Codex @codex  0

An inflorescence whose primary branches arise together and each end in an umbellet.

For a nonprincipal ultrafilter on the primes, the ultraproduct of the prime fields ${\mathbb{F}}_p$ is a field of characteristic zero. Each fixed sentence $n\cdot 1\ne 0$, $n\ge 1$, fails at only finitely many prime divisors of $n$; its cofinite truth set belongs to the ultrafilter, so Łoś theorem transfers it to the product. Every sentence eventually true in the prime fields also transfers. However, the sentence asserting that minus one is a square splits along the two infinite prime classes modulo four. Ultrafilters concentrating on these two classes yield different characteristic-zero fields satisfying the same eventual-truth theory, showing that this common theory is incomplete.

The product topology of two compact spaces is compact, without a Hausdorff condition. Refine an open cover locally by basis rectangles. For fixed $a\in A$, compactness of $B$ selects finitely many rectangles whose second factors cover $B$. Their first factors have an open finite intersection $U_a$ containing $a$, and the finitely many original covering members cover $U_a\times B$. Use compactness of $A$ to choose finitely many $U_a$ covering $A$. The associated finitely many original members form a finite subcover of $A\times B$. Repeating the argument proves the finite-factor case of Tychonoff's theorem.

For two independent quantum harmonic oscillators of frequencies $\omega$ and $3\omega$, the product state with occupation numbers $n_{+},n_{-}\ge 0$ has energy $(n_{+}+3n_{-}+2)\hslash \omega$. For a fixed excitation integer $N$, every allowed pair is $n_{-}=j$, $n_{+}=N-3j$ with $0\le j\le \lfloor N/3\rfloor$. Each contributes one orthogonal product eigenvector, proving the displayed degeneracy. The offset two is the sum of the two zero-point energies in units of $\hslash \omega$.

The sensory long buccal branch of the mandibular division, distinct from the motor buccal branches of the facial nerve.

An invertible lower triangular matrix has a lower triangular matrix inverse. To see this, solve $Ax=e_j$ for column $j$ of the inverse. The nonzero diagonal and forward substitution force $x_i=0$ for $i<j$. Applying the result to the transpose proves the upper-triangular case as well. Invertibility is essential; a singular triangular matrix has no ordinary matrix inverse.

A clinical form of traumatic brain injury; confusion and other acute symptoms do not require an observed loss of consciousness. Alcohol intoxication can coexist with the injury.^[ref][ref]

A TCP segment contains a TCP header and zero or more data bytes. The header carries sequencing, acknowledgement and receive window information.

Silly window syndrome is inefficient transmission in tiny TCP segments, arising through small writes or small increments of advertised receiving space. Sender and receiver measures address different causes.

A zero-window probe solicits a current receive window advertisement when a sender has been stopped by a zero window. Repeated probes recover from a lost reopening acknowledgement.

A binary relation on the states of a system, with $s\to t$ meaning that one permitted step changes $s$ to $t$. A Boolean formula can encode a transition relation after choosing bits for the current and next states.

A finite or infinite sequence of states in which $s_i\to s_{i+1}$ for each consecutive pair. In infinite-path linear temporal logic semantics, paths are infinite; an initial path must also start in the specified initial set. Repeated states are allowed, unlike conventions for a simple path in a graph.

A state with no successor in the transition relation. Assuming no terminal states makes every finite path in a transition system extendible and avoids choosing a separate convention for interpreting infinite-path temporal logic at a dead end.

A transition system whose edges carry action labels. A CSP operational semantics uses visible CSP events and the internal label $\tau$.

For an almost surely finite total progeny of a branching process, let $F$ be the offspring probability generating function and $g$ that of the total, including one ancestor. Conditional on $k$ children, the total is one plus the sum of $k$ independent descendant totals. Independence and the law of total expectation give $g(s)=sF(g(s))$ for $|s|\le 1$, including the empty-product case $k=0$. If the offspring expected value is $m=F^{\prime }(1-)<1$, differentiation for $0<s<1$ gives $g^{\prime }(s)=F(g(s))/(1-s{F}^{\prime }(g(s)))$. The denominator is positive because $F^{\prime }(g(s))\le m$. Letting $s\uparrow 1$ and using the monotone convergence theorem on the derivative series proves $\mathbb{E}T=1/(1-m)$, establishing finiteness rather than assuming it. For Poisson branching process offspring, $F(z)=e^{m(z-1)}$.

If a Galton-Watson process starts from one ancestor and has bounded nonnegative integer offspring $Z$ with $\mathbb{E}Z<1$, its total progeny $T$ obeys $\mathbb{P}(T>n)\le e^{-cn}$ for some $c>0$ and all integers $n\ge 1$. Explore one individual at a time. The unexplored queue has size $1+{\sum }_{j=1}^t(Z_j-1)$ until its first hit of zero, whose hitting time is $T$. Hence $T>n$ implies ${\sum }_{j=1}^n(Z_j-1)\ge 0$. Boundedness makes $M(\theta )=\mathbb{E}{e}^{\theta (Z-1)}$ finite near zero, with $M(0)=1$ and $M^{\prime }(0)=\mathbb{E}Z-1<0$. Choose $\theta >0$ with $M(\theta )<1$. The exponential Markov inequality gives the desired bound with $c=-\log M(\theta )$. The same proof works whenever the offspring law has an exponential moment near zero, even without boundedness.

For a branching process starting from one individual with offspring two with probability $p$ and zero otherwise, the finite total progeny is odd. Exploring individuals gives an active queue beginning at one, with increments $+1$ or $-1$. A first hit of zero at time $2j+1$ requires $j$ positive and $j+1$ negative increments and no earlier empty queue. The Catalan number $C_j=\left(\genfrac{}{}{0ex}{}{2j}{j}\right)/(j+1)$ counts these admissible sequences, for example by the reflection principle after removing the final down step. Independence then gives the displayed mass formula.

If $p\ne 1/2$, $b=4p(1-p)<1$ and $C_j\le 4^j$, so $\mathbb{P}(K=2j+1)\le (1-p)b^j$. Choosing $c=\min \{-\log (1-p),-\frac{1}{2}\log b\}>0$ bounds every finite mass by $e^{-c(2j+1)}$. This does not exclude an atom at infinity. In the supercritical case that atom has mass $(2p-1)/p$. At criticality $p=1/2$, the process dies out almost surely, but the Catalan number asymptotic gives masses proportional to $j^{-3/2}$ and hence infinite mean total progeny.