (a -> [b] -> b) -> RTree a -> b: recursively fold the children, then combine their results with the current label. Unlike a binary tree, a rose tree has no fixed pair of child positions.

A root arising from a non-root organ, such as a plant stem.

Ringworm of the body skin.

A dermatophyte fungal infection involving beard-area hair and follicles. It differs from every other inflammatory beard eruption, making sampling and organism identification useful when appearance overlaps.^[ref]

A rng has an abelian group under addition, an associative multiplication, and the distributive property; a multiplicative identity element is not required. Every ring under this wiki's unital convention is a rng. A nonunital ring may happen to have an identity: the terminology removes an axiom rather than imposing the absence of an identity. For example, the even integers with their usual operations have no multiplicative identity.

An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

An orientation-preserving rigid transformation, composed of a proper rotation and a translation. This solid-modelling convention distinguishes continuous rigid motion from orientation-reversing distance-preserving maps such as a reflection.

For a smooth hyperbolic system $U_t+A(U)U_x=S(U)$, suppose $\nabla R$ is a left eigenvector of $A$ with eigenvalue $\lambda$. The chain rule gives $({\partial }_t+\lambda {\partial }_x)R=\nabla R\cdot S$. Hence the homogeneous invariant remains an exact Riemann invariant when its gradient annihilates the source. For $U=(u,v)$, $A=\left(\begin{array}{cc}v& u\\ u^{-3}& v\end{array}\right)$, $S=(-u^3,u)$ and $u>0$, the function $R=v-1/u$ has $\nabla R=(u^{-2},1)$, eigenvalue $v+1/u$, and $\nabla R\cdot S=0$. On its zero level, the other characteristic speed vanishes and $v_t=1/v$ along constant spatial positions. Source terms need not destroy every invariant.

For $U_x+A(U)U_y=(1,\sqrt{v}{)}^T$ with $A=\left(\begin{array}{cc}u& 2\\ 2v& u\end{array}\right)$ and $v>0$, the variables $R=u-2\sqrt{v}$ and $S=u+2\sqrt{v}$ decouple into an unforced and a uniformly forced Inviscid Burgers equation. Smooth data for $u,v$ need not be smooth data for these variables when $v$ vanishes. For example $u(0,y)=2y$, $v(0,y)=y^2$ gives $R_0=4\min (y,0)$ and $S_0=4\max (y,0)$. Their derivative corners travel along different characteristic families. In the positive quadrant, $R=0$ and $S=2x$ below $y=x^2$, whereas $S=2(x+2x^2+2y)/(1+4x)$ above. Reconstructing $u=(R+S)/2$, $v=(S-R{)}^2/16$ gives a continuous piecewise classical continuation whose $y$ derivative jumps at the parabola. Treating $\sqrt{y^2}$ as $y$ for all real $y$ would miss this obstruction. The degeneracy at zero density and the regularity class are essential parts of the Cauchy problem.

Richardson extrapolation cancels a specified leading term in an asymptotic expansion, but an unquantified remainder does not certify a numerical error at two finite parameter values. For any positive distinct parameters $h_1,h_2$ and prescribed observations $v_1,v_2$, every candidate limit $L$ can be made consistent with a model $E(h)=L-c{h}^2-d{h}^3$: choose $c,d$ to solve $c{h}_i^2+d{h}_i^3=L-v_i$ for $i=1,2$. The coefficient matrix has nonzero determinant $h_1^2{h}_2^2(h_2-h_1)$. Thus the observations alone cannot identify $L$, even though every model has exactly the advertised $O(h^3)$ remainder in Big O notation. Quantitative remainder bounds or additional information are needed for an error certificate.

The ratio of outward carbon dioxide transfer to inward oxygen transfer measured at the mouth. It can differ from the cellular respiratory quotient during transient changes in body gas stores.

Ventilation insufficient to eliminate the carbon dioxide being produced. It can raise arterial carbon dioxide even when supplemental oxygen improves oxygenation, distinguishing ventilation from oxygen supply.

In the centre-of-momentum frame with total energy $\mathcal{E}$, four-momentum conservation for a final massive particle of rest mass $m>0$ and one photon gives equal and opposite three-momenta. The energy–momentum relation and $|{\mathbf{p}}_{\gamma }|=E_{\gamma }/c$ give the displayed energies, andThe photon has positive energy when $\mathcal{E}>m{c}^2$. Kinematics fixes the speed and the opposition of the two final directions, while an angular distribution requires information about the interaction.