Moment-free ends require ${\zeta }^{\prime \prime }={\kappa }_0$ there, so a highly stretched finite filament can retain curvature in boundary layers of width $\ell =\sqrt{B/F}$. For constant intrinsic curvature $\kappa$, zero transverse end forces give slope ${\zeta }^{\prime }(x)=\kappa \ell \sinh [(x-L/2)/\ell ]/\cosh [L/(2\ell )]$. Direct integration yieldsIts large-force deficit is ${\kappa }^2{\ell }^3/2$, of order $F^{-3/2}$, even though the smooth bulk curvature derivative vanishes. This shows why endpoint assumptions matter in force-extension of an intrinsically curved filament.

For spherical thermal hydrostatic equilibrium, $dP/dr=-{\rho }_{g}GM(<r)/r^2$. Inserting the ideal gas law $P={\rho }_g{k}_{B}T/(\mu m_p)$ with constant mean molecular weight gives the displayed estimator. The density ${\rho }_g$ is gas density, while $M$ includes every gravitating component. A varying $\mu$ adds $-d\log \mu /d\log r$ inside the logarithmic bracket. Nonthermal pressure requires differentiating the total supporting pressure rather than thermal pressure alone.

A sequence of nested intervals keeps every new enclosure inside the previous one:Each step adds a restriction without discarding an earlier restriction. This is useful when an exact real number is described by increasingly precise bounds rather than by a finite decimal expansion. The bisection method, for instance, produces nested intervals that keep enclosing a zero of a function.

For a simple example, the closed intervals $I_n=[1-1/n,1+1/n]$ are nested and all contain $1$. Their lengths $2/n$ tend to zero, so the nested interval theorem leaves exactly that point. The family $J_n=[1-1/n,2+1/n]$ is also nested, but its intersection is the whole closed interval $[1,2]$. Nesting alone does not force uniqueness.

The open intervals $K_n=(0,1/n)$ illustrate a different possibility. Every finite collection has a nonempty intersection, but the intersection of all of them is empty: any positive candidate eventually exceeds $1/n$, and zero is excluded from every member. Each finite row in the diagram contains points; the full infinite family need not.

For a continuous function from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The intermediate value theorem gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I{J}^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.

The interval square range treats squaring as one function of one input. For $X=[a,b]$ it is exactlyIt retains the relationship that ordinary interval multiplication loses: for $Z=[-1/2,1/2]$, the interval square range is $[0,1/4]$, whereas the independent-input product $ZZ$ is $[-1/4,1/4]$.

Consequently, completing the square for $x(1-x)$ and using the interval square range givesThis is the exact image of a function on $X=[0,1]$. Merely rewriting the expression and evaluating $[1/4,1/4]-ZZ$ gives $[0,1/2]$, so it does not achieve the same sharpening. A mathematical $Z^2$ must be interpreted explicitly here: a square range operation and an independent-input interval product have different meanings. Choosing an operation that preserves known structure can save many steps of interval subdivision.

A natural interval extension evaluates a chosen expression by replacing its arithmetic operations with interval arithmetic operations, its constants with enclosing closed intervals, and any additional functions with valid interval enclosures. Each intermediate result must remain defined on its inputs. The resulting $F(X)$ encloses the actual image of a function $f(X)$, but its width depends on the expression used. Rump's INTLAB tutorial describes this inclusion principle and warns that repeated inputs can produce severe overestimation.^[ref]

For a proper morphism of varieties $f:X\to Y$, a Cartier divisor $D$ on $Y$, and an algebraic cycle $\alpha$ on $X$, pullback of $D$, intersection with $\alpha$, and proper pushforward satisfy the displayed identity in the Chow group. In particular, for a generically finite morphism of surfaces of degree $r$, $(f^{\ast }D{)}^2=r{D}^2$. For a curve $C$ contracted to a point, $f_{\ast }C=0$, so $f^{\ast }D\cdot C=0$.

For a proper integral Cartier curve $C$ on a smooth algebraic surface, its self-intersection is $\mathrm{deg}{\mathcal{O}}_C(C)$. If $C$ is smooth, this is the degree of its normal line bundle. It is defined over any algebraically closed field, without requiring an associated real or complex manifold.

// Target: geometry-and-topology.bigb

For $\dim L=2n-1$ and lower middle perversity, the suspension with two distinct vertices has $I{H}_i(\mathrm{Susp}L)=I{H}_i(L)$ for $i<n$, zero for $i=n$, and $I{H}_{i-1}(L)$ for $i>n$. Cover it by its two open cones. The Mayer–Vietoris sequence compares their truncated groups with $I{H}_{\ast }(L)$; below $n$ the overlap map is the injective signed diagonal, and above $n$ both cone groups vanish.

Here $H$ is the hyperbolic form with matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The degree-two cohomology consists of the two factor orientation classes and $4gh$ products of degree-one classes. The factor classes pair once and have zero squares. In symplectic bases $a_i,b_i$ and $c_j,d_j$, $(a_i{c}_j)(b_k{d}_l)=-{\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$ and $(a_i{d}_j)(b_k{c}_l)={\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$. Each pair $(i,j)$ contributes two hyperbolic forms; the fiber classes contribute one more. The sign in the first formula comes from interchanging two odd-degree factors.

For a closed oriented smooth four-manifold, $b_2^{+}$ is the dimension of a maximal positive-definite subspace of its real intersection form. A symplectic form forces $b_2^{+}\ge 1$, because its cohomology class has positive square. Reversing orientation exchanges the positive and negative indices.

For a simply connected closed four-manifold with second Betti number one, Poincare duality and the universal coefficient theorem for cohomology make $H_2$ torsion-free of rank one. The integral intersection form is unimodular, so its one-by-one matrix is $[1]$ or $[-1]$. Hence the square of a generator $u\in H^2$ generates $H^4$, giving the displayed cohomology ring. This is a ring isomorphism; it need not preserve a preassigned orientation class.

The degree-zero normalized Deligne intersection complex extends the constant sheaf from the regular stratum by successive derived direct images and cohomological truncations at degree $\bar{p}(k)$ when codimension-$k$ strata are added. Its hypercohomology defines intersection cohomology. On a cone with vertex of codimension $k$, its local cohomology is the link intersection cohomology in degrees at most $\bar{p}(k)$ and zero above.

Let a compact oriented $(2m+1)$-manifold $M$ have collared boundary $B$, and cap $B$ by its closed cone. In the lower-middle Deligne convention the result has $I{H}^i=H^i(M)$ for $i<m$, $I{H}^m=\mathrm{im}(H^m(M,B)\to H^m(M))$, and $I{H}^i=H^i(M,B)$ for $i>m$. Excision identifies the relative intersection groups with $H^{\ast }(M,B)$; the cone neighbourhood has $H^i(B)$ only for $i<m$, so its long exact sequence gives the formula. The cap is Witt exactly when $H^m(B;\mathbb{Q})=0$.

A set family is t-intersecting if every pair of its members has at least $t$ common elements. For $r$-sets on an $n$-point ground set, the whole level is t-intersecting when $n\le 2r-t$. Ordinary intersecting families are the case $t=1$.

A maximal intersecting family is an up-set. It cannot contain both a set and its complement. If neither were present, maximality would supply a member disjoint from the first set, and upward closure would then include its complement. Thus precisely one member of each complementary pair occurs, so the family has $2^{n-1}$ members for $n\ge 1$.

For $1\le r\le n/2$, the Erdős-Ko-Rado theorem bounds an intersecting family by the displayed size. The first that many rank-$r$ sets in lexicographic order all contain coordinate one. Therefore the lexicographic initial segment of the same size as any intersecting family is also intersecting. This statement concerns ordinary one-element intersection and does not extend to higher intersection requirements.

On five points, the four sets $12,13,14,15$ are an intersecting family. The first four two-element sets in colexicographic order are $12,13,23,14$, containing the disjoint pair $23,14$. Thus replacing a family by an equally large colexicographic initial segment can destroy intersection, even though it minimizes the lower shadow.

For intersecting families of subsets of $[n]$, take each upward closure of a set family. It remains intersecting and contains at most one set in each complementary pair, so its complement as a family has density at least one half. Those complements are decreasing set families. Repeated Harris-Kleitman inequality gives intersection density at least $2^{-k}$, proving the displayed bound. For $k\le n$, the families of sets containing coordinate $i$, for $1\le i\le k$, attain equality.

Choose $k,m>0$, $K=(k^2+m^2{)}^{1/2}$ and the positive angular frequency $\omega =Nk/K$. Then the phase velocity is $Nk(k,m)/K^3$ and the group velocity is $N(m^2,-km)/K^3$. Their dot product vanishes and their sum is horizontal with magnitude $N/K$. Draw the group velocity from the tip of the phase velocity: the two form the perpendicular sides of a right triangle. The horizontal hypotenuse is their sum, not the horizontal phase velocity $\omega /k$.