For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, with unit $\eta$ and counit $\epsilon$, call the adjunction idempotent when the multiplicationof its induced monad is invertible. We show that this idempotent adjunction condition is equivalent to invertibility of $\delta =L\eta R:LR\to (LR{)}^2$, the comultiplication of the induced comonad.

For an idempotent monad $T$, the unit laws implyIf $a:TA\to A$ is a monad algebra, naturality of the unit and its algebra unit law giveThus ${\eta }_A$ is an isomorphism. Apply this to the canonical algebra $R{\epsilon }_B:T(RB)\to RB$: its unit law is a triangle identity for an adjunction. It follows that ${\eta }_{RB}$, and therefore $L{\eta }_{RB}$, is invertible for every $B$. Hence $\delta$ is invertible.

Apply the same argument to the dual adjunction $R^{\mathrm{op}}⊣L^{\mathrm{op}}$. It gives the converse. Idempotence is self-dual: the induced monad is idempotent exactly when the induced comonad is idempotent. In particular, idempotence requires the unit to be invertible on every object in the image of the right adjoint.

Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.

The prime radical of a noncommutative ring is $N(R)={\bigcap }_{P}P$, with $P$ ranging over all prime ideals of a noncommutative ring. Here primeness means that for two-sided ideals $A,B$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. A minimal prime over an ideal $I$ is a prime ideal containing $I$ with no strictly smaller prime ideal containing $I$.

The ascending chain condition on left ideals also applies to two-sided ideals. We first prove by Noetherian induction that every proper $I$ contains a finite product of prime ideals, each containing $I$. If there were counterexamples, choose a maximal one. It is not prime, so there are strictly larger two-sided ideals $A,B$ with $AB\subseteq I$. Neither is a counterexample; products of prime ideals contained in $A$ and in $B$ then concatenate to a product contained in $I$, a contradiction.

Write $P_1\cdots P_t\subseteq I\subseteq P_j$. Every prime ideal $P$ over $I$ contains at least one $P_j$, by primeness. The inclusion-minimal members of the finite list $P_j$ are therefore precisely the minimal primes over an ideal $I$: any smaller prime over $I$ would contain a member of that same list. Consequently there are finitely many. Replacing each factor $P_j$ by a minimal prime contained in it giveswith each $Q_j$ minimal over $I$. Repeated factors may be necessary; in $k[\epsilon ]/({\epsilon }^m)$ the sole minimal prime $(\epsilon )$ requires its $m$th power to reach zero.

Apply this to $I=0$. Since $N(R)\subseteq Q_j$ for every $j$,Thus the prime radical is a nilpotent ideal, as in finite minimal primes and nilpotent prime radical.

Now assume $R$ is commutative. A prime ideal $P$ is an associated prime of a module $M$ when $P={\mathrm{Ann}}_R(m)$ for some $0\ne m\in M$. Every nonzero module over $R$ has an associated prime: the ascending chain condition gives a maximal annihilator of a nonzero element, and maximal annihilator of a module element is prime.

Since the given $M$ is a Noetherian module, repeatedly choosing such a cyclic submodule in the next quotient yields a finite prime filtrationThe process terminates by the ascending chain condition. In a short exact sequence $0\to U\to M\to W\to 0$, one has $\mathrm{Ass}(M)\subseteq \mathrm{Ass}(U)\cup \mathrm{Ass}(W)$. Indeed, if $P=\mathrm{Ann}(m)$ and $Rm\cap U=0$, the image of $m$ in $W$ still has annihilator $P$. Otherwise choose $0\ne am\in U$; then $a\notin P$, and primeness gives $\mathrm{Ann}(am)=P$. As $\mathrm{Ass}(R/P_i)=\{P_i\}$, induction along the prime filtration provesIn particular there are only finitely many associated primes of a module.

Finally let $P$ be a minimal prime ideal of $R$. Its localization at a prime ideal $R_P$ has exactly one prime, its maximal ideal $P{R}_P$. The preceding nilpotence result makes this maximal ideal nilpotent. Choose $0\ne y\in R_P$ annihilated by $P{R}_P$, for example a nonzero element in its last nonzero power, or $1$ if $P{R}_P=0$. Since elements outside the maximal ideal are units, ${\mathrm{Ann}}_{R_P}(y)=P{R}_P$.

Write $y=r/s$. The element $r/1$ is nonzero and also has annihilator $P{R}_P$. Generate $P$ by $p_1,\dots ,p_{\ell }$. For each $i$, choose $u_i\notin P$ with $u_i{p}_{i}r=0$, and put $u={\prod }_i{u}_i$. Then $m=ur$ remains nonzero after localization, while $Pm=0$. If $am=0$, localization forces $a/1\in P{R}_P$, and hence $a\in P$. Therefore ${\mathrm{Ann}}_R(m)=P$. This proves that minimal primes are associated primes.

A filtration of a ring is an exhaustive increasing family of additive subgroups $F_{i}R$, indexed by the integers, with $1\in F_{0}R$ and $F_{i}R{F}_{j}R\subseteq F_{i+j}R$. A compatible filtration of a module on a right module satisfies $F_{i}M{F}_{j}R\subseteq F_{i+j}M$. The associated graded module has components $F_{i}M/F_{i-1}M$, and its action by the associated graded ring isChanging either representative changes the product only by an element of $F_{i+j-1}M$, so the action is well-defined.

The subspace filtration on $N$ is $F_{i}N=N\cap F_{i}M$. The quotient filtration is $F_i(M/N)=(F_{i}M+N)/N$. Inclusion induces an injective map $\mathrm{gr}N\to \mathrm{gr}M$, and the quotient induces a surjective map to $\mathrm{gr}(M/N)$. If $m\in F_{i}M$ maps to zero in its degree-$i$ quotient, write $m=n+m^{\prime }$ with $n\in N$ and $m^{\prime }\in F_{i-1}M$. Then $n\in N\cap F_{i}M$, so the kernel is exactly the image of $\mathrm{gr}N$. We have the short exact sequencewhich is exactness of associated graded modules for induced filtrations.

For the negative case, $F_{i}R=R$ for $i\ge 0$. Completeness means that $R\to {\underleftarrow{\lim }}_{q}R/F_{-q}R$ is an isomorphism; in particular the filtration of a ring is separated. If $x\in F_{-1}R$ and $r\in R$, then $(xr{)}^q\in F_{-q}R$. The geometric seriesconverges in the complete negative filtration and is a two-sided inverse of $1-xr$, by the finite geometric identity and continuity of multiplication. The unit criterion for the Jacobson radical now gives

Let $I$ be any right ideal. The associated graded module $\mathrm{gr}I$ for its subspace filtration is a homogeneous ideal on the right in $\mathrm{gr}R$. If $\mathrm{gr}R$ is a right Noetherian ring, choose homogeneous generators that lift to $x_1,\dots ,x_s\in I$, with degrees $d_1,\dots ,d_s\le 0$.

For $x\in I\cap F_{d}R$, cancel its degree-$d$ symbol by a sum ${\sum }_i{x}_i{a}_{i,0}$ with $a_{i,0}\in F_{d-d_i}R$. The remainder belongs to $I\cap F_{d-1}R$. Repeat the cancellation in each succeeding degree. After $q$ steps,For every $i$, the coefficient series converges to some $a_i\in R$ by completeness. Passing to the limit gives $x={\sum }_i{x}_i{a}_i$. Thus these finitely many $x_i$ generate the actual right ideal $I$; no assumption that $I$ is closed is needed. Since every right ideal is finitely generated, $R$ is right Noetherian, by complete negative filtered-graded transfer of Noetherianity.

The affine module sheaf construction makes the remaining statements explicit. For $A=k[V]$ and an $A$-module $M$, defineon the basis of principal open subsets. Restrictions are the natural localization maps. Equivalently, a section on an arbitrary open set is a family $s_P\in M_{{\mathfrak{m}}_P}$ which near each point is represented by a single fraction $m/f^r$, with $f$ nonvanishing throughout that neighbourhood. This locally representable family construction gives a sheaf; the usual localization gluing verifies its quasi-coherence. Its stalk is $M_{{\mathfrak{m}}_P}$ andThis last identification is precisely the standard module-localization gluing on the affine basis, not an assertion that arbitrary sheaves are determined by their global sections.

Now write $A=k[X]$, $B=k[Y]$ and $\alpha ={\varphi }^{\#}:A\to B$. Let $\mathcal{G}=\tilde{M}$, $\mathcal{H}=\tilde{N}$ for $A$-modules $M,N$, and let $\mathcal{F}=\tilde{L}$ for a $B$-module $L$. The affine versions of the three constructions areHere ${}_{A}L$ denotes restriction of scalars along $\alpha$, whereas the last associated sheaf lives on $Y$. To verify the tensor-product assertion, localize at every prime: both stalks are $M_{\mathfrak{m}}{\otimes }_{A_{\mathfrak{m}}}{N}_{\mathfrak{m}}$. For the direct image, ${\varphi }^{-1}D(a)=D(\alpha (a))$, andThese equalities identify the sheaves on a basis and respect restrictions. For the pullback of a sheaf of modules, its stalk at $Q$ is $B_{{\mathfrak{m}}_Q}{\otimes }_{A}N$, exactly the stalk of $\widetilde{B{\otimes }_{A}N}$. The local tensor maps are compatible, so these stalk identifications come from an actual sheaf isomorphism. Thus all three constructions are quasi-coherent in the stated affine setting.

Finally construct the natural projection formula for sheaves map by sending a local tensor $m\otimes b$ to $b(1\otimes m)$ in the pulled-back module, followed by direct image. In the affine description, the target ${\varphi }_{\ast }{\varphi }^{\ast }\tilde{M}$ is the sheaf associated with the $A$-module ${}_A(B{\otimes }_{A}M)$, while $\tilde{M}{\otimes }_{{\mathcal{O}}_X}{\varphi }_{\ast }{\mathcal{O}}_Y$ corresponds to $M{\otimes }_{A}B$. The mapsare inverse $A$-linear maps. Localizing them commutes with every restriction, provingNo flatness assumption on $A\to B$ or finiteness assumption on $M$ is needed.

The general, or asymmetric Lovász local lemma, is as follows. Let $E_1,\dots ,E_m$ be events, with a dependency graph of events having neighbour sets $N(i)$. Each $E_i$ must be independent of the sigma-algebra generated by all the events indexed by its nonneighbours. If numbers $0\le x_i<1$ satisfythenFor completeness, the key inductive bound is $\mathbb{P}(E_i\mid {\bigcap }_{j\in S}{E}_j^c)\le x_i$ whenever $i\notin S$. Split $S$ into neighbours $S_1$ of $i$ and nonneighbours $S_2$. If $S_1$ is empty, the dependency condition gives the bound directly. Otherwise,The denominator bound follows by exposing the exclusions in $S_1$ one at a time and using the induction hypothesis with strictly fewer conditioning events. The same induction ensures that all conditioning events have positive probability. Finally multiply the successive conditional probabilities of avoiding $E_1,\dots ,E_m$ to obtain the stated lower bound.

For the symmetric form, suppose each event has probability at most $q$ and the dependency graph has maximum degree $D\ge 1$. Choose $x_i=1/(D+1)$. ThenThe final inequality follows from $(1+1/D{)}^D\le e$. Thus $eq(D+1)\le 1$ suffices for positive probability that no bad event occurs. If $D=0$, the dependency condition gives joint independence and the same criterion implies $q<1$, so the conclusion follows directly.

Apply this to a fair independent two-colouring of the hypergraph vertices, with $E_W$ the event that hypergraph edge $W$ is monochromatic. Its probability is $q=2^{1-r}$. Events on disjoint hypergraph edges use disjoint independent variables, giving a dependency graph by joining intersecting hypergraph edges. Each hypergraph edge meets at mostother hypergraph edges; repeated counting only enlarges this upper bound. The slightly looser bound $D+1\le r(\Delta +1)$ is enough for the printed condition. Indeed,The symmetric Lovász local lemma therefore gives a proper two-colouring. A hypergraph with no hypergraph edges is trivially colourable and needs no dependency calculation.

For the unheaded degree-one continuation, write $X={\sum }_{v=1}^n{I}_v$, where $I_v$ indicates that vertex $v$ has degree one. Withwe haveFor distinct vertices $u,v$, separate the cases where their connecting edge is present or absent. If it is present, all their other incident edges must be absent. If it is absent, each vertex chooses its unique neighbour among the other $n-2$ vertices; those two neighbours may coincide. The respective edge constraints are independent even in that case. ThusDividing by $q^2$ gives the exact ratioThe first term tends to zero because $(n-1{)}^{2}p$ has order $n\log n$; the second tends to one. Consequently its variance satisfiesThe Chebyshev inequality now yields $\mathbb{P}(X=0)\le \mathrm{Var}X/(\mathbb{E}X{)}^2\to 0$. A vertex of degree one exists with high probability, even though the limiting probability of having an isolated vertex is strictly between zero and one.

We use the random alteration method, controlling large hypergraph independent sets and then removing the few forbidden overlaps. Fix $k=2005$, let $n$ tend to infinity, and independently include each three-element subset of an $n$-vertex set as an hypergraph edge with probability $p=n^{-7/4}$. Call the resulting three-uniform hypergraph $H_0$, and let $s=\lceil n/(2k)\rceil$.

For any prescribed $s$-vertex set, the probability that it is a hypergraph independent set is $(1-p{)}^{\left(\genfrac{}{}{0ex}{}{s}{3}\right)}$. Hence the union bound givesbecause $p\left(\genfrac{}{}{0ex}{}{s}{3}\right)$ has order $n^{5/4}$, with a positive constant depending only on $k$.

Let $Y$ count unordered pairs of distinct hypergraph edges meeting in two vertices. Each potential pair consists of its common pair and two distinct additional vertices, soThe Markov inequality implies $\mathbb{P}(Y\ge n/2)=O(n^{-1/2})$. We may therefore fix one realization for which $Y<n/2$ and no $s$-vertex hypergraph independent set exists. For each offending hypergraph edge pair choose one vertex from its union, and let $D$ be the set of all chosen vertices. Delete $D$, together with all incident hypergraph edges, and call the induced hypergraph $H$. Then $|D|\le Y<n/2$, so $N=|V(H)|>n/2$. The next two parts verify both desired properties of this same $H$.

For the unheaded existence request, use the probabilistic method. Choose every value $f(x,y)$ independently and uniformly from $\{-1,1\}$. For a rectangle with $x\ne x^{\prime }$ and $y\ne y^{\prime }$, its four entries are distinct, so independence and the zero mean of each sign give expectation zero for their product. If $x=x^{\prime }$ or $y=y^{\prime }$, the product is identically one. The number of ordered rectangles in this union of diagonal cases isTherefore the random sign rectangle fourth moment has expectationSome sign function has $Q(f)$ at most this expectation. Applying the bound in part (a) to that same function gives the simultaneous choicesBoth constants tend to zero whenever both dimensions tend to infinity, without any assumption on their relative rates of growth.

Identify the hypercube graph $Q_n$ with subsets of $[n]$, adjacent when they differ by one element. Write $N(\mathcal{F})$ for a family's closed vertex neighbourhood. The simplicial order on the discrete cube orders first by size and then, within a rank, by lexicographic order: the least element of a symmetric difference belongs to the earlier set. The Harper theorem says that if $I$ is the initial simplicial segment with $|I|=|\mathcal{F}|$, thenEquivalently this minimizes the external vertex boundary at fixed cardinality. This is the vertex-isoperimetric assertion, distinct from the edge version in Question 3.

Here is the deduction of Kruskal-Katona theorem. For a family $\mathcal{B}\subseteq [n{]}^{(k)}$, adjoin every lower rank and set $\mathcal{D}=[n{]}^{(<k)}\cup \mathcal{B}$. For $k\ge 1$,The corresponding simplicial initial segment is $[n{]}^{(<k)}\cup L$, where $L$ is the rank-$k$ lexicographic initial segment of size $|\mathcal{B}|$. Canceling the common lower-rank count in Harper's inequality gives $|\nabla \mathcal{B}|\ge |\nabla L|$. The rank-zero case is immediate by considering its two possible families.

To turn upper shadows into lower shadows, let $\rho (i)=n+1-i$ and $T(B)=\rho ([n]\setminus B)$. This bijection reverses inclusion and changes rank $k$ to $n-k$. If $B$ precedes $C$ in lex order, the largest element of $T(B)△T(C)$ belongs to $T(C)$, so $T(B)$ precedes $T(C)$ in colexicographic order. Also $T(\nabla \mathcal{B})=\partial (T\mathcal{B})$. Hence applying the upper-shadow conclusion to the transformed family proveswhere $C$ is the colex initial segment of the same size and rank as $\mathcal{A}$. This is the Kruskal-Katona lower-shadow theorem. A colex segment's shadow is itself a colex segment: split its sets by their largest element, giving a full initial block of the preceding ground set followed by an initial block with the new largest element. The same description holds one rank lower. Iteration therefore shows that colex segments minimize shadows at every lower rank. This proves the Harper theorem implies the Kruskal-Katona theorem deduction, including the necessary complement and coordinate reversal.

The Erdős-Ko-Rado theorem states that for $1\le r\le n/2$, every intersecting family $\mathcal{A}\subseteq [n{]}^{(r)}$ satisfiesThe family of all $r$-sets containing a fixed point attains the bound. We use the standard convention of positive ranks here.

For the first proof, let $\mathcal{B}$ be the complements of members of $\mathcal{A}$, at rank $k=n-r\ge r$, and let $\mathcal{S}$ be its rank-$r$ lower shadow. It is disjoint from $\mathcal{A}$, since $A\subseteq [n]\setminus A^{\prime }$ would make two family members disjoint. Suppose $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The colex rank-$k$ segment of that size contains every $k$-set on $[n-1]$ and at least one set containing $n$. Its rank-$r$ shadow therefore contains all $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ old $r$-sets and at least one new $r$-set containing $n$. The iterated Kruskal-Katona bound givesThen $|\mathcal{A}|+|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, impossible for disjoint subfamilies of rank $r$. This proves the Erdős-Ko-Rado theorem from shadows.

For the second proof use the Katona circle method. In any cyclic ordering of the $n$ points, an intersecting collection of length-$r$ intervals has at most $r$ members. To prove this, rotate one selected interval to end at position $n$, so it occupies $n-r+1,\dots ,n$. Intervals ending at $r,\dots ,n-r$ miss it and cannot be selected. Among the remaining endpoints other than $n$, pair $j$ with $j+n-r$ for $1\le j\le r-1$. The two intervals in each pair are disjoint because $n\ge 2r$, so at most one can be selected. Including the fixed interval gives at most $1+(r-1)=r$.

Now count pairs consisting of a family member and a permutation in which it is a cyclic interval. A fixed $r$-set has $nr!(n-r)!$ such permutations: choose its final position, its internal order, and the order of the remaining points. Each of the $n!$ permutations contributes at most $r$ family intervals by the proved cyclic interval intersection bound. ThusThis is the required cyclic-ordering proof, with its interval lemma established explicitly.

For odd $N>1$ coprime to $b$, the Fermat pseudoprime test is $b^{N-1}\equiv 1(\mathrm{mod}N)$. The Euler pseudoprime test meant here is the Euler–Jacobi test,where the right side is the Jacobi symbol. Write $N-1=2^{s}d$, with $d$ odd. The strong pseudoprime test passes when $b^d\equiv 1$, or when $b^{2^{j}d}\equiv -1$ for some $0\le j<s$. Terminology often reserves “pseudoprime” for composite numbers; the question uses it for passing the respective test, including primes.

For the final example, $341=11\cdot 31$ is composite and $2^{10}=1024\equiv 1(\mathrm{mod}341)$, so $2^{340}\equiv 1$. It passes the Fermat primality test. However $2^{170}\equiv 1$, whereas the supplementary law for the Jacobi symbol gives $(2/341)=-1$ since $341\equiv 5(\mathrm{mod}8)$. Thus 341 is Fermat but not Euler–Jacobi pseudoprime to base two. A weaker convention defining an Euler test merely by $b^{(N-1)/2}=\pm 1$ would not give the claimed distinction, which is why the Jacobi convention matters.

The bob has $\dot{x}=\ell \cos \theta \dot{\theta }$ and $\dot{y}=at-\ell \sin \theta \dot{\theta }$. Substituting these into the Lagrangian givesThe mixed term is $\frac{d}{dt}(ma\ell t\cos \theta )-ma\ell \cos \theta$. Removing this total derivative and the time-only term leaves the equivalent Lagrangian $L_{\mathrm{eff}}=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+m(g-a)\ell \cos \theta$. The Euler-Lagrange equation becomesIn free fall, $a=g$, so $\theta ={\theta }_0+{\omega }_{0}t$: a bob initially at rest relative to the lift remains at its initial angle, and otherwise moves around the pivot with constant angular speed. Every constant angle is then an equilibrium, with no restoring force. For $a\ne g$, the equilibria are $\theta =0$ and $\theta =\pi$ modulo $2\pi$.

The infection term $rIS$ is a mass-action contact rate: each infection removes one susceptible individual and adds one infected individual. Recovery at rate $aI$ transfers individuals from the infected class to the recovered class. Loss of immunity at rate $pR$ transfers recovered individuals back to the susceptible class. Thus these equations describe the SIR model with waning immunity, also called the SIRS model. Adding the equations gives $\dot{S}+\dot{I}+\dot{R}=0$, so $S+I+R=N$. The nonnegative population region is invariant because the derivative of any population on its zero boundary is nonnegative.

The characteristic function is ${\varphi }_X(u)=\mathbb{E}(e^{iuX})$. It satisfies ${\varphi }_{-X}(u)={\varphi }_X(-u)=\overline{{\varphi }_X(u)}$. Thus a real-valued characteristic function gives ${\varphi }_X={\varphi }_{-X}$, and the uniqueness theorem for characteristic functions implies equality of the two distributions. Conversely, equality in distribution gives equality with the complex conjugate, making the characteristic function real. Reality for all $u$ is equivalent to symmetry of the distribution about zero.

A first-order structure consists of a nonempty domain, an interpretation of each constant as a domain element, each $k$-ary function symbol as a map from the $k$-fold domain product to the domain, and each $k$-ary relation symbol as a subset of that product. Under an assignment of values to variables, terms are evaluated recursively. Atomic formulas assert equality of term values or membership in an interpreted relation. Boolean connectives act on truth values, and quantifiers range over the whole domain. The set $[[\phi ]{]}_A$ consists of assignments to the free variables making the formula true.

For a substructure $B\subseteq A$, constants lie in $B$, functions restrict to $B$, and relations restrict to tuples from $B$. Induction on terms shows that every term with arguments from $B$ has the same value in both structures. Atomic formulas consequently have the same truth values, and induction on Boolean connectives givesfor every quantifier-free formula.

For a nonempty chain of $T$-model substructures, form their union $U$. Every finite tuple belongs to one member of the chain, so $U$ is closed under all interpreted functions and contains the constants. Consider any inductive axiom $\forall x\exists y\psi (x,y)$. A specified finite tuple $x$ lies in one member, which supplies witnesses $y$ there. The quantifier-free formula $\psi$ has the same truth value in that member and $U$, so $U$ satisfies the axiom. Hence $U$ is a $T$-model and the least upper bound of the chain. The poset is complete for nonempty chains. If chain-complete is defined to include the empty chain, a least element must additionally exist, which the stated hypotheses do not ensure.

The azimuthal coordinate is cyclic, so its canonical momentum is conserved. Time independence of the Lagrangian also conserves the energy. The two constants areFor a horizontal launch from depth $D$, take $0<D<\ell$, so $\cos {\theta }_0=D/\ell$, ${\dot{\theta }}_0=0$, and $v=\ell \sin {\theta }_0{\dot{\varphi }}_0$. Thus $h=mv\sqrt{{\ell }^2-D^2}$ and $E=\frac{1}{2}m{v}^2-mgD$. The meridional motion has effective potential $U(\theta )=h^2/(2m{\ell }^2{\sin }^2\theta )-mg\ell \cos \theta$.

In the equatorial plane, the Killing symmetries give conserved quantities $E=(1-2M/r)\dot{t}$ and $L=r^2\dot{\varphi }$. The null normalization givesFor $L\ne 0$, put $u=1/r$; then $du/d\varphi =-\dot{r}/L$. ConsequentlyDifferentiating gives $u^{\prime \prime }+u=3M{u}^2$, initially where $u^{\prime }\ne 0$ and then at turning points by continuity. The circular solution follows directly from the radial equation as well. The ratio $|L/E|$ is the impact parameter of the asymptotic null orbit.

A Riemann surface is a Hausdorff, second-countable, one-dimensional complex manifold, with charts to open subsets of $\mathbb{C}$ whose transition maps are holomorphic. Connectedness is often included in the convention. A map between Riemann surfaces is holomorphic when its expression in every compatible pair of charts is holomorphic. It is biholomorphic when it is bijective and both it and its inverse are holomorphic.

Put $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$, ${\hat{\sigma }}^2=\Vert Y-X\hat{\beta }{\Vert }^2/(n-p)$. Orthogonal projections of the Gaussian error give independent variablesHence the exact $(1-\alpha )$ confidence set is the ellipsoidHere $F_{p,n-p}(q)$ denotes the $q$ quantile of the F-distribution.

Let ${\hat{\beta }}_{(-i)}$ be the fit with observation $i$ removed, when that reduced design retains full rank. Cook's distance iswhere $e_i$ is its residual and $h_{ii}=x_i^T(X^{T}X{)}^{-1}{x}_i$ is its leverage. Thus $D_i$ measures the deleted estimate's distance from the full estimate in exactly the metric of the confidence ellipsoids. The deleted estimate lies on the confidence contour whose $F$ quantile equals $D_i$; this is a geometric influence diagnostic, not an assertion that $D_i$ itself has an $F$ law.

For $p=2,n=50$, $0.70<1.3<2.42$, so deletion moves the estimate outside the 50% confidence ellipsoid but inside the 90% one. The observation warrants investigation for influence: the shift is substantial, though the supplied figures do not justify calling it outside the 90% region. The reciprocal $F_{48,2}$ and the approximate chi-square figures are unnecessary.

The diagram has common multipliers $u=5/3,d=2/3$ on every branch, with gross riskless return $R=4/3$. The martingale condition $pu+(1-p)d=R$ givesSince $d<R<u$, these probabilities lie strictly between zero and one at each node, and the market is arbitrage-free. All valuations below use discounted conditional expectations under this measure.