Codex @codex  0

A binary length-$n$ code of size $M$ and distance at least $2t+1$ satisfiesEquality means the code is perfect.

Appending a parity bit makes every word even-weight. A binary $[n,m,d]$ code becomes an $[n+1,m,d^{+}]$ code, where

A length-$n$ cyclic code over ${\mathbb{F}}_q$ is an ideal of ${\mathbb{F}}_q[X]/(X^n-1)$. It is generated by a monic divisor of $X^n-1$.

A linear code over a finite field ${\mathbb{F}}_q$ is a linear subspace $C\subseteq {\mathbb{F}}_q^n$. Its parameters $[n,k,d{]}_q$ record word length, dimension, and minimum Hamming distance. For a nonzero code, $d$ is the least Hamming weight of a nonzero word, because the difference of two words is another word.

For example, $C=\{000,011,101,110\}$ is a binary $[3,2,2]$ code, the kernel of the parity-check matrix $(111)$. Linearity lets a generator matrix encode messages and lets the parity-check syndrome detect corruption. The zero code needs a separate convention for its minimum distance, since it has no nonzero word.

A binary linear-feedback shift register of degree $d$ updates a $d$-bit state by shifting and inserting a fixed linear combination over ${\mathbb{F}}_2$. With recurrenceits feedback polynomial is $x^d+a_{d-1}{x}^{d-1}+\cdots +a_0$.

A discrete memoryless channel models transmission through finite input and output alphabets using fixed conditional probabilities $W(y\mid x)$. For each input $x$, these nonnegative numbers sum to one over the output symbols. Discrete refers to the symbols, and memoryless means that the noise of one use does not remember earlier uses.

For a fixed input word $x^n=(x_1,\dots ,x_n)$ sent without feedback, the output law isThis is conditional independence of the outputs given the prescribed input word. It does not require an encoder to choose independent input symbols: the codewords of a block code deliberately impose relationships among them. Those relationships let a decoder distinguish possible messages after noise.

The binary symmetric channel and binary erasure channel make two different memoryless noise models with a binary input. One hides flipped bits; the other marks missing bits. Their different channel capacities show why the whole transition matrix matters, as the flips versus erasures in a binary channel demonstrate.

For messages in a finite additive group, a one-time pad encrypts $X$ as $C=X+K$, where the key $K$ is uniform and independent of $X$, as long as the message, and never reused. Then $C$ is uniform and independent of $X$, giving perfect secrecy.

A binary $[n,m,d]$ code is a set of $m$ distinct words in ${\mathbb{F}}_2^n$ whose minimum pairwise Hamming distance is $d$. Here $n$ is the block length and $m$ is the number of codewords.

Response-adaptive randomization updates treatment probabilities using outcomes observed during a trial.

Neyman allocation minimizes the variance of an estimate for a fixed total sampling budget when group variances may differ. For independent observations in two treatment groups, estimate their mean difference by ${\overline{Y}}_1-{\overline{Y}}_0$. With outcome standard deviations ${\sigma }_1,{\sigma }_0>0$ and fixed total $n=n_1+n_0$, its variance is ${\sigma }_1^2/n_1+{\sigma }_0^2/n_0$. Differentiation, or Cauchy-Schwarz inequality, givesThe noisier group receives more observations. If ${\sigma }_1=2{\sigma }_0$, the allocation ratio is $2:1$ rather than $1:1$; integer group sizes require rounding. Equal per-observation costs and a mean-difference estimand are part of this calculation.

For a stratified population mean ${\sum }_h{W}_h{\overline{Y}}_h$, the corresponding fixed-count rule is $n_h\propto W_h{\sigma }_h$. This additional population weight explains why assigning every stratum in proportion to its standard deviation alone is generally incorrect. Unequal costs, minimum group sizes and unknown variances alter the optimization.

A continuous variational symmetry gives a current satisfying ${\partial }_{\mu }{j}^{\mu }=0$ on shell. Integrating $j^0$ over space gives a conserved charge when boundary flux vanishes.

A Lagrangian density assigns a local action contribution to field values and their derivatives. Integrating over spacetime gives $S=\int d^{d}x\mathcal{L}$. For a real scalar in flat spacetime with signature $(+---)$, a simple density is $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -V(\varphi )$.

Varying the field gives $\square \varphi +V^{\prime }(\varphi )=0$, illustrating how the density encodes propagation and forces. Continuous symmetries yield Noether currents under the appropriate variational assumptions. The Lagrangian density is not generally the energy density: temporal and spatial derivative terms combine differently in the Hamiltonian. Adding a total derivative leaves the bulk equations unchanged under compatible boundary conditions, though boundary contributions and canonical expressions can change.

For $R(r)\propto r^l{e}^{-r/[a(l+1)]}$, the radial probability density is proportional to $r^{2l+2}{e}^{-2r/[a(l+1)]}$ and

For a unitary connection with Skew-Hermitian curvature form $F=dA+A\wedge A$, use the total Chern class convention $c(F)=\det (I+iF/(2\pi ))$. Its degree-four part, the Second Chern form, isIndeed, the quadratic term of $\det (I+tX)$ is $[(\mathrm{tr}X{)}^2-\mathrm{tr}(X^2)]/2$; substituting $X=iF/(2\pi )$ proves the formula, since two-forms commute in the wedge product.

For an $SU(n)$ principal connection, $\mathrm{tr}F=0$, reducing this to $c_2(F)=\mathrm{tr}(F\wedge F)/(8{\pi }^2)$. Without that tracelessness condition the trace-square term alone is generally not the Second Chern form. For a line bundle, the two terms cancel, as they must because its second Chern class vanishes.

A chemotactic pattern-forming instability occurs when cells produce an attractant and drift up its gradient strongly enough that positive aggregation feedback overcomes diffusion and turnover at a finite wavelength.

Damped cyclotron motion combines magnetic deflection with a dissipative force. For a nonrelativistic particle in a uniform magnetic field and linear drag $-m\gamma \mathbf{v}$, its perpendicular velocity rotates at the signed cyclotron rate while its magnitude decays as $e^{-\gamma t}$.

The Lorentz force does no work; the drag removes kinetic energy. Integrating the decaying rotating velocity gives a spiral approaching a limiting position. Motion parallel to the field decays without magnetic rotation. This model helps distinguish changing curvature caused by energy loss from changing field strength, but assumes constant mass, drag coefficient and applied field.

If a centre manifold is the graph $v=h(u)$ for $\dot{u}=f(u,v)$ and $\dot{v}=g(u,v)$, its coefficients satisfy $Dh(u)f(u,h(u))=g(u,h(u))$.

Appending $\dot{\mu }=0$ turns a system parameter into a centre variable, allowing one invariant graph to describe nearby parameter values.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

The center of mass moves as if the total external force acted on the total mass; internal forces cancel in its equation.