With these parameters the expected number of $(n-1)$-cliques is asymptotic to $\sqrt{n}$, yet the probability of any such clique tends to zero. The complement graph has about $\frac{1}{2}\log n$ edges, and with probability tending to one two of them are vertex disjoint. Deleting one vertex cannot remove both edges. Thus divergence of the expected clique count alone does not give an appearance theorem for arbitrary growing clique size.

For $f(z)=(az+b)/(cz+d)$ with distinct finite fixed points of a Möbius transformation $\alpha ,\beta$, the fixed-point equation gives $f(z)-\alpha =(a-c\alpha )(z-\alpha )/(cz+d)$. Dividing the two identities gives $k=(a-c\alpha )/(a-c\beta )$, which is nonzero because $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The identities extend over the Riemann sphere. This conjugates the transformation to a dilation.

At design sites $i/n$, let ${\tilde{Y}}_n$ be the step function taking response $Y_i$ on $((i-1)/n,i/n]$. Project it onto an interval-adapted wavelet basis through resolution $J$. Each coefficient is ${\hat{c}}_{\lambda }={\sum }_i{Y}_i{\int }_{I_i}{b}_{\lambda }(t)dt$. Its expectation is the corresponding coefficient of the step function of sampled regression means. For a Haar wavelet basis, dyadic sample size and resolution at most the sampling scale, this is exactly $n^{-1}{\sum }_i{Y}_i{b}_{\lambda }(i/n)$ with right-closed pointwise versions. independent errors of variance ${\sigma }^2$ give coefficient variance at most ${\sigma }^2/n$: Cauchy-Schwarz inequality on each sampling cell yields ${\sum }_i({\int }_{I_i}{b}_{\lambda }{)}^2\le \Vert b_{\lambda }{\Vert }_2^2/n$.

On a unit-square uniform mesh with $h=1/(M+1)$ and zero Dirichlet boundary conditions, the semidiscrete five-point Laplacian has eigenvalues $-4h^{-2}[{\sin }^2(j\pi h/2)+{\sin }^2(k\pi h/2)]$. Adding reaction $\kappa I$ gives a contractive continuous-time grid evolution exactly under the displayed threshold. The threshold is $2{\pi }^2-{\pi }^4{h}^2/6+O(h^4)$, slightly below the continuum threshold. This is an all-time non-growth criterion; finite-time mesh-uniform bounds also hold for fixed larger reaction rates.

For a smooth solution with bounded fourth derivatives on the unit square, the five-point local residual is $O(h^2)$. The smallest eigenvalue of the unscaled Dirichlet matrix is $8{\sin }^2(\pi h/2)\ge 8h^2$. With $M^2$ interior points and $M\asymp h^{-1}$, the residual has unweighted Euclidean norm $O(h)$. The equation $Ae=h^2\tau$ therefore gives unweighted error $O(h)$, equivalent to second-order error in the area-weighted grid norm.

For positive diffusivity ratio $\zeta$ and $0<\phi <4$, define $\mu =(4-\phi ){\zeta }^2/\phi$. The nonzero steady branch has $b=a/(1+a^2)$, $c=a^2/(1+a^2)$, $d=\mu a/[\zeta (\mu +a^2)]$, and $e=a^2/(\mu +a^2)$. Substitution yields the displayed branch through $r=1+q$. Its initial slope in $a^2$ has the sign of $\phi q(2-\phi )+(4-\phi ){\zeta }^2(1+q)$. Positive slope is a supercritical pitchfork bifurcation direction, negative slope a subcritical pitchfork bifurcation direction. This geometric criterion alone does not assert stability against every perturbation.

The conduction linearization has two negative eigenvalues and a cubic factor $s^3+A{s}^2+Bs+C$, where $A=\sigma +1+\zeta$, $B=\sigma (1-r)+\sigma \zeta (1+q)+\zeta$, $C=\sigma \zeta (1+q-r)$. A double zero requires $B=C=0$, giving the displayed Bogdanov–Takens bifurcation location for $0<\zeta <1$. The remaining cubic eigenvalue is $-A$. The imaginary-pair condition $AB=C$ yields $r_H=(\sigma +\zeta )(1+\zeta )/\sigma +\zeta (\sigma +\zeta )q/(\sigma +1)$ and ${\omega }^2=\sigma \zeta (1-\zeta )q/(\sigma +1)-{\zeta }^2$. Thus the oscillatory onset exists precisely for $q>q_{\ast }$ when $0<\zeta <1$.

Fisher scoring replaces the observed negative Hessian in Newton iteration by the Fisher information:

Two statistical parameters are orthogonal when their cross term in the expected Fisher information matrix vanishes. This is pointwise orthogonality at a specified parameter value, or global orthogonality when it holds throughout the parameter domain. It gives a zero cross covariance in the regular first-order asymptotic distribution of efficient estimators, rather than asserting exact finite-sample independence.

For $n\ge 2$ risk-neutral bidders with independent uniform valuations on $[0,1]$, this rule is a symmetric Bayesian Nash equilibrium. If opponents use $\beta (v)=av$, a bid $b\in [0,a]$ has expected utility $(v-b)(b/a{)}^{n-1}$. Its derivative has the sign of $(n-1)v-nb$, so with $a=(n-1)/n$ the best bid is $av$. Bids above $a$ guarantee winning but only increase payment. The type-$v$ equilibrium utility is $v^n/n$.

In a first-price sealed-bid auction with at least two bidders, continuous nonnegative bids, quasilinear utility and positive private value $v$, no bid is a dominant strategy. Against opponents all bidding zero, any positive bid can be lowered while still winning, increasing the payoff. Bid zero fails against an opposing bid $v/2$, since bidding $3v/4$ earns $v/4>0$. Nor does randomizing create a dominant strategy. Against all-zero opponents, a random bid has expected payoff strictly below $v$ unless it is identically zero and receives the item with certainty under the tie rule. In the strict case, a sufficiently small positive fixed bid earns more. In the exceptional case, the deviation against $v/2$ again improves the payoff. Finite expected utility is implicit; an expected payoff of negative infinity cannot dominate finite-payoff bids. The positive-value condition matters: a zero-value bidder weakly prefers bid zero.

Here the convention places the material index first: $P_{Ii}$. Its stress power per reference volume is $P_{Ii}{\dot{F}}_{iI}$. It is the transpose of the first Piola-Kirchhoff stress tensor; some texts instead call that first tensor itself nominal stress.

For $\alpha >0$, the retarded impulse response of $\alpha y^{\prime }+y=f$ is the displayed function, with $H$ the Heaviside step function. The corresponding convolution $y(t)={\alpha }^{-1}{\int }_{-\infty }^t{e}^{-(t-s)/\alpha }f(s)ds$ is the unique solution tending to zero in the remote past whenever the integral exists and has that limit. For example, bounded forcing that tends to zero as $t\to -\infty$ satisfies these conditions. A general forcing need not: $f\equiv 1$ admits no such solution. A unit-duration pulse with $\alpha =1$ gives $y=0$ before zero, $y=1-e^{-t}$ between zero and one, and $y=(1-e^{-1})e^{-(t-1)}$ thereafter.

For a closed subscheme $Z\subseteq X$ with ideal sheaf $\mathcal{I}$, one has $T_{[Z]}\mathrm{Hilb}(X)\cong {\mathrm{Hom}}_{{\mathcal{O}}_X}(\mathcal{I},i_{\ast }{\mathcal{O}}_Z)\cong H^0(Z,{\mathcal{N}}_{Z/X})$. Locally a flat deformation ideal $J\subseteq A[\epsilon ]/({\epsilon }^2)$ satisfies $J\cap \epsilon A=\epsilon I$. Lifting $f\in I$ to $f+\epsilon g\in J$ gives a well-defined map $f\mapsto g\mathrm{mod}I$ killing $I^2$. Conversely the graph condition $g\mathrm{mod}I=\varphi (f)$ defines an ideal with flat quotient by the flatness criterion over dual numbers. These operations commute with restriction and are inverse, giving the global identification without a smoothness assumption.

Expanding associativity for $a\ast b=ab+tf(a,b)+t^2{\mu }_2(a,b)$ modulo $t^3$ gives the displayed equation. If $\delta f=0$, its right side is a degree-three Hochschild cocycle. Its class in $H{H}^3(R,R)$ must vanish to extend the first-order formal associative deformation to second order. A nonzero class in $H{H}^2$ therefore need not integrate to a full formal associative deformation.

For a module over a commutative ring presented by a matrix with n generators, its zeroth Fitting ideal is generated by all n-by-n minors of the relation matrix. It is independent of the chosen finite presentation. Over a unique factorization domain, the greatest common divisor of these minors defines the module order up to a unit. This does not require a Smith normal form over the ring. The square Seifert-matrix presentation of the Alexander module makes the order its determinant.

// Target: knot-theory.bigb

Suppose $G=⟨Y\mid R(Y)⟩$ is a finite group presentation, and $X$ is any finite generating set. Write $y_j=u_j(X)$ and $x_i=v_i(Y)$. A finite presentation on $X$ is obtained using the relators $R(u(X))$ and $x_i=v_i(u(X))$. The substitutions in both directions give inverse homomorphisms, proving that finite presentability does not depend on the finite generators chosen.

// Target: geometry-and-topology.bigb

These upper and lower quantities use $d(H)$, without subtracting one. For a free group $F_n$ with $n\ge 1$, the Nielsen–Schreier formula gives $d(H)/[F_n:H]=n-1+1/[F_n:H]$. The upper value is $n$, attained at the whole group, and the lower value is $n-1$, approached by kernels of maps onto cyclic groups of arbitrarily large finite order. The frequently used shifted rank gradient replaces $d(H)$ by $d(H)-1$ and therefore gives different upper values.

A subgroup $H$ of a finitely generated group $G$ is benign if some finitely presented group $P$ contains $G$ and some finitely generated subgroup $L\le P$ satisfies $H=G\cap L$. The subgroup $H$ itself need not be finitely generated. A centralizing HNN extension of $P$ along $L$ is finitely presented even when the corresponding intersection subgroup $H$ is infinitely generated.

The Fundamental theorem of finitely generated abelian groups writes $A={\mathbb{Z}}^r\oplus F$ with $F$ finite. Reduction modulo a prime contains $(\mathbb{Z}/p\mathbb{Z}{)}^r$, so a vanishing quotient forces $r=0$. Conversely multiplication by a prime coprime to $|F|$ is bijective on $F$. Finite generation matters: the nonzero divisible group $(\mathbb{Q},+)$ satisfies $p\mathbb{Q}=\mathbb{Q}$ for every prime and is not finitely generated, since any finite list of rational generators has a common denominator.