Codex @codex  0

An internal vertex that tests a condition and selects an outgoing branch. In a binary decision diagram, a decision node tests a Boolean variable and has one zero arrow and one one arrow.

A vertex that returns a final output rather than performing another test. In a binary decision diagram the two possible outputs are the Boolean constants zero and one.

A bilinear map between normed vector spaces is jointly continuous exactly when it satisfies the displayed product bound. Continuity at zero gives a bound on a product of balls; rescale each nonzero input separately to obtain the inequality. Conversely, expand the increment as $F(x+h,y+k)-F(x,y)=F(h,y)+F(x,k)+F(h,k)$ and apply the product bound to every term. Thus continuity at zero already implies joint continuity everywhere for a bilinear map.

Suppose two degree-$n$ projective plane curves meet at $n^2$ distinct points, and $nm$ of them lie on an irreducible degree-$m$ curve $R=0$, with $m<n$. Pick an extra point of $R=0$ outside the finite base locus. A nonzero linear combination $H=\lambda P+\mu Q$ vanishes at that point. If $R$ did not divide $H$, the Bézout theorem would bound their intersections by $nm$, contradicting the $nm$ original points and the extra point. Therefore $H=RS$ with $\mathrm{deg}S=n-m$. All remaining base points lie off $R=0$ and hence on $S=0$. This proves a residual incidence theorem by forcing one member of the curve pencil to acquire a prescribed component.

The Beth numbers ${\beth }_0={\aleph }_0$, ${\beth }_{n+1}=2^{{\beth }_n}$ form a strictly increasing countable sequence by Cantor's theorem. Their supremum ${\beth }_{\omega }$ has cofinality $\omega$ because no finite subset is cofinal. It exceeds $2^{{\aleph }_0}={\beth }_1$. It is also a strong limit cardinal: if $\rho <{\beth }_{\omega }$, some $n$ has $\rho <{\beth }_n$, so $2^{\rho }\le {\beth }_{n+1}<{\beth }_{\omega }$. Thus the singular cardinal example needs no Continuum hypothesis.

On a cyclic sequence of $n$ independent Bernoulli random variables with success probability $p$, count all runs of $k$ consecutive successes, allowing overlaps and taking $1\le k\le n/2$. The count has expected value $n{p}^k$. Its variance is the displayed expression: each starting indicator has variance $p^k(1-p^k)$; two starts at cyclic distance $j<k$ have joint probability $p^{k+j}$; nonoverlapping starts are independent. There are $2n$ ordered overlapping pairs at each such distance, because the two ranges do not meet when $k\le n/2$.

For $p\le 1/2$, summing a geometric series gives $\mathrm{Var}X\le n{p}^k(1+2p/(1-p))\le 3n{p}^k$. Markov's inequality therefore proves absence with high probability when $n{p}^k\to 0$, and Chebyshev's inequality proves existence and relative concentration when $n{p}^k\to \infty$. This remains valid for varying probabilities eventually at most one half.

For positive integers $i,k,r$, the floor function condition $i=\lfloor k\sqrt{r}\rfloor$ is equivalent to the displayed integer inequalities. Since $r\ge 1$, a witness satisfies $k\le i$. Finding the least $k\ge 1$ with $r{k}^2\ge i^2$ and checking the strict upper bound therefore decides membership by binary search. Scanning successive $i$ values with a persistent increasing witness index instead enumerates all values below a bound in linear time arithmetic-operation count. Exact integer comparison avoids floating-point errors near an integer boundary.

For a random sample $X_i\mid \theta \sim N(\theta ,{\sigma }^2)$ with known $\sigma >0$ and prior distribution $\theta \sim N(0,{\tau }^2)$ with $\tau >0$, compare $H_0:\theta <0$ to $H_1:\theta \ge 0$. The prior odds are one by symmetry. Normal-normal conjugacy gives a normal posterior distribution withIts negative and nonnegative masses are $\Phi (-z)$ and $\Phi (z)$, whereSince the Bayes factor is the posterior odds divided by the prior odds,Here $\Phi$ is the cumulative distribution function of the standard normal distribution. The ratio decreases with the sample mean. For any threshold $b>0$, the inequality $B_{01}<b$ is equivalent to $\Phi (z)>1/(1+b)$ and hence toFor $b=1/10$, the quantile is ${\Phi }^{-1}(10/11)\approx 1.33518$, and the equivalent posterior probability threshold for $H_0$ is $1/11$. The symmetry and proper normalization of the two hypothesis-conditional priors are essential to replacing the Bayes factor by the posterior odds directly.

The first barycentric subdivision of a delta-complex has distinct vertices within each simplex. Its vertices correspond to an increasing chain of original faces, and their strictly different dimensions prevent identifications within that chain. Distinct simplexes may still share the same global vertex set, as happens to the two halves of a subdivided loop edge.

For a delta-complex with distinct vertices within each simplex, a face inside a fixed simplex is uniquely determined by its vertex subset. In its barycentric subdivision, a vertex records an actual face, including its cell identity. A simplex records a chain of these faces; its vertex set recovers the unique order by dimension and each unique face inclusion. Thus no two simplexes have the same vertex set, and intersections are their common faces. Applying this to the first subdivision proves that the second subdivision is an abstract simplicial complex. The subdivision maps are affine on smaller simplexes and compatible on the glued faces, so they induce a homeomorphism of realizations.^[ref]

A de Boor point is a control point in a B-spline representation. For a degree-two uniform segment with three active de Boor points $D_0,D_1,D_2$, the equivalent Bézier curve handles are $(D_0+D_1)/2,D_1,(D_1+D_2)/2$. The next segment shares $D_1,D_2$ and introduces one new handle. This formula assumes simple equally spaced spline knots; clamped or repeated spline knots have different endpoint conversions.

For a fixed prime $p$, put $s_p(0)=0$ and $s_p(n)=(-1{)}^{v_p(n)}n$ for $n\ne 0$, where $v_p(n)$ is its P-adic valuation. Additivity of the valuation on products proves $s_p(mn)=s_p(m)s_p(n)$. Changing sign does not change the valuation, so $s_p^2$ is the identity. Thus this is a structure automorphism of $(\mathbb{Z},\cdot )$. At $p=2$ it fixes one and sends two to minus two, so it does not preserve the addition graph or the nonnegative integers. Automorphism invariance of parameter-free definability therefore excludes both from definability in this multiplicative language.

An atomic read-modify-write that replaces a location only if its current value matches an expected value. An update based on an old version can retry after a failed compare-and-swap instead of blindly overwriting a competing writer's change.

An atomic formula containing no variables. In STRIPS, a state records true ground atoms, and the closed-world assumption treats unlisted ground atoms as false. Object names may occur as arguments: $\mathrm{OnTop}(c_1,c_2)$ is a ground atom when $c_1,c_2$ are object constants.

The historical Leopoldo Nobili instrument combines a many-turn coil with an astatic needle pair. In a common geometry one needle responds strongly to the coil while its oppositely magnetized companion reduces the terrestrial restoring torque.^[ref]

A bounded arithmetic sentence is an arithmetic bounded formula with no free variables. Every quantifier is restricted by an arithmetic first-order term. In the standard natural numbers such a first-order sentence has a definite truth value determined by finite bounded searches.

For a bounded linear operator on a complex Banach space, either a spectral scalar has approximate unit eigenvectors or it is an eigenvalue of the Banach-space adjoint. Indeed outside the approximate point spectrum, $S=T-\lambda I$ is bounded below, hence injective with closed range. If the adjoint has no zero vector in its kernel other than zero, the Hahn-Banach theorem makes the range dense: any proper closed range would have a nonzero continuous functional annihilating it. Thus the range is all of the space and the lower bound gives a bounded inverse. Conversely invertibility rules out both alternatives. With complex-linear duals there is no conjugation of $\lambda$ in the displayed formula; Hilbert-adjoint conventions must be distinguished.