For $G=⟨a,t\mid t{a}^2{t}^{-1}=a^3⟩$, the surjective endomorphism $\theta (a)=a^2$, $\theta (t)=t$ has nontrivial kernel by Britton's lemma. The commutator subgroup is generated by $a_i=t^{i}a{t}^{-i}$ for $i\in \mathbb{Z}$. For $j>i$, its defining relations give $a_j^{2^{j-i}}=a_i^{3^{j-i}}$. Hence ${\theta }^k([a_i,a_j])=1$ whenever $k\ge j-i$. The union of the increasing kernels is normal, and the images of all $a_i$ commute in its quotient, proving the first inclusion. The second follows from finite residual contains surjective endomorphism kernels applied to every iterate.

Represent $y$ by translation $t\mapsto t+1$ and $x$ by $t\mapsto t/3$; the relator holds and $y$ has infinite order. Base-three digits $d_j\in \{0,1,2\}$ give words recursively by $W_0=1$, $W_{j+1}=x^{-1}{W}_{j}x{y}^{d_j}$. These have length at most $4k$ and represent all $y^m$, $0\le m<3^k$. Thus the growth function of a finitely generated group obeys $\beta (4k)\ge 3^k$. Counting freely reduced words in four letters gives $\beta (n)\le 2\cdot 3^n-1$. The growth is therefore exponential up to coarse equivalence.

The null space property relative to $S$ is $\Vert h_S{\Vert }_1<\Vert h_{S^c}{\Vert }_1$ for every nonzero $h\in \ker A$. It is equivalent to exact basis pursuit recovery of every vector supported in $S$. Sufficiency follows from the triangle inequality; for necessity, compare $-h_S$ and $h_{S^c}$, which have the same image under $A$. Recovery of a single fixed signed vector can hold without this uniform property.

For an invariant Cartier divisor $D={\sum }_{\rho }{a}_{\rho }{D}_{\rho }$ on a complete smooth toric variety, let $m_{\sigma }$ satisfy $⟨m_{\sigma },v_{\rho }⟩=-a_{\rho }$ for rays in a maximal cone $\sigma$. The divisor is basepoint-free exactly when every $m_{\sigma }$ lies in the lattice polytope of a toric divisor $P_D$. Its character section then trivializes the line bundle on the entire chart $U_{\sigma }$.

// Target: geometry-and-topology.bigb

Add diffusivities $1,p$ to the basally forced quadratic activator-inhibitor model. A Fourier mode with $s=k^2$ has matrix $J-s\mathrm{diag}(1,p)$ and determinant $D(s)=p{s}^2+(q-pa)s+q$. When $q>a$, its trace is negative for every $s\ge 0$. A positive-growth mode therefore occurs precisely when $D(s)<0$ for some positive $s$. The parabola has a positive minimizer only if $pa>q$, and its minimum is negative exactly when $(pa-q{)}^2>4pq$. Thus the ordinary Turing instability requires $a>0$, $q>a$ and $p>q/(\sqrt{1+a}-1{)}^2$. At onset, $k_c^2=(pa-q)/(2p)=\sqrt{q/p}$ and the critical wavelength is $2\pi (p/q{)}^{1/4}$. A bounded spatial domain further requires an allowed mode inside the unstable band.

At $q=a=(r-1)/(r+1)>0$, the reaction eigenvalues are $\pm i\sqrt{a}$. This linear marginal case is a weakly attracting focus, not a nonlinear center. To check it directly, put $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and $X=x$, $Y=(hy-ax)/\omega$. The equations become $\dot{X}=-\omega Y+F_2+F_3+O(4)$, $\dot{Y}=\omega X+G_2+G_3+O(4)$, withOn the unit circle write $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, $C=\cos \theta G_2-\sin \theta F_2$. In polar coordinates, $d\varrho /d\theta =(A/\omega ){\varrho }^2+(B/\omega -AC/{\omega }^2){\varrho }^3+O({\varrho }^4)$. The quadratic angular average vanishes. Integrating the quadratic correction over a whole cycle contributes its final squared integral, also zero, leaving the cubic average. Using the even trigonometric moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map is therefore $\varrho \mapsto \varrho +[2\pi /\omega ](a-1){\varrho }^3/8+O({\varrho }^4)$. Since $0<a<1$, it contracts sufficiently small positive radii and proves nonlinear asymptotic stability at the threshold.

A down-and-in claim with lower barrier $b$ pays a specified terminal payoff only if the stock reaches $b$ before maturity. Before activation its value solves a killed-domain pricing problem; at activation its value equals the ordinary terminal-payoff claim value.

A down-and-out claim expires worthless if the stock reaches a lower barrier before maturity. A matching down-and-in claim plus a down-and-out claim pays the ordinary terminal claim on every path; this is in-out parity.

An up-and-out power claim pays a power of the terminal stock price provided an upper barrier has never been reached. In the Black-Scholes model, set $A=\log (c/S_0)/\sigma$ and $m_p=(\rho +(p-\frac{1}{2}){\sigma }^2)/\sigma$, where $\sigma >0$, $c>S_0$ and $T>0$. The risk-neutral pricing value isHere $\Phi$ is the standard normal distribution function. Weighting the Brownian motion endpoint by $e^{p\sigma W_T-p^2{\sigma }^{2}T/2}$ changes the logarithmic drift to $m_p$; the finite-horizon maximum of Brownian motion with drift then supplies the survival factor.

An up-and-in claim activates when the stock reaches an upper barrier $b>S_0$ during its lifetime. After activation its terminal payoff is the prescribed function $f(S_T)$; without activation it pays zero. Under continuous stock paths the barrier event is described by the displayed running maximum.

On a nonradial zero-energy Kepler orbit, conserved specific angular momentum $h$ gives periapsis $r_p=h^2/(2GM)$ and ${\dot{r}}^2=2GM(r-r_p)/r^2$. Integrating with $s=\sqrt{r-r_p}$ yields $t-t_p=\pm \sqrt{2/(GM)}(r+2r_p)\sqrt{r-r_p}/3$. Squaring gives the displayed version of the Barker equation, valid on both incoming and outgoing branches. The time origin $t_p$ is passage through periapsis, not an arbitrary point on the orbit.

Suppose continuous functions $f_n:\mathbb{R}\to \mathbb{R}$ converge pointwise to $f$. For each $ϵ>0$, some nonempty open interval $I$ and integer $N$ satisfy $|f_n(x)-f(x)|\le ϵ$ for every $x\in I$ and $n\ge N$. Indeed the closed sets $Q_N={\bigcap }_{n,m\ge N}\{|f_n-f_m|\le ϵ\}$ cover $\mathbb{R}$. The Baire category theorem gives an interval inside one $Q_N$, and passing $m\to \infty$ proves the assertion. The interval may depend on $ϵ$; this does not assert uniform convergence on one fixed interval.

Every proper affine hyperplane in ${\mathbb{R}}^n$ is closed and has empty interior: a small displacement along its nonzero normal vector leaves it. The Baire category theorem therefore makes the complement of a countable collection of such hyperplanes dense. In particular, avoiding the hyperplanes ${\sum }_{j=1}^n{k}_j{x}_j=k_{n+1}$ for nonzero integer tuples yields a vector for which $1,x_1,\dots ,x_n$ are linearly independent over $\mathbb{Q}$. Tuples with all the first $n$ entries zero define an empty forbidden set.

For $f\in C^2({\mathbb{R}}^2)$ and standard Brownian motion $B$, the drift in Itô formula for $f(B_t,t)$ vanishes exactly when the displayed equation holds for every spatial point and every $t\ge 0$. Necessity follows because a zero finite-variation integral has continuous integrand $f_t(B_t,t)+\frac{1}{2}{f}_{xx}(B_t,t)$ equal to zero, and at every positive time the normal distribution of $B_t$ has full support. Continuity extends the equation to time zero. There is no condition at negative times except compatibility with the global smoothness of $f$.

For strictly increasing spline knots, $M_i(t)=k[t_i,\dots ,t_{i+k}](\cdot -t{)}_{+}^{k-1}$ is a nonnegative integral-normalized B-spline. Integrating the truncated power function over its support replaces its spline knot value by $(u-t_i{)}^k/k$. The order-$k$ divided difference of this degree-$k$ polynomial is its leading coefficient $1/k$. Multiplying by the prefactor $k$ in $M_i$ gives integral one. The corresponding partition-normalized B-spline is $N_i=(t_{i+k}-t_i)M_i/k$.

For degree $p$, knots $t_i$ and span $[t_k,t_{k+1})$, initialize $d_j^{(0)}=P_{k-p+j}$ for $0\le j\le p$. At level $r$, update $j=p,p-1,\dots ,r$ by $d_j^{(r)}=(1-{\alpha }_{j,r})d_{j-1}^{(r-1)}+{\alpha }_{j,r}{d}_j^{(r-1)}$, where ${\alpha }_{j,r}=(t-t_{k-p+j})/(t_{k+1+j-r}-t_{k-p+j})$. The point is $d_p^{(p)}$. This evaluates a B-spline by local affine combinations.

// Target: analysis.bigb

Knot insertion refines the spline knot sequence and changes the control points without changing the represented B-spline. Repeated insertion exposes local Bézier curves and provides subdivision-based evaluation.

// Target: analysis.bigb

A non-uniform rational B-spline surface isHere the $N_{i,p}$ and $M_{j,q}$ are B-spline basis functions for possibly nonuniform spline knot sequences, and the $w_{ij}$ are weights. Positive weights make each evaluated point a convex combination of the active control points, supplying bounding volumes for geometric searches. The denominator must remain nonzero for the parametrization to be defined.

For a uniform quadratic B-spline with interior control points $P_i$, label its $i$th span $[i,i+1]$. A quartic B-spline with interior spline knot sequence ${\tau }_{3i}={\tau }_{3i+1}={\tau }_{3i+2}=i$ represents the identical curve whenTo verify this, raise each quadratic Bézier curve span twice by degree elevation of Bernstein coefficients. The three interior quartic controls are the displayed points. One additional knot insertion at each interior knot supplies the shared endpoint as the mean of its two neighboring controls, recovering all five quartic Bézier curve controls. Thus the knot multiplicity is three, giving $C^1$ joins rather than imposing unwanted extra smoothness.

Let the real spline $s={\sum }_{i=p}^q{c}_i{N}_i$ have distinct knots, order $k\ge 2$, nonzero endpoint coefficients, and no identically zero knot cell in its support. It and its first $k-2$ derivatives vanish at both support endpoints. If $Z$ is its number of interior distinct zeros, there are initially $Z+2$ zero components. Each differentiation through order $k-2$ increases this count by at least one: a nonzero gap between zero components has an interior extremum, and the two endpoint zeros remain separate. The final derivative is continuous piecewise linear on $q-p+k$ cells, and each cell can meet at most one zero component unless it is identically zero, in which case all its zeros belong to the same component. It therefore has at most $q-p+k$ zero components. Thus $Z+k\le q-p+k$, proving the bound.