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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.

Parametrize the circle by $z=e^{it}$, $0\le t\le 2\pi$. Choose a continuous argument $\vartheta (t)$ of $\varphi (e^{it})$, so $\varphi (e^{it})=|\varphi (e^{it})|e^{i\vartheta (t)}$. Its winding number about zero isThe two facts proved below determine the requested calculation without needing differentiability or contour integration. Put $P(z)=(3z-2)(z-3)(2z+1)$. On $|z|=1$, $|P(z)|\ge 1\cdot 2\cdot 1=2$, so adding one does not change its winding. The first and third factors have winding one, since their constant terms have modulus smaller than their linear terms; the second has winding zero, since its $z$ term has modulus smaller than its constant term. Nonzero constants have winding zero and $z$ has winding one. Thus

Uniform convergence of $f_n$ to $f$ on an interval $I$ means that for every $ϵ>0$ there is $N$ such that $|f_n(x)-f(x)|<ϵ$ for all $n\ge N$ and every $x\in I$. Equivalently ${\sup }_{x\in I}|f_n(x)-f(x)|\to 0$. The same $N$ must work throughout the interval, unlike pointwise convergence.

For completeness, continuity is preserved: choose $N$ with uniform error below $ϵ/3$, and use continuity of $f_N$ at a given $x$ to make $|f_N(y)-f_N(x)|<ϵ/3$. The triangle inequality then gives $|f(y)-f(x)|<ϵ$. Thus each function involved in the integral is continuous on the closed interval and Riemann integrable. The estimateproves convergence of the integrals by uniform convergence. When $a=b$ both integrals are zero.

An equivalence relation on a set $A$ is a binary relation that is reflexive, symmetric and transitive: every element is related to itself, related pairs may be reversed, and two consecutive related pairs imply the pair of endpoints is related.

Write $T=R\circ S$, using the printed left-to-right composition of relations convention. Reflexivity implies $R\subseteq T$ and $S\subseteq T$, and hence $T$ is reflexive. Reversing a two-step chain givessince $R,S$ are symmetric. Also $R\circ R=R$ and $S\circ S=S$: one inclusion is transitivity, and the reverse follows by inserting a reflexive step. These elementary identities supply the equivalences below, without presuming that a composite of equivalence relations is automatically an equivalence relation.

An equivalence relation on $A$ is a binary relation that is reflexive, symmetric and transitive: $xRx$ for every $x$; $xRy\Rightarrow yRx$; and $xRy,yRz\Rightarrow xRz$. Write $T=R\circ S$, using the stated order for composition of relations. Composition is associative because both possible bracketings mean that there is a two-step sequence of intermediate witnesses. Reflexivity showsand also that $T$ is reflexive. Transitivity and reflexivity of the original relations give $R\circ R=R$, $S\circ S=S$. Since both original relations are symmetric,The four parts below show that each condition is equivalent to the commuting equivalence relations condition $R\circ S=S\circ R$.

For the integer example, put $d=\gcd (m,n)$. Membership of $(x,z)$ in the composite means that some $y$ satisfies $y-x=ma$, $z-y=nb$ for integers $a,b$. Consequently $z-x=ma+nb$ is divisible by $d$. Conversely, if $z-x$ is divisible by $d$, the Bezout identity expresses it as $ma+nb$; setting $y=x+ma$ provides the required witness. ThusReversing the roles of $m,n$ gives exactly the same modular congruence relation, hence $R\circ S=S\circ R$. This verifies all four conditions and identifies their common smallest equivalence relation explicitly.

When $\alpha =2\beta$, all three values reduce to $-\beta /3$. To prove path independence for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ givesIndeed ${\Phi }_x=2\beta (x^2+xy)=\alpha (x^2+xy)$ and ${\Phi }_y=\beta (x^2+y^2)$. The integrand is the exact differential $d\Phi$ throughout ${\mathbb{R}}^2$. The fundamental theorem for line integrals therefore gives, for every such path from $A$ to $B$,

Let $c=1/\sqrt{2}$. A concrete real matrix isIts first coordinate has eigenvalue $-1$. The lower block has characteristic polynomial $(\lambda -c{)}^2+c^2$, whose roots are $c\pm ic=(1\pm i)/\sqrt{2}$. Thus this example has all three required eigenvalues. It is also an orthogonal matrix, but orthogonality is not forced by its spectrum, as the second part shows.

A Cartesian isotropic tensor is unchanged by every proper rotation matrix. For a rank-two tensor this means $B^{\prime }=RB{R}^{\mathsf{T}}=B$, or $B_{ij}^{\prime }=R_{ik}{R}_{jl}{B}_{kl}=B_{ij}$. No symmetry of $B$ is assumed. For a quarter-turn about the $z$ axis,Comparing entries yields $B_{11}=B_{22}$, $B_{12}=-B_{21}$, $B_{13}=-B_{23}=-B_{13}$, and $B_{31}=-B_{32}=-B_{31}$. In particular,The remaining matrix has the form $\left(\begin{array}{ccc}b& c& 0\\ -c& b& 0\\ 0& 0& d\end{array}\right)$. A quarter-turn about the $x$ axis givesInvariance now forces $c=0$ and $d=b$, proving that an isotropic second-rank tensor isa scalar multiple of the Kronecker delta.

For the final unheaded conductivity identity, let $P=n{n}^{\mathsf{T}}$, the orthogonal projection onto the unit vector $n$. Then $P^2=P$ and the uniaxial conductivity tensor is $\sigma =\alpha I+\gamma P$. The antisymmetric tensor $D$ sends a vector $v$ to $v\times n$. Applying it twice and using the vector triple product givesConsequentlyThis equals $-\sigma =-\alpha I-\gamma P$ exactly when $(\alpha +\gamma )P=0$. Since $P$ has rank one and is nonzero, the requested parameter isAt this value conductivity vanishes along $n$; it is precisely the singular case excluded in part (ii), so there is no conflict with that part's conclusion.