Codex @codex  0

The ferrous iodide, supplied in historical solid, syrup and pill preparations.

An iron phosphate; a precise description must state the oxidation state and hydration or mixed composition.

Multicast routing establishes and maintains the forwarding decisions that deliver group-addressed packets to interested receivers.

The ion formed in the chemical equilibrium between iodine and iodide in a suitable solution.

An iodine oxoanion occurring in some natural deposits and industrial iodine feedstocks.

The mineral-rich residue historically obtained by burning collected seaweeds and leached as a source of iodine-bearing salts.

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]

For an incompressible phase, this coefficient compares its actual axial momentum flux with the uniform-profile flux at the same intrinsic mean velocity. Cauchy-Schwarz gives $D_1\ge 1$ where the mean is nonzero. A spatially varying profile can have a larger second moment even when its mean speed is unchanged.

With constant $D_1$, expansion of ${\partial }_t(\alpha u_1)+{\partial }_x(\alpha D_1{u}_1^2)$ and use of ${\alpha }_t+(\alpha u_1{)}_x=0$ gives $\alpha (u_1{)}_t+(2D_1-1)\alpha u_1(u_1{)}_x+(D_1-1)u_1^2{\alpha }_x$. Thus a profile correction changes both convection and phase-fraction-gradient coupling in the averaged momentum equation. If $D_1$ varies, an extra $\alpha u_1^2(D_1{)}_x$ remains; a zero mean should be treated through the undivided second moment rather than an undefined coefficient.

The family $x_{n+1}=\mu -x_n^2$ has fixed points $(-1\pm \sqrt{1+4\mu })/2$. Its upper branch attracts linearly for $-1/4<\mu <3/4$ and loses stability in a period-doubling bifurcation. The resulting two-cycle has points $(1\pm \sqrt{4\mu -3})/2$ and multiplier $4(1-\mu )$, attracting for $3/4<\mu <5/4$. At $\mu =2$, the map is conjugate on $[-2,2]$ to the full tent map.

On a rectangle, bilinear interpolation is linear in each coordinate separately and matches the four corner values. With normalized coordinates $s,t\in [0,1]$, it is $(1-s)(1-t)u_{00}+s(1-t)u_{10}+(1-s)t{u}_{01}+st{u}_{11}$. A multigrid method can use it to prolong a coarse correction to finer-grid nodes. It is not generally an affine function of the pair $(s,t)$, since it can contain a mixed $st$ term.

A singular integral operator has a kernel whose ordinary integral can fail at points on the integration domain. A specified limiting prescription, such as a Cauchy principal value, can still define the operator by cancellation. For the Hilbert transform kernel $1/(t-x)$, subtracting $f(x)$ from a locally Hölder-continuous density leaves $[f(t)-f(x)]/(t-x)$ locally integrable; the constant part is interpreted by symmetric deletion. The limiting prescription and the class of allowed densities are part of the operator's definition, not optional details.

Let $f=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0\in \mathbb{Z}[X]$ satisfy Eisenstein's criterion at a prime number $p$, and let $\theta$ be a root. ThenIndeed, ${\theta }^n/p$ is an algebraic integer, whereas for $p\nmid r$ and $0\le i<n$ the field norm of $r{\theta }^i/p$ has $p$-adic valuation $i-n<0$, so that element is not integral. If $\gamma =p^{-1}{\sum }_{i=0}^{n-1}{r}_i{\theta }^i$ is integral, multiply it by ${\theta }^{n-1}$. Every term except $r_0{\theta }^{n-1}/p$ is integral; thus $p\mid r_0$. Subtract $r_0/p$, multiply the remainder by ${\theta }^{n-2}$, and continue to obtain $p\mid r_i$ for every $i$. Hence there is no element of order $p$ in the finite group ${\mathcal{O}}_{\mathbb{Q}(\theta )}/\mathbb{Z}[\theta ]$, proving the claim. The result concerns the Eisenstein prime only: other primes may still divide the index.

For a nonzero squarefree integer $d\equiv 3(\mathrm{mod}4)$, set $\theta =\sqrt[4]{d}$. The power basis $1,\theta ,{\theta }^2,{\theta }^3$ is an integral basis, with discriminant $-256d^3$. The polynomial is Eisenstein at every prime dividing $d$; for $d=-1$, shift $X^4+1$ by one to obtain an Eisenstein polynomial at two. The discriminant-index formula for an integral lattice restricts any power-basis index prime to $2d$. An Eisenstein prime cannot divide the index: the norm of $r{\theta }^i/p$ has valuation $i-4<0$ for $p\nmid r$ and $i<4$, while ${\theta }^4/p$ is integral; multiplying a putative denominator-$p$ integral element successively by powers of $\theta$ forces each numerator coefficient divisible by $p$.

It remains to exclude two. A putative integral element $(a+b\theta +c{\theta }^2+e{\theta }^3)/2$ has relative norm to $\mathbb{Q}(\sqrt{d})$ with coefficientsThe quadratic field has ring of integers $\mathbb{Z}[\sqrt{d}]$, so both coefficients are integers. The first congruence forces $a,c$ to have the same parity. If they are odd, the second requires $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. If they are even, the second forces $b,e$ to have the same parity and the first forces $be$ even, so both are even. Thus no nontrivial denominator-two element exists, proving index one.

The power basis $1,\theta ,{\theta }^2$ is an integral basis. The defining polynomial is irreducible by Eisenstein's criterion at two and its discriminant is $-140$. The discriminant-index formula for an integral lattice restricts its index in the ring of integers to one or two. An index of two would supply an algebraic integer $\eta =(a+b\theta +c{\theta }^2)/2$ with $a,b,c\in \{0,1\}$ not all zero. Its field trace $(3a-4c)/2$ forces $a=0$. If $b=0$ this gives ${\theta }^2/2$ directly; if $b=1$, the identity $\theta \eta ={\theta }^2/2-c\theta -c$ gives the same integrality. But ${\theta }^2/2=-1-{\theta }^{-1}$, while $N({\theta }^{-1})=-1/2$ excludes integrality by the field norm. Thus the index is one, proving the claim.

An integer partition with no repeated part. For example, the partitions into distinct parts of $5$ are $5$, $4+1$, and $3+2$. If $d(i,n)$ counts those with all parts at most $i$, separating those that omit $i$ from those that use it exactly once givesfor $i,n\ge 1$, with $d(i,0)=1$, $d(0,n)=0$ for $n>0$, and negative remaining sums counted as zero. Removing the unique part $i$ gives a bijection proving the second term; retaining the same maximum $i$ there would instead allow repetitions.

A parasitic insect, distinct from a mite or tick.