Codex @codex  0

If a property of an ordinal follows whenever it holds for every smaller ordinal, then it holds for every ordinal.

Transfinite recursion defines a value at each ordinal from the function of all earlier values.

A nonzero ordinal is a limit ordinal when it is not a successor ordinal. Equivalently, it is the supremum of all smaller ordinals.

The order type of a well-order is the unique ordinal isomorphic to it.

A countable ordinal has a countable set as its underlying set.

These are the ordinals strictly below the first uncountable ordinal.

The successor of an ordinal $\alpha$ is $\alpha +1=\alpha \cup \{\alpha \}$.

Independent and identically distributed random variables with a continuous probability distribution realize each of their possible strict orderings with equal probability.

A conserved scalar order parameter evolves by transporting it rather than creating or destroying its total amount. A simple relaxational model has chemical potential $\mu =\delta F/\delta \varphi$, current $\mathbf{J}=-M\nabla \mu$ andwith compatible boundary conditions and optional conserved thermal noise. Integrating gives constant total $\int \varphi$ when boundary flux vanishes.

Linearizing a quadratic Landau-Ginzburg theory gives a decay rate $\Gamma (q)=M{q}^2(a+\kappa q^2)$. Away from criticality, positive $a$ yields ordinary diffusive $q^2$ scaling at small wavevector. At the Gaussian critical point $a=0$, the leading rate is instead proportional to $q^4$. Conservation forces the zero-wavevector mode to remain unchanged but does not by itself fix every critical decay exponent.

Nonconserved relaxational order-parameter dynamics has ${\partial }_t\psi =-\Gamma \delta F/\delta \psi$ plus thermal noise. Its zero-wavenumber mode may decay.

An involution is a group element $s$ satisfying $s^2=1$. Some conventions require $s\ne 1$.

A mean-motion resonance has $j{n}_2-k{n}_1\simeq 0$ for small integers $j,k$. Its slow resonant angle combines mean longitudes with apsidal or nodal longitudes.

For an optical ray in cylindrical coordinates, a helical curve has $r=r_0>0$ and $\theta (z)=az+{\theta }_0$. It is a helical extremal only if it satisfies the Euler-Lagrange equations of the optical action. In a radial-index medium $n=n(r)>0$, the angular and axial first integrals are constant on this ansatz, while the radial equation reduces toIndeed the derivative with respect to $r\prime$ vanishes along the constant-radius ansatz, leaving the partial derivative of $n(r)\sqrt{1+r^2{a}^2}$ equal to zero.

A nonzero real winding rate therefore requiresFor example, the smooth positive profile $n(r)=n_0{e}^{-\alpha r^2/2}$ with $n_0,\alpha >0$ admits $a^2=\alpha /(1-\alpha r_0^2)$ for $r_0<{\alpha }^{-1/2}$. In a constant-index medium, the equation instead forces $a=0$, giving a straight axial line. These are stationary-ray conditions on the chosen segment, not a proof that every helical segment minimizes optical length.

The open disc with centre $z$ and radius $r>0$ is $\{w:|w-z|<r\}$.

In a metric space, the open ball of radius $r>0$ about $x$ is $\{y:d(x,y)<r\}$; it is the open-boundary convention for a ball.

This is the open-boundary case of an interval.

An open interval in the real line is a set $(a,b)=\{x:a<x<b\}$, allowing either endpoint to be infinite.

A bounded bijective linear map between Banach spaces has a bounded inverse. This follows because an open bijection has continuous inverse.

Suppose one source message is transmitted with its endpoint supplied externally, and allow the empty word. List source symbols in decreasing order of probability, $p_1\ge p_2\ge \cdots$, and assign distinct binary words in increasing length order. Then the $i$th word has length $\lfloor {\log }_{2}i\rfloor$, minimizing expected codeword length among one-to-one source codes.^[ref]

There are $1+2+\cdots +2^{\ell }=2^{\ell +1}-1$ binary words of length at most $\ell$. This proves the length formula. If a more probable symbol had a longer word than a less probable one, exchanging the two would decrease or preserve the expected value of the length, proving optimality.

For probabilities $(1/2,1/4,1/4)$, use the empty word, 0 and 1. The average is $1/2$ bit, compared with $3/2$ bits for an optimal binary prefix code. The difference uses the supplied endpoint: the three-word assignment cannot decode an arbitrary concatenated message stream without framing information.

If an interceptor knows that an additive one-time-pad ciphertext $C$ encrypts plaintext $P$ and wants it to decrypt as $P^{\prime }$, replacing it byworks without knowing the key, since $C^{\prime }=P^{\prime }+K$.

An encryption scheme has perfect secrecy when the ciphertext and plaintext are independent random variables, so observing the ciphertext does not change the plaintext's probability distribution.

If the left and right travelling-wave coefficients are related bythen