Codex @codex  0

A flow on a directed graph satisfying capacity constraints and a specified net inflow $d_v$ at each vertex. Positive $d_v$ denotes consumption and negative $d_v$ supply. Summing all balances cancels each edge once as inflow and once as outflow, so ${\sum }_v{d}_v=0$ is necessary. With balanced demands and no lower bounds, add a source edge of flow network edge capacity $-d_v$ into every supply vertex and an edge of flow network edge capacity $d_v$ from each demand vertex to a new sink. A maximum flow saturating all auxiliary edges exists exactly when the demands are feasible.

A circular arch of chord length $2a$ described as a smooth single-valued graph has radius $R>a$ and length $2R\arcsin (a/R)$, ranging strictly between $2a$ and $\pi a$. The straight-line limit has length $2a$. At the semicircle limit the endpoint slopes are infinite. Longer circular arcs require overhangs and cannot be graphs in this class.

For real $A,C$ and complex $B$, if $A\ne 0$, completing the square gives $|z+B/A{|}^2=(|B{|}^2-AC)/A^2$. A positive right side gives a circle; zero gives a single point; a negative value gives the empty set. If $A=0$ and $B\ne 0$, the equation is the straight line $2\mathrm{Re}(\overline{B}z)+C=0$. If $A=B=0$, it describes the whole complex plane when $C=0$, and the empty set otherwise. Thus unrestricted coefficients require both the single-point and whole-plane degeneracies.

Every continuous injective map $g:S^1\to S^1$ is a homeomorphism. After normalizing $g(1)=1$ by a rotation, it is homotopic either to the identity or to complex conjugation, and has degree $1$ or $-1$.

To see this without assuming surjectivity, lift $t\mapsto g(e^{2\pi it})$ on $[0,1]$ through the covering map $u\mapsto e^{2\pi iu}$ to a path $G$ with $G(0)=0$. Injectivity makes $G$ injective on $[0,1)$, hence strictly monotone. Its endpoint $d=G(1)$ is an integer. Strict monotonicity excludes $d=0$; if $|d|>1$, the intermediate value theorem gives an interior integer value, contradicting injectivity of $g$. Thus $d=\pm 1$, and $\exp (2\pi i((1-s)G(t)+sdt))$ descends to a homotopy from $g$ to $z\mapsto z^d$. If $g$ omitted a point, a continuous branch of the complex argument on the punctured circle would make it null-homotopic, contradicting this nonzero degree. Hence it is surjective. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

When $\alpha =\kappa =\eta =1$, the instability threshold of a spatial mode with squared wave number $s$ is $\beta =s+2+1/s$. On a rectangle $L_1=1$, $L_2=1/2$, modes varying in both coordinates have $s\ge 5{\pi }^2$. Since the threshold increases for $s>1$, the first such mode is $(n,m)=(1,1)$, at ${\beta }_{crit}=5{\pi }^2+2+(5{\pi }^2{)}^{-1}$. At equality it is neutral; strict instability begins above this value.

An unbiased coefficient estimator with known variance $v$ gives a conservative level-$\alpha$ test of a zero coefficient by rejecting when its absolute value is at least $\sqrt{v/\alpha }$. Chebyshev's inequality bounds the null rejection probability by $\alpha$. This test needs the variance assumptions but no Gaussian error distribution; a normal-reference test requires additional distributional justification.

For $0<a<b$, the displayed polynomial minimizes ${\max }_{t\in [a,b]}|p(t)|$ among real polynomials of degree at most $k$ with $p(0)=1$. Its $k+1$ alternating extremal values prove the assertion: a strictly smaller competitor would make the difference change sign in each consecutive extremal interval and also vanish at zero outside $[a,b]$, giving too many roots. The normalizing factor makes either the monic or the usual Chebyshev polynomial convention give the same ratio. For $\kappa =b/a$ and $q=(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$, its maximum is $2q^k/(1+q^{2k})$. This yields an error bound for polynomial acceleration of Richardson iteration.

The positive characteristic of a ring which is an integral domain is a prime number. The characteristic cannot be one because $1\ne 0$. If its least positive value were $n=ab$ with $1<a,b<n$, minimality would make $a1$ and $b1$ nonzero, while $(a1)(b1)=n1=0$ would give zero divisors.

The analogous result holds for a nonzero commutative nonunital ring without zero divisors, using the characteristic of a rng. If $n=ab$ were composite and minimal, choose $r$ with $ar\ne 0$. For every $s$, $(ar)(bs)=n(rs)=0$, so cancellation forces $bs=0$ for every $s$, contradicting minimality of $n$.

In a finite unital ring $R$, some positive multiple of $1$ is zero, because two of the finitely many successive multiples coincide. If $n=\mathrm{char}R$, the permutation $r\mapsto r+1$ has every cycle of a permutation of length exactly $n$: returning to $r$ after $k$ steps means $k1=0$, whose least positive solution is $n$. These disjoint cycles partition the finite set $R$, so $n$ divides its cardinality. The zero ring has cardinality and characteristic both one.

The radial transport equation has projected characteristic curves $(x,y)=e^s(x_0,y_0)$, which remain on positive rays. Data on $x+y=1$ reach precisely the half-plane $x+y>0$, since $x+y=e^s$. A datum $u(a,1-a)=h(a)$ therefore gives $u(x,y)=(x+y{)}^{k}h(x/(x+y))$ in that half-plane. The opposite half-plane is not determined by these data. Extending a characteristic line through the origin would join distinct flow orbits across a point where the projected vector field vanishes.

Expurgation discards messages with unusually high conditional decoding error. If a codebook with $2M$ equally likely messages has average error at most $ϵ$, at least $M$ messages have individual error at most $2ϵ$, by Markov's inequality. Retaining those messages and restricting the decoder's candidates gives a codebook with maximal error at most $2ϵ$, provided removal of candidates cannot spoil a previously successful decoding. This is the case for bounded-distance decoding with errors on ambiguity, and for a decoder that retains its former decisions whenever the decoded message remains.

For a sequence of block lengths $n$, halving a book changes its logarithmic rate by $1/n$, which tends to zero. Thus a vanishing average-error bound from a random coding construction can be converted into a vanishing maximal decoding error probability at the same limiting rate. The probabilistic construction first supplies a deterministic book with sufficiently small average error; expurgation is a subsequent deterministic step.

Random coding constructs a channel code by sampling its codewords from a specified probability distribution and bounding the expected decoding error over that ensemble. If the expected average error is at most $ϵ$, at least one deterministic codebook has average error at most $ϵ$; otherwise every book would exceed the expectation bound. This is the probabilistic method applied to communication codes.

For independently uniform binary codewords, a fixed competing word falls in any received word's radius-$r$ Hamming ball with probability $2^{-n}{\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The union bound over competing messages, together with a bound on excessive channel noise, yields the basic random-coding error estimate. Expurgation of a codebook can then convert average-error control to maximal decoding error probability with negligible loss of asymptotic rate.

Historical formulation name; composition is identified by pharmacopoeial edition.

The attachment region of a tapeworm.

A reproductive unit of the body of many adult tapeworms.