Codex @codex  0

A pole of a scattering matrix arising from the numerator $F(-k)$ rather than a zero of the incoming Jost function $F(k)$ can fail to correspond to a bound state. For $F(k)=(k-i\kappa )/(k+i\alpha )$, the pole at $i\kappa$ is associated with the incoming zero; the pole at $i\alpha$ comes from the numerator. The regular Eckart realization has $\alpha >\kappa$. The distinction and this explicit example are discussed in dipot.ulb.ac.be/dspace/bitstream/2013/373712/4/2306.12216.pdf.

This one-step family interpolates between the Forward Euler method at $a=0$, the trapezoidal rule at $a=1/2$, and the Backward Euler method at $a=1$. Its stability function is $R(z)=[1+(1-a)z]/[1-az]$. It is first order except at one half, where it is second order. Applying it to spatial diffusion gives the theta diffusion stability criterion.

An $s$-stage Runge-Kutta method has stability function $R(z)=1+z{b}^T(I-zA{)}^{-1}1$, a rational function with numerator and denominator degrees at most $s$. Order $2s$ forces $R(z)-e^z=O(z^{2s+1})$, so uniqueness of the diagonal Padé approximant identifies $R$ with $[s/s{]}_{e^z}$. That approximant has no poles in the closed left half-plane and modulus at most one there. Hence every such maximal-order method is A-stable. Since the Padé approximant numerator and denominator are coprime and both have degree $s$, the original stage determinant cannot have hidden canceled poles in that half-plane.

For sufficiently regular vector fields, for example bounded fields with bounded derivatives through order three when $2<p<3$, the solution map of a rough differential equation exists uniquely and is locally continuous in the rough path metric. Ordinary differential equations driven by smooth approximations therefore converge whenever their enhanced drivers converge. The theorem makes the enhancement, rather than just the original path, the correct input for a continuous solution map.

For a finite time grid, suppose a nonnegative error is zero on neighboring points and obeys $a_{s,t}\le a_{s,u}+a_{u,t}+L\omega (s,t{)}^{1/p}{a}_{s,u}+A\omega (s,t{)}^{\theta }$. For a control function for rough paths and $\theta >1$, sufficiently small control intervals have the displayed bound with constants independent of the grid. Splitting near half the control size makes the higher-order errors contract, and handles a crossing single grid interval by its zero error. Applying this to local Euler composition errors yields compactness of approximate pathlevel solutions.

A pathlevel solution is the ordinary state path satisfying a local rough Taylor expansion with a remainder of order $\theta >1$ in a control function for rough paths. It specifies the first level of the solution, rather than requiring the full lifted solution as part of the definition. With vector fields of regularity ${\mathrm{Lip}}^{\gamma }$, $\gamma >p$, the Davie lemma bounds discrete Euler errors uniformly; compactness and passage to the limit give such a path.

Write $r_3(N)$ for the maximum cardinality of a subset of $[N]$ containing no nonconstant three-term arithmetic progression. There is an absolute constant $C$ such that every three-term-progression-free subset $A\subseteq [N]$ satisfies $|A|\le CN/\log \log N$ for $N\ge 3$. A quantitative Roth density-increment step gives, for subset density $\delta$ and $N\ge 4\cdot 10^6{\delta }^{-4}$, a arithmetic progression of length at least ${\delta }^2\sqrt{N}/2000$ and increased subset density at least $\delta +{\delta }^2/64$. The increment decreases reciprocal subset density by at least $1/65$, so fewer than $\lceil 65/\delta \rceil$ iterations are possible. If $N_j$ is the successive length, then $\log N_{j+1}\ge \frac{1}{2}\log N_j-\log (2000/{\delta }^2)$, using the initial subset density to bound all later ones. Hence $\log N_j\ge 2^{-j}\log N-2\log (2000/{\delta }^2)$. If $\delta \log \log N$ exceeds a sufficiently large absolute constant, all these intervals remain above the density increment threshold, a contradiction. This proves the displayed bound. The result is a classical quantitative version of Roth theorem on three-term arithmetic progressions, without asserting an optimal bound.

Embed $A\subseteq [N]$ in the odd cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$, so additive three-term relations in $A$ cannot wrap around. Form three copies of $G$, putting edges $x\sim y$ when $y-x\in A$, $y\sim z$ when $z-y\in A$, and $x\sim z$ when $(z-x)/2\in A$. A triangle in a graph encodes $a+b=2c$ in $A$. If $A$ has no nonconstant arithmetic progression, every triangle in a graph has $a=b=c$; there are exactly $|G||A|$ triangles in a graph, and they are edge-disjoint. Thus at least $|G||A|$ edges must be removed to eliminate them. For fixed positive subset density of $A$, this is a positive fraction of the squared vertex count, whereas the triangle in a graph count is only quadratic. The triangle removal lemma permits deletion of an arbitrarily small quadratic fraction when the cubic triangle in a graph subset density is sufficiently small. This contradiction establishes Roth theorem on three-term arithmetic progressions.

If a density-$\delta$ subset of a sufficiently long integer interval has no nonconstant three-term arithmetic progression, Fourier expansion of the progression count gives a large coefficient of its balanced indicator. Dirichlet approximation supplies a step at most $\sqrt{N}$ on which that character varies slowly. Partitioning the interval into equal-length progressions of that step, and controlling the short tails, produces the displayed density increment. Iteration must terminate since density cannot exceed one. The same argument treats the three-point pattern with coefficients $2,-3,1$.

A proper three-dimensional rotation matrix has an axis eigenvalue one and perpendicular-plane eigenvalues $e^{i\theta },e^{-i\theta }$. Their sum gives the displayed trace identity. It determines the angle cosine but does not by itself choose the oriented axis or angle sign.

Apply $R_1$ to a column coordinate vector, then apply $R_2$ in the same fixed basis. The function composition isThus the rightmost matrix acts first. The product is again a rotation matrix, since $(R_2{R}_1{)}^T(R_2{R}_1)=I$ and $\det (R_2{R}_1)=1$. For planar rotations about the same origin, the sine addition formula and cosine addition formula giveFor a concrete composition of rotation matrices, first rotate $(2,0{)}^T$ by $\pi /3$ radians, reaching $(1,\sqrt{3}{)}^T$, then by $\pi /6$ radians:The total angle is $\pi /2$. Reversing these two planar rotations gives the same answer, but noncommuting rotations about fixed axes show why one cannot assume that property for rotations in three dimensions.^[ref]

Rotate an orthonormal basis counterclockwise by $\theta$ while keeping the vector $v$ fixed. The new basis vectors, written in the old basis, are the columns of $P=R_{\theta }$. The change-of-basis matrix convention therefore givesFor example, the fixed vector with old coordinate vector $(2,0{)}^T$ has new coordinate vector $(1,-\sqrt{3}{)}^T$ after the basis turns by $\pi /3$ radians. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an active planar rotation through that same angle moves the arrow to old coordinates $(1,\sqrt{3}{)}^T$. The sign difference follows from which object moves. The passive component formulas are $x^{\prime }=x\cos \theta +y\sin \theta$, $y^{\prime }=-x\sin \theta +y\cos \theta$.^[ref]

If coordinate vectors are instead written as rows, transposing the active equation gives $(v^{\prime }{)}^T=v^T{R}_{\theta }^T$. Switching to rows changes the side and the transpose; it does not by itself turn an active planar rotation into passive rotation of coordinates.

For homogeneous, incompressible, inviscid flow with constant rotation vector and negligible inertia, taking the curl of $2\mathbf{\Omega }\times \mathbf{u}=-\nabla p$ gives $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus a geostrophic flow is independent of distance along the rotation axis. Body forces modify the relation by their curl; it applies unchanged in unforced exterior regions. A nontrivial exterior column cannot both obey this leading balance and decay along the rotation axis.

For a convective star, the activity convention compares its rotation period with a convective turnover time. It differs by fixed factors from the fluid Rossby number estimated from speed, length and angular velocity. Small ${\mathrm{Ro}}_{\ast }$ denotes strong rotational influence on convection. Magnetic activity usually rises as this ratio decreases in the unsaturated regime, but levels off in a saturated rapid-rotation regime.

A vertex whose parent in a rooted tree is the specified vertex.

For a nonroot vertex, its adjacent predecessor on the unique graph path from the root of a tree. The root of a tree has no parent.

The distinguished vertex that specifies a rooted tree. Its choice determines the parent in a rooted tree and children in a rooted tree relationships.

With $m$ the characteristic-prime-free part of $n$, all roots of unity over a finite field lie in ${\mathbb{F}}_{q^s}$ exactly when $m\mid q^s-1$. This follows from the cyclicity of the multiplicative group of a finite field. For $m=1$ the least positive degree is one. A primitive root of order $m$ has the same field degree, because its Frobenius orbit closes exactly when $m\mid q^s-1$.

A round body of radius $R_0$ in a fixed cylinder of radius $R_1$ has center of mass path radius $a=R_1-R_0$. With orbital angle $\theta$ and spin angular velocity $\Omega$ both positive about the same axis, rolling without slipping gives $a\dot{\theta }+R_0\Omega =0$. For moment of inertia $I=\gamma m{R}_0^2$, its kinetic energy is $\frac{1}{2}m{a}^2(1+\gamma ){\dot{\theta }}^2$. Release from rest at angle $\alpha$ gives, by conservation of energy, a descent time $\sqrt{1+\gamma }$ times that of frictionless sliding along the same center of mass path. The spin sign reverses if its positive direction is chosen opposite to the orbital angle.

For a uniform solid cylinder, take $s$ positive uphill and $\omega$ positive for uphill rolling without slipping. The velocity of the contact point is $w=\dot{s}-a\omega$. Uphill kinetic friction gives $\ddot{s}=-g(\sin \alpha -\mu \cos \alpha )$ and $a\dot{\omega }=-2\mu g\cos \alpha$. Starting from $\dot{s}(0)=a\omega (0)$, one obtains $w=-gt(\sin \alpha -3\mu \cos \alpha )$. Thus $\tan \alpha >3\mu$ makes the assumed uphill kinetic friction self-consistent for the entire ascent and descent. Without slip the required uphill static friction is $mg\sin \alpha /3$; it is available exactly when $\tan \alpha \le 3\mu$ under the same limiting coefficient.