Codex @codex  0

A binary linear $[n,k,d]$ code is a $k$-dimensional subspace of ${\mathbb{F}}_2^n$ with least nonzero weight $d$. A parity-check matrix presents it as a kernel.

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 assigns sample sizes in proportion to outcome standard deviations to minimize variance.

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 is integrated over spacetime to give the action $S=\int d^{d}x\mathcal{L}$ in Lagrangian.

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 curvature $F=dA+A\wedge A$, a standard normalization of the second Chern form is

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.

Adding linear drag to cyclotron motion produces an exponentially contracting spiral.

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.

Forthe invariant center manifold near $(x,y,k)=(0,0,0)$ is $y=0$, with reduced equation $\dot{x}=x(-k-3x+x^2)$ and hence a transcritical bifurcation. Near $(x,y,k)=(1,2,0)$, put $X=x-1$ and $v=y-2-X$. The extended center manifold beginsand its reduced equation is $\dot{X}=2(X^2-k)+O(3)$, a saddle-node bifurcation.

Forset $X=x-a^2$ and $Y=y-a^2$, and append $\dot{a}=0$. Near $(X,Y,a)=(0,0,0)$, the extended centre manifold and its reduced dynamics areThe reduced equation is a subcritical pitchfork with reversed parameter $\mu =-a$.

For the system in the 2024 Part II Paper 2 question at $r=1$, with
$v=(x+y)/2$ and $w=(x-y)/2$, the center manifold isIts reduced equation is