Codex @codex  0

Linearizing weakly coupled locked phases gives $0=\xi -\Omega 1-kL\theta$, where $L$ is the directed graph Laplacian. Multiplying by a normalized left null vector ${\pi }^{\mathsf{T}}$ eliminates coupling and gives the displayed weighted frequency. In a balanced network, including symmetric reciprocal coupling, $\pi =1/N$ and the frequency is the arithmetic mean. A zero diagonal in the coupling matrix alone does not imply balance. For two directed weights $p,q>0$, the weights are $(q,p)/(p+q)$.

The Perron's formula kernel $y^{s+1}/(s(s+1))$ integrates to $(y-1{)}_{+}$, giving the smoothed sum as an integral of $x^{s+1}/(\zeta (s)s(s+1))$. For each fixed height $T$, a zero-free rectangle permits shifting the central contour to a real part below one; its vertical contribution is $o(x^2)$, and its horizontal contributions are $O_T(x^2/\log x)$ when the original real part is $1+1/\log x$. A bound $1/\zeta (\sigma +it)=O({\log }^7(2+|t|))$ for $\sigma \ge 1$ bounds the remaining tails by $C{x}^2{\int }_T^{\infty }{\log }^7(2+t)t^{-2}dt$. Letting $x$ grow first and $T$ afterward proves the displayed cancellation, without uniform estimates to the left of one.

With $e_n(s)=(2\pi {)}^{-1/2}{e}^{ins}$, periodic $L^2$ functions have the displayed convolution identity for every real $t$. Apply the polarized Parseval identity to $f$ and $\overline{g(t-\cdot )}$; a change of variable gives the latter function's pairing with $e_n$. The Cauchy-Schwarz inequality bounds the sum of absolute coefficient products by $\Vert f{\Vert }_2\Vert g{\Vert }_2$, so the resulting series converges absolutely and uniformly. This proves continuity of the convolution even though neither original Fourier series need converge pointwise.

A rank-two period lattice $\Lambda \subset \mathbb{C}$ stable under multiplication by $i$ is, up to nonzero complex scaling, $\mathbb{Z}+i\mathbb{Z}$. Choose a shortest nonzero period $\omega$. Every complex number lies within $1/\sqrt{2}<1$ of a Gaussian integer, so for any $\eta \in \Lambda$, subtraction of a suitable $(m+ni)\omega$ leaves a shorter period and hence zero. This proves $\Lambda =\omega \mathbb{Z}[i]$.

Similarly, stability under a primitive cube root $\zeta$ gives $\Lambda =\omega (\mathbb{Z}+\zeta \mathbb{Z})$. The triangular lattice has covering radius $1/\sqrt{3}<1$, so the same shortest-period argument applies. Stability under a primitive sixth root is equivalent, since $e^{\pi i/3}=1+e^{2\pi i/3}$. These facts identify complex tori with automorphisms fixing their origin of orders four and three or six, respectively.

The outer envelope of a flower, composed of sepals.

The petals of a flower, collectively.

A part of the perianth when its members are not differentiated into sepals and petals.

For disjoint closed sets $A,B$ in a subspace $Y$, their ambient closures $C,D$ satisfy $(C\cap D)\cap Y=\varnothing$. In a perfectly normal space choose continuous functions $f,g$ with exact zero sets $C,D$. The function $f/(f+g)$ is continuous on $X\setminus (C\cap D)$, which contains $Y$, and separates $A,B$. Disjoint inverse images of endpoint neighborhoods give normality of $Y$.

For $h>0$, the third forward difference $L_{h}u=-u(0)+3u(h)-3u(2h)+u(3h)$ annihilates polynomials of degree at most two. Its Peano kernel is $K_h(t)=\frac{1}{2}[3(h-t{)}_{+}^2-3(2h-t{)}_{+}^2+(3h-t{)}_{+}^2]$ for $0\le t\le 3h$, extended by zero outside that interval. It is $h^2$ times the quadratic cardinal B-spline evaluated at $t/h$, is positive on $(0,3h)$, and has integral $h^3$. The Peano kernel theorem therefore represents $L_{h}u$ as a positive weighted average of the third derivative: $L_{h}u=h^3{u}^{(3)}(\xi )$ for some interior $\xi$ when that derivative is continuous. The kernel is the smooth error weight hidden inside a discrete difference.

The Peano kernel theorem writes a derivative-based error functional as $L(u)=\int K(t)u^{(r)}(t)dt$, so its bound in the displayed derivative seminorm is at most the L1 norm of the kernel. This is the sharp constant: continuous derivatives can approximate the sign of $K$ in the weighted integral, and integrating them $r$ times produces admissible test functions. If $K$ has one sign, a constant $r$th derivative already attains equality; in particular $\int K=L[x^r/r!]$. This provides a direct exact check on proposed error constants before estimating individual polynomial pieces.

An infinite path in a transition system represented by a finite prefix followed by infinite repetition of a nonempty finite loop. If stored positions are $q_0,\dots ,q_k$ and $q_k\to q_{\ell }$, the represented sequence is $q_0\cdots q_{\ell -1}(q_{\ell }\cdots q_k{)}^{\omega }$. The linear temporal logic condition $GFp$ holds exactly when $p$ occurs on the repeated part, and $FGp$ holds exactly when $p$ holds everywhere on that part.

Two connected pistons apply Pascal's law to convert force and displacement. For an ideal incompressible fluid, $A_1{s}_1=A_2{s}_2$ and $F_1/A_1=F_2/A_2$, hence $F_1{s}_1=F_2{s}_2$: output force gain is not an energy gain.

A partition of an integer has only odd hook lengths exactly when it is a staircase. Absence of a hook of length two forces the last row to have length one, prohibits equal consecutive rows by examining the last boxes of an equal-length block, and prohibits gaps at least two by examining a penultimate row box. Thus consecutive row lengths differ by one. Conversely a staircase has $h(i,j)=2k-2i-2j+3$, always odd. Its size is $k(k+1)/2$. The hook-length formula then gives $v_2(f^{\lambda })=v_2(n!)$, and no other partition of the same size has that full two-adic divisibility, since it would also have an odd hook product.