For each vertex triple $S$, let $I_S$ indicate that its three edges are present. Then $X={\sum }_S{I}_S$ andEach variance is $p^3(1-p^3)$. Two distinct triples have independent indicators unless they share an edge; in that case their covariance is $p^5-p^6$. There are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)=6\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ unordered pairs sharing an edge. Therefore the triangle count in a binomial random graph hasIf $np\to 0$, Markov inequality gives $\mathrm{Pr}(X>0)\le \mathbb{E}X\to 0$. If $np\to \infty$, then $\mathbb{E}X\to \infty$ andThe second term tends to zero since $n^{2}p=n(np)\to \infty$. Chebyshev inequality now gives $\mathrm{Pr}(X=0)\le \mathrm{Var}X/(\mathbb{E}X{)}^2\to 0$, proving the stated threshold. Here the random-graph phrase “almost surely” means asymptotically almost surely, that is, with probability tending to one.

At $p=n^{-1/2}$, let $Y$ count edges. Direct subtraction givesSome graph has $Y-X$ at least this expectation. Successively deleting one edge from each remaining triangle destroys all triangles with at most $X$ deletions, so a triangle-free graph remains with at least $Y-X$ edges. This triangle deletion in the alteration method proves the requested existence bound.

It is not within a constant factor of the optimum. A balanced complete bipartite graph has $\lfloor n^2/4\rfloor$ edges and no triangle. Conversely, in a triangle-free graph each edge $uv$ satisfies $d(u)+d(v)\le n$. Summing over edges gives ${\sum }_{v}d(v{)}^2\le nM$, where $M$ is its edge count. Cauchy-Schwarz inequality gives $4M^2/n\le \sum d(v{)}^2$, hence $M\le n^2/4$. The optimum is therefore $\boxed{\lfloor n^2/4\rfloor }$, quadratic rather than the obtained $n^{3/2}$ bound.

A transitive class $M$ satisfies $x\in M\Rightarrow x\subseteq M$. Its membership structure therefore does not omit elements of its members. In particular bounded membership statements are absolute between a transitive model and the ambient universe, and its membership relation is the actual well-founded relation, rather than an arbitrary externally ill-founded coding.

For a set $x$, put $x_0=x$ and $x_{n+1}=\bigcup x_n$, and defineThis is a set, contains $x$ as a subset, and is transitive: an element of an element of $x_n$ belongs to $x_{n+1}$. Any transitive set containing $x$ as a subset contains every $x_n$ by induction, so this construction is the least such set. Shifting the union index givesThese formulas use the specified subset convention for transitive closure.

Work in the usual ambient set theory with Choice. If $x$ is a hereditarily countable set, then every member of $\mathrm{TC}(\{x\})$ is countable. Iterated countable unions show that this entire closure is countable. Conversely a countable transitive closure makes each of its members countable. This equivalent criterion makes the following closure arguments explicit.

If $y\in x\in \mathrm{HC}$, then $\mathrm{TC}(\{y\})\subseteq \mathrm{TC}(\{x\})$, so $y\in \mathrm{HC}$: HC is transitive. If $x\in \mathrm{HC}$, its union is countable and its descendant sets are descendants of $x$, so $\bigcup x\in \mathrm{HC}$. The actual union therefore witnesses the Axiom of union inside $(\mathrm{HC},\in )$.

For axiom schema of separation, take any formula with parameters in HC and a set $x\in \mathrm{HC}$. Form the subset of $x$ consisting of the elements that satisfy that formula with all quantifiers relativized to HC. Ambient Separation supplies this set. It is a subset of a countable set, and all its descendants are inherited from $x$, so it belongs to HC. No claim that the formula itself is absolute is needed.

The axioms satisfied by hereditarily countable sets also include Extensionality, Empty Set, Pairing, Infinity, Foundation and Replacement. Extensionality and Foundation follow from transitivity and ambient well-foundedness. Empty Set and $\omega$ belong to HC, and a pair of HC sets has countable transitive closure. For Replacement, the domain is countable; a definable image of its elements is a countable family of HC sets, whose combined transitive closures form a countable union of countable sets, so the range again belongs to HC. Ambient Choice additionally gives Choice internally: a well-order of a countable HC set has a hereditarily countable relation and hence lies in HC.

Power Set fails. Every subset of $\omega$ is hereditarily countable, so an internal power-set witness would have to contain all ambient subsets of $\omega$. That collection is uncountable by Cantor theorem, whereas every HC set is countable. Thus HC satisfies the other ZF axioms and, in this ambient setting, Choice, but not the Axiom of power set.

Put $\pi =p-eA/c=m\dot{q}$. The Hamilton equations areDifferentiate the mechanical momentum and substitute:With $B=\nabla \times A$ and $E=-c^{-1}{\partial }_{t}A$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot{q}=eE+(e/c)\dot{q}\times B}$, the Lorentz force in the units of the Hamiltonian.

For $A=(-y{B}_0(z,t),0,0)$, $x$ is cyclic, so $p_x$ is conserved. Choose $p_x=0$, giving $\dot{x}=\Omega y$, where $\Omega =e{B}_0/(mc)$. Since $p_y=m\dot{y}$, its Hamilton equation gives ${\dot{p}}_y=-(e/c)B_0\dot{x}=-m{\Omega }^{2}y$, henceFor nonzero constant $B_0$, the $(y,p_y)$ orbit is an ellipse of semiaxes $\sqrt{2E^{\prime }/(m{\Omega }^2)}$ and $\sqrt{2m{E}^{\prime }}$. Its enclosed phase-space area is $2\pi E^{\prime }/|\Omega |$, so the action variable isThe absolute value keeps the action positive if the signed charge or magnetic field is negative.

When the field varies slowly along the trajectory, with $|d\Omega /dt|\ll {\Omega }^2$ and no zero-frequency crossing, adiabatic invariance of the action makes $I$ approximately constant. Thus $E^{\prime }\simeq I|\Omega (z,t)|$ and the slow motion has effective HamiltonianThe effective potential is the second term. Direct averaging checks its force: the exact equation is $m\ddot{z}=-m\Omega {\partial }_z\Omega y^2$, while the oscillator has $⟨y^2⟩=E^{\prime }/(m{\Omega }^2)$. Hence $m\ddot{z}=-I{\partial }_z|\Omega |$. Spatial field variation along the slow motion must satisfy the same adiabatic condition; slow explicit time variation alone does not control passage through a sharp spatial gradient. A time-dependent effective potential need not conserve total energy.

An interval covering relation $I\to J$ means $F(I)\supseteq J$. The associated directed graph has one vertex for each interval and an edge $i\to j$ for each covering. Its transition matrix is $A_{ij}=1$ when that edge exists, and zero otherwise. Matrix multiplication counts walks; thus $\boxed{\mathrm{tr}(A^n)}$ counts closed length-$n$ itineraries, with a distinguished starting vertex.

To see how itineraries force periodic orbits, a continuous map covering a closed interval has a closed subinterval mapping onto it: choose successive first/last crossings of its endpoints. Pull these subintervals back along a closed walk. This yields $K\subseteq I_{i_0}$ with $F^n(K)\supseteq I_{i_0}\supseteq K$. At the two preimages of the ends of $K$, $F^n(x)-x$ has opposite weak signs; the intermediate value theorem gives a fixed point of $F^n$.

A graph alone does not give an exact count of periodic points for a general continuous map: extra oscillations or a fixed interval can create additional points. For example on $I=[0,1]$, both $F(x)=x$ and $F(x)=1-x$ have matrix $A=(1)$; the first has infinitely many fixed points and the second only one. Distinct itineraries give distinct points when their interval interiors are disjoint and the periodic orbit avoids shared endpoints. In that setting, $\mathrm{tr}(A^n)$ is a guaranteed lower bound, not a universal equality. Likewise, the primitive-word count $(1/n){\sum }_{d\mid n}\mu (d)\mathrm{tr}(A^{n/d})$ counts guaranteed least-period itineraries under this separation condition. With overlapping intervals, even distinctness needs qualification. This is the scope of counting cycles in an interval covering graph.

For the first five-cycle, put $I_j=[x_j,x_{j+1}]$, $0\le j\le 3$. The graph and matrix areThe four traces are $1,3,7,15$. Shared endpoints have least period five, so they do not spoil the distinctness argument for the requested periods. Primitive cyclic itineraries are: $3$ for period one; $(3,2)$ for period two; $(3,1,2)$ and $(3,3,2)$ for period three; $(0,1,2,3)$, $(3,3,1,2)$ and $(3,3,3,2)$ for period four. Consequently the monotone five-cycle interval covering pattern guaranteesThese words specify the successive intervals containing their points; cyclic rotations denote the same orbit.

There is a horseshoe for an interval map for $F^2$: both $I_2$ and $I_3$ cover $I_2\cup I_3$ under that iterate, and their interiors are disjoint. Repeated binary choices produce at least $2^n$ separated length-$n$ itineraries for the second iterate, hence positive topological entropy, at least $(\log 2)/2$ for $F$. Thus the dynamics are chaotic in this standard interval-map sense. No conclusion of global transitivity on all of $\mathbb{R}$ follows.

A one-step horseshoe is not guaranteed if that is the intended definition. For example the unimodal continuous map $F(x)=x+1$ for $x\le 3$ and $F(x)=16-4x$ for $x\ge 3$ has cycle $0,1,2,3,4$. A left-branch interval cannot cover itself, since its image is shifted to the right. Two right-branch intervals cannot each cover their union because that branch is injective. If one interval straddles the maximum and covers itself, its right endpoint must map below its left endpoint; every disjoint interval farther to the right then maps below it and cannot self-cover. A disjoint interval to the left also cannot self-cover. So this example has no one-step two-branch horseshoe, although its second iterate does. It also proves that $F$ can be unimodal.

For the second order, write $J_0=[x_3,x_1]$, $J_1=[x_1,x_0]$, $J_2=[x_0,x_2]$, $J_3=[x_2,x_4]$. Endpoint images giveThe traces are $1,3,1,7$. The guaranteed primitive words are $1$, $(0,3)$ and $(1,2,0,3)$, giving $\boxed{\text{periods }1,2,4\text{ but not necessarily }3}$. To prove the last qualification, take ordered points $0,1,2,3,4$ and linearly interpolate the map values $4,3,1,0,2$ on $[0,4]$, extending by $G(x)=4-x$ for $x<0$ and $G(x)=2$ for $x>4$. Both exterior regions enter $[0,4]$ after at most two steps, so they introduce no extra periodic points. Its only closed length-three itinerary is $111$; on $J_1$ it is $x\mapsto 5-2x$, whose third iterate has only the same fixed point. Endpoints have least period five, so this example has no three-cycle.

The distinct fourth-iterate return walks $1111$ and $1203$ yield two subintervals of $J_1$ that return over $J_1$, giving a horseshoe for $G^4$ and positive entropy at least $(\log 2)/4$. Thus $G$ too must have chaotic interval dynamics and a horseshoe in some iterate. The affine example has no one-step horseshoe: its increasing branch lies strictly below the diagonal, while on the decreasing branch injectivity prevents two self-covering intervals; a minimum-straddling self-covering interval excludes a self-covering interval on either side by the same endpoint argument. The distinction between a map and its iterate is essential for both patterns.

Put $M=X^{T}X$, which is positive definite because $X$ has full column rank. The log likelihood differs from $-\Vert Y-X\beta {\Vert }^2/(2{\sigma }^2)$ by a term independent of $\beta$. Differentiation gives the normal equations $M\hat{\beta }=X^{T}Y$, henceThe distribution follows by writing $\hat{\beta }=\beta +M^{-1}{X}^T\epsilon$ and applying the linear transformation rule for a multivariate normal distribution.

Let $s={\sigma }^2$ and $R=\Vert Y-X\hat{\beta }{\Vert }^2$. Orthogonality of the residual to the columns of $X$ gives the exact decompositionThe prior is density proportional to $s^{-1}$ with respect to $d\beta ds$. Thus the joint posterior distribution isCompleting the quadratic identifiesThe sampling distribution treats the true $\beta$ as fixed and varies the data, whereas the conditional posterior fixes the observed data and distributes $\beta$. Their covariance formulas agree, but their centres and their interpretations differ.

For a new row vector $x^{\ast }$, let $h^{\ast }=x^{\ast }{M}^{-1}(x^{\ast }{)}^T$ and suppose its new error is independent. With known ${\sigma }^2$, the maximum-likelihood plug-in prediction uses mean $x^{\ast }\hat{\beta }$ and noise variance ${\sigma }^2$, while a calibrated frequentist prediction must account for the estimator: $y^{\ast }-x^{\ast }\hat{\beta }\sim N(0,{\sigma }^2(1+h^{\ast }))$. If variance is estimated, the maximum-likelihood estimator is $R/n$, whereas the exact frequentist prediction pivot uses $\hat{s}=R/(n-p)$ and has Student t-distribution with $n-p$ degrees of freedom.

In the Bayesian calculation, integrating $\beta$ first gives $y^{\ast }\mid s,X,Y\sim N(x^{\ast }\hat{\beta },s(1+h^{\ast }))$. Integrating $\beta$ out of the joint posterior gives $s\mid X,Y\sim \mathrm{InvGamma}((n-p)/2,R/2)$. For $\nu =n-p$, integration of the predictive normal against this inverse-gamma distribution has density proportional toThe substitution $t=[R+(y^{\ast }-x^{\ast }\hat{\beta }{)}^2/(1+h^{\ast })]/(2s)$ reduces the integral to a Gamma function. Normalizing identifies the Student t-distribution, givingwhere the second parameter is a scale, not its variance. These Bayesian intervals coincide with the usual exact Gaussian-regression prediction intervals for this prior, although their probability interpretations differ. Posterior propriety requires $R>0$ and $n>p$; the former holds almost surely under the nondegenerate Gaussian model. Prediction of the noiseless conditional mean omits the added one in $1+h^{\ast }$.

Define the Möbius function by $\mu (1)=1$, $\mu (n)=(-1{)}^r$ for a product of $r$ distinct primes, and $\mu (n)=0$ if a prime square divides $n$. For $n>1$, summing over square-free divisors gives ${\sum }_{d\mid n}\mu (d)=(1-1{)}^r=0$, while the sum is one at $n=1$.

Both Dirichlet series $\sum n^{-s}$ and $\sum \mu (n)n^{-s}$ converge absolutely for $\sigma =\mathrm{Re}s>1$. Their product may therefore be regrouped by the product of the indices:This simultaneously proves $\boxed{\zeta (s)\ne 0\text{ for }\sigma >1}$ andIn particular $|1/\zeta (\sigma +it)|\le \sum |\mu (n)|n^{-\sigma }\le \zeta (\sigma )$.

Under the assumed analytic continuation, $\zeta$ is holomorphic near $s_0=1+i{t}_0$ when $t_0\ne 0$, and $\zeta (\sigma )=1/(\sigma -1)+O(1)$ as $\sigma \downarrow 1$. If $s_0$ were a zero of order $m\ge 2$, local factorization would give $|1/\zeta (s_0+h)|\ge c{h}^{-m}$ for sufficiently small positive $h$. The preceding absolutely convergent bound gives instead $|1/\zeta (1+h+i{t}_0)|=O(h^{-1})$, a contradiction. Thus $\boxed{m\le 1}$. This reciprocal Dirichlet-series bound on boundary-zero multiplicity does not assert that any such boundary zero exists.

First suppose the integral of $g(\xi )/(\xi -z)$ converges, for example if $g\in L^1(\mathbb{R})$. Substitute the complexified coordinates $x=(z+\bar{a})/2$, $y=(z-\bar{a})/(2i)$ into the Poisson integral. The factorization of its denominator givesThe given Schwarz integral formula therefore writes $f(z)$ as $(i\pi {)}^{-1}\int g(\xi )/(\xi -z)d\xi$ plus a constant. For $z=x+iy$, the real part of this integral is exactly the Poisson integral, sinceThe remaining constant has zero real part, giving $\boxed{f(z)=(i\pi {)}^{-1}{\int }_{\mathbb{R}}\frac{g(\xi )}{\xi -z}d\xi +ic}$ with $c\in \mathbb{R}$. The complex substitution is initially understood where the real-analytic harmonic function has its complexification; the holomorphic integral then continues the identity throughout the upper half-plane.

Decay of $u$ alone does not guarantee convergence of the printed unregularized integral. For example, choose a continuous $g$ which vanishes on $(-\infty ,e]$ and equals $1/\log \xi$ for $\xi \ge e^2$, with continuous interpolation. It is bounded and tends to zero at both ends. Its Poisson integral tends to zero as $|z|\to \infty$ in the closed upper half-plane: split the boundary function into a small uniform tail and a compactly supported part. Nevertheless ${\int }^{\infty }g(\xi )/(\xi -z)d\xi$ diverges like $\log \log \xi$, even as a symmetric principal value. With only the stated decay assumptions, the always convergent version isThe subtracted kernel is real on the integration line, so it changes only the imaginary constant and leaves the harmonic function $u$ unchanged. The kernel difference is $O({\xi }^{-2})$, which ensures convergence for bounded $g$.

The fixed points satisfy $x[rx(1-x)-1]=0$. There is always $x=0$, and the other branches areThey coincide at $(r,x)=(4,1/2)$. For $0<x<1$, $F(x)>0$ and its maximum occurs at $x=2/3$, with value $4r/27$. Consequently $\boxed{F((0,1))\subset (0,1)⟺0<r<27/4}$; equality at the upper endpoint fails because the maximum is attained inside the interval.

The fixed-point multiplier at zero is zero, so zero is attracting for every $r>0$. On either nonzero branch,The lower branch has multiplier greater than one for $r>4$ and is repelling. The upper branch is attracting for $4<r<16/3$, reaches multiplier $-1$ at $x=3/4$, and is repelling for $r>16/3$. At $r=4$ the multiplier is $+1$, and the two branches are created in a saddle-node bifurcation. The fixed point there is attracting from one side and repelling from the other: $F(x)-x=-4x(x-1/2{)}^2$.

The second change is a supercritical period-doubling bifurcation at $\boxed{(r,x)=(16/3,3/4)}$. To check its type, expand at that fixed point. The local map has quadratic coefficient $a=-20/3$ and cubic coefficient $b=-16/3$ when the multiplier is $-1$. Its second iterate has cubic coefficient $-2(a^2+b)=-704/9<0$. The multiplier decreases through $-1$ as $r$ increases, so a small attracting two-cycle is born on the $r>16/3$ side. These exhaust the fixed-point bifurcations, since their multipliers have no other crossings of $\pm 1$.

Let $p_j^n$ be the probability of occupying site $ja$ after $n$ steps, with $p_j^0=1_{\{j=0\}}$ and $\alpha +\beta =1$. The master equation isWrite $p_j^n\approx ap(ja,n\tau )$ so that the limiting $p$ is a probability density. A Taylor expansion of this master equation givesThe weakly biased diffusion limit of a weakly biased random walk keeps $a^2/\tau$ and $(\alpha -\beta )a/\tau$ finite as $a,\tau \to 0$, so $\alpha -\beta =O(a)$. The limiting advection-diffusion equation isIts initial condition is the Dirac delta function at the origin. At finite spacing, the exact variance gained per step is $a^2[1-(\alpha -\beta {)}^2]=4a^2\alpha \beta$. Thus a finite-step central limit theorem gives the variance-matched coefficient $2a^2\alpha \beta /\tau$; its difference from the displayed $D$ vanishes in the stated weak-bias diffusion limit of a weakly biased random walk. Keeping a fixed nonzero bias while taking diffusive space-time scaling would instead make $V$ diverge.

Partition the design matrix accordingly as $X=(X_1\cdots X_k)$. The blocks ${\beta }_i,{\beta }_j$ are orthogonal when $X_i^{\mathsf{T}}{X}_j=0$, meaning that their column spaces are orthogonal vectors. Mutual orthogonality means that this holds for every pair of distinct blocks, so $X^{\mathsf{T}}X$ is block diagonal.

For the three scalar coefficients, the corresponding columns are $1,x_1,x_2$. The necessary and sufficient conditions areUnder the assumed full rank, neither explanatory column is zero. The maximum-likelihood estimator is $\hat{\beta }=(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}Y$, and the normal distribution of $Y$ impliesThus the three coefficient estimators are independent normal random variables, not merely uncorrelated. This is a sampling-distribution statement with the true ${\sigma }^2$ fixed.

Use the binary Hamming code $C=\ker H\subset {\mathbb{F}}_2^{15}$, where the $j$th column $h_j$ of the $4\times 15$ parity-check matrix $H$ is the four-digit binary expansion of $j$, for $1\le j\le 15$. Thus the columns are precisely the nonzero vectors of ${\mathbb{F}}_2^4$. They include the four coordinate vectors, so $\mathrm{rank}H=4$ and $\dim C=15-4=11$. Its code rate is therefore $\boxed{11/15}$.

For a received word $y$, compute its syndrome $s=Hy$. If $s=0$, report no error. Otherwise identify the unique column $h_j=s$ and flip bit $j$. Indeed, if $y=c+e_j$ with $c\in C$, then $Hy=Hc+H{e}_j=h_j$, so this procedure recovers $c$ exactly. No column is zero and no two columns coincide; hence no nonzero codeword has Hamming weight one or two. Some three columns sum to zero, for example those indexed by $1,2,3$, so the minimum distance is exactly three. Thus every single-bit error is corrected. The assertion requires at most one error; arbitrary multiple errors need not be detected by this procedure.

Consider the Möbius transformationsTheir group lies in ${\mathrm{PSL}}_2(\mathbb{Z})$, which is a discrete subgroup of the Möbius group: integer entries cannot approach the identity matrix except by eventually equalling it, and passing to the quotient by $\{I,-I\}$ preserves this property.

For freeness let $X=\{x\in \mathbb{R}:|x|>1\}$ and $Y=\{x\in \mathbb{R}:|x|<1\}$. For every nonzero integer $n$, $A^n(Y)=(2n-1,2n+1)\subset X$ andA reduced word involving both generators can be conjugated to a cyclically reduced word beginning with $A^n$ and ending with a nonzero power of $B$, after interchanging the generators if necessary. Repeated use of these inclusions shows that this word sends $X$ into the proper subset $A^n(Y)$ of $X$, so it cannot be the identity. Nonzero pure powers of either generator are also nonidentity. This is the ping-pong lemma in this particular action, and proves that $\boxed{⟨A,B⟩\cong F_2}$ is a free group of rank two as well as a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$.