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For the reciprocal-Jacobian quadratic energy under $F=I+e_1\otimes e_2$, $T_R/\mu$ has off-diagonal entries $(1,2)$ and $(2,1)$ equal to one and all others zero. Every unit reference normal in the first two coordinate directions therefore maximizes nominal traction. The Cauchy stress tensor has upper-left block $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$. Its largest absolute eigenvalue is $\phi =(1+\sqrt{5})/2$, with spatial normal proportional to $(\phi ,1,0)$. Pulling this back by $F^T$ gives reference normal proportional to $(1,\phi ,0)$ and plane $x_1+\phi x_2=0$.

For a nontrivial simple soluble group, the commutator subgroup is either the identity or the whole group. It cannot be the whole group because the derived series would never terminate. The group is therefore Abelian. Every nonidentity element then generates a normal cyclic subgroup, which must be the whole group. An infinite cyclic group has proper nontrivial subgroups; a finite cyclic group is simple exactly when its order is prime. Thus the soluble simple groups are precisely the cyclic groups of prime order.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

Nagle's algorithm coalesces small writes while data are unacknowledged, permitting another send when a full TCP segment can be formed or the outstanding data have been acknowledged. It reduces tiny transmissions but can add delay.

For a measurable space $(\Omega ,\mathcal{F})$ and any subset $E\subseteq \Omega$, the trace sigma-algebra is the displayed family of subsets of $E$. It contains $E=E\cap \Omega$, is closed under relative complements since $E\setminus (E\cap A)=E\cap (\Omega \setminus A)$, and is closed under countable unions. If $E\in \mathcal{F}$, it equals $\{B\in \mathcal{F}:B\subseteq E\}$. Without that measurability assumption, the traces need not be measurable sets in the original sigma-algebra.

The sigma-algebra generated by a collection is the smallest sigma-algebra containing all sets in that collection. The displayed intersection is over a nonempty family, since the power set is one candidate. An intersection of sigma-algebras contains the whole space and is closed under complements and countable unions, proving it is a sigma-algebra and establishing minimality.

For example, the half-lines $(-\infty ,x]$ generate the Borel sigma-algebra on the real line. They are Borel, and conversely $(-\infty ,b)={\bigcup }_n(-\infty ,b-1/n]$ and $(a,b)=(-\infty ,b)\setminus (-\infty ,a]$ recover every open interval. Preimages of these half-lines similarly generate the sigma-algebra generated by a random variable.

An arithmetic theory is Sigma-1 complete when it proves every true Sigma-1 formula that is a sentence. A genuine finite proof gives a true existential assertion that a code is a proof; Sigma-1 completeness therefore supplies the external necessitation clause for the usual arithmetical provability predicate. Its formalized version says that, for each Sigma-1 sentence $\sigma$, $T⊢\sigma \to {\mathrm{Pr}}_T(┌\sigma ┐)$. Applying this to a provability assertion, which is itself Sigma-1, gives the third of the Hilbert-Bernays-Löb derivability conditions.

If a total single-valued numerical function has a Sigma-1 graph, the complement of that graph is also Sigma-1. An input-output pair is outside the graph exactly when some different output is in the graph. Totality and uniqueness are essential to that equivalence. Stage-comparison separation of enumerations of the graph and its complement then gives weak definability of a numerical function in a bounded-complete arithmetic first-order theory.

In a triangle with root $v$, assign $w(va)=w(vb)=2$ and $w(ab)=-1$. Among graph paths, the unique shortest path from $v$ to $a$ is $v,b,a$, and that from $v$ to $b$ is $v,a,b$, both of weight one. A shortest-path tree would have to contain all three edges, impossible for a tree. The example uses shortest simple paths; arbitrary walks can have unboundedly negative weight by repeatedly traversing the negative edge.

In a finite connected graph with strictly positive edge weights, choose any shortest path from a root $v$ to every other vertex $u$, and let $p(u)$ be its penultimate vertex. Every prefix is shortest: a better prefix followed by the final edge would give a cheaper walk, whose positive-weight cycles can be erased to give a cheaper path. Thus the displayed distance identity holds.

Following predecessors strictly decreases the shortest-path distance, so no predecessor cycle is possible. Since every nonroot vertex has a predecessor, every chain reaches the root. Opposite predecessor choices cannot select the same undirected edge, again by strict distance decrease. The selected graph is connected and has exactly one edge per nonroot vertex, so is a spanning tree. Along its root paths the distance differences telescope, proving that it is a shortest-path tree even if the original shortest paths were chosen independently.

A smooth solution undergoes a gradient catastrophe when a spatial derivative becomes unbounded while the solution itself remains bounded. For $u_t+q^{\prime }(u)u_x=0$, characteristics have $x=s+t{q}^{\prime }(u_0(s))$ and $u=u_0(s)$; thus $u_x=u_0^{\prime }(s)/[1+t{q}^{\prime \prime }(u_0(s))u_0^{\prime }(s)]$. Vanishing of the characteristic projection derivative gives the first such time. A weak solution with an entropy condition can continue after this breakdown of the smooth description.

Roger Fletcher's method uses an L1 exact penalty on linearized constraints and a bounded trust region. It therefore admits a model minimizer even when hard linearized constraints are inconsistent or the quadratic model is indefinite.^[ref]

On ${\ell }^1$, $q$ is a norm, bounded above by $\Vert x{\Vert }_1$. The partial-sum map $Jx=({\sum }_{n\le k}{x}_n{)}_k$ is an isometric linear embedding into the space $c$ of convergent sequences with the supremum norm. Its range contains every eventually constant sequence: take successive differences of such a sequence to obtain a finitely supported element of ${\ell }^1$. These sequences are dense in $c$, so the completion of a normed space is $c$.

The range is proper: partial sums of the alternating harmonic series converge but their successive differences do not belong to ${\ell }^1$. The finite truncations of these differences are $q$-Cauchy and cannot have a limit in ${\ell }^1$. Thus this normed version of ${\ell }^1$ is incomplete, despite the completeness of its usual norm.

A topological space is totally separated when any two distinct points are separated by a clopen set. Such a space has no nontrivial connected subset. A zero-dimensional space with its usual clopen-base definition and $T_1$ is totally separated. Conversely these conditions agree for compact Hausdorff spaces by the Šura-Bura lemma and compactness, although the definitions should not be conflated for arbitrary spaces. Any closed totally separated subset of the unit square is nowhere dense, since every nonempty relatively open subset of the square contains a nontrivial connected line segment. Restricting a clopen set separating the segment's endpoints would disconnect that segment.

A topological space is regular when a point and a closed set not containing it have disjoint open neighborhoods. Equivalently, for every point $x$ in an open set $U$, there is an open $W$ with $x\in W\subseteq \overline{W}\subseteq U$. To obtain shrinking, separate $x$ from the closed complement of $U$; the closure of the point's neighborhood avoids the other neighborhood. Conversely use such a shrinking and the complement of its closure. This definition alone does not require $T_1$.

Suppose the $S_i$ are pairwise nonisomorphic simple modules. Any submodule $W$ of their finite direct sum that projects onto every summand is the entire sum. Induct on $k$. Projection onto the first $k-1$ factors is onto by induction. Its kernel is a submodule of $S_k$, hence zero or all of $S_k$. If zero, $W$ is isomorphic to the first $k-1$ factors but also has quotient $S_k$, contradicting their distinct composition factors and the Jordan–Hölder theorem. Therefore the kernel is $S_k$, proving the assertion. Consequently, if $S_i=A{x}_i$, the map $a\mapsto (a{x}_i{)}_i$ is onto and gives $A/({\bigcap }_i{\mathrm{Ann}}_A(x_i))\cong {⨁}_i{S}_i$ as left modules. The annihilators of these vectors need only be left ideals.

For a finite-dimensional module, it suffices that every maximal proper submodule have a submodule complement in order that every submodule have one. Let $W$ be the sum of all simple modules contained in the given module. If $W$ is proper, choose a maximal submodule $M$ containing it. A complement $C$ is isomorphic to the simple quotient $V/M$ and hence is itself simple. This forces $C\subseteq W\subseteq M$, contradicting the complement. Thus $W=V$. Greedily adding simple summands not already in the sum gives a direct sum in finitely many steps. Finally, given any $U$, choose a maximal sum of these simple summands disjoint from $U$. A missing simple summand has zero intersection with their sum plus $U$, contradicting maximality unless their sum complements $U$. This proves the semisimple module criterion, including the zero module.

If the left regular module of a unital ring is semisimple, then every left module is semisimple. Every module is a quotient of a direct sum of copies of the regular module, hence a quotient of a sum of simple modules. Their simple or zero images span the quotient. A maximal independent direct sum of simple submodules must span: otherwise one of these simple images has zero intersection with it and can be added. Thus the quotient is a direct sum of simples. For an infinite sum the maximality argument uses Zorn's lemma.

If $M$ is symmetric positive definite and $K$ symmetric negative semidefinite, the update $(M-\Delta tK/2)U^{k+1}=(M+\Delta tK/2)U^k$ is uniquely solvable for every positive step. Multiply its difference form by $(U^{k+1}+U^k{)}^T$ to obtainFor finite element heat flow, the mass norm is precisely the spatial $L^2$ norm, giving unconditional stability of the Crank-Nicolson method.