Codex @codex  0

Let $x$ be the parameter-contrast vector for the specified game: it has entries $+1$ for ${\beta }_i,{\beta }_j,{\beta }_{\{i,j\}}$, entries $-1$ for ${\beta }_k,{\beta }_{\ell },{\beta }_{\{k,\ell \}}$, and zero elsewhere. PutUnder the stated regularity assumptions, asymptotic normality of a maximum likelihood estimator giveswhere $i_N(\beta )/N\to I(\beta )$. Since $g^{\prime }(\eta )=p(1-p)$, the delta method givesEquivalently, for large $N$,with $p=e^{x^T\beta }/(1+e^{x^T\beta })$.

Whenever players 1 and 2 occur, the linear predictor contains their parameters only throughFor any constants $a,b$, replacingleaves every fitted probability and hence the likelihood function unchanged. The parameterization is therefore nonidentifiable: at most ${\gamma }_{12}$ can be estimated uniquely. Hence there is no unique maximum-likelihood estimate of the three separate parameters.

No intercept should be included unless the ordering of the two teams has an external meaning such as home advantage. Ifthen reversing the teams gives ${\eta }_{k\ell ,ij}=-{\eta }_{ij,k\ell }$, so the logistic model automatically givesAn intercept would not change sign when the teams are swapped and would violate this required symmetry. Thus an intercept does not make sense in the stated model.

The converted source omits $r$ from the product; the PDF asks about $N=pqr$. Fix the odd prime $r$, and suppose $pqr$ is Carmichael. Its three prime factors are distinct by the Korselt criterion. Relabel $p,q$ so that $p<q$.

The divisibility conditions includeWritefor a positive integer $b$. Since $p<q$,so $1\le b<r$. Moreover,Multiplying the second divisibility by $b$ and reducing modulo $p-1$ givesThe positive integer on the right is at most $(r-1)(2r-1)$. ConsequentlyThere are only finitely many possible primes $p$ and integers $b\in \{1,\dots ,r-1\}$, and each pair determines at most one $q=(pr+b-1)/b$. Therefore

An odd composite $N$ coprime to $b$ is a Fermat pseudoprime to base $b$ when $b^{N-1}\equiv 1(\mathrm{mod}N)$. It is a Carmichael number when this congruence holds for every $b$ coprime to $N$.

Suppose a Carmichael number has $p^a\Vert N$ with $a\ge 2$. The group $(\mathbb{Z}/p^a\mathbb{Z}{)}^{\ast }$ is cyclic, so choose a residue of orderBy the Chinese remainder theorem, extend it to an integer $b$ that is congruent to $1$ modulo every other prime-power factor of $N$. Then $\gcd (b,N)=1$, and the Carmichael congruence forces the order of $b$ modulo $p^a$ to divide $N-1$. In particular $p\mid N-1$, contradicting $p\mid N$. Therefore every Carmichael number is square-free.

Now assume $N$ is square-free. If $p-1\mid N-1$ for every $p\mid N$, then for every $b$ coprime to $N$, Fermat little theorem givesfor each $p\mid N$. The Chinese remainder theorem then gives the congruence modulo $N$, so $N$ is Carmichael.

Conversely, if $N$ is Carmichael, fix $p\mid N$. Choose a primitive root $g$ modulo $p$ and use the Chinese remainder theorem to choose $b\equiv g(\mathrm{mod}p)$ and $b\equiv 1(\mathrm{mod}N/p)$. Then $b^{N-1}\equiv 1(\mathrm{mod}p)$, so the order $p-1$ of $g$ divides $N-1$. This proves the Korselt criterion in the required form:

A Carmichael number cannot have one prime factor because it is composite. If $N=pq$ with distinct primes $p<q$, the condition $q-1\mid pq-1$ implies $q-1\mid p-1$, which is impossible. Hence every Carmichael number has at least three prime factors.

The Fermat-Euler theorem states thatIndeed, if $r_1,\dots ,r_{\phi (n)}$ are the reduced residue classes modulo $n$, multiplication by $a$ permutes them. ThereforeThe product ${\prod }_i{r}_i$ is invertible modulo $n$, so cancellation gives the theorem.

Now let $p$ be prime. If $k\equiv 1(\mathrm{mod}p-1)$, then for $p\nmid b$, Fermat little theorem gives $b^{k-1}\equiv 1(\mathrm{mod}p)$, while $p\mid b$ makes $b^k\equiv b\equiv 0$. Thus $b^k\equiv b$ for every integer $b$.

Conversely, assume this congruence holds for every $b$. Choose a primitive root $g$ modulo $p$. Then $g^{k-1}\equiv 1(\mathrm{mod}p)$. Since $g$ has order $p-1$, one has $p-1\mid k-1$. Hence

No classical one-query algorithm can determine $x_0$ with certainty. If it queries a string $x$ and receives $1$, then it knows $x_0=x$. But if it receives $0$, each of the other three strings remains compatible with the response. Randomization cannot remove this ambiguity when certainty is required. ThusIndeed, three adaptive classical queries are necessary and sufficient in the worst case: after three negative answers, the unqueried fourth string must be $x_0$.

The marked-state phase oracle changes only the amplitude of $|x_0⟩$. Since that amplitude in $|\psi ⟩$ is $1/2$,Also,It follows that $A|\psi ⟩=-|\psi ⟩$ and $A|x_0⟩=|x_0⟩-|\psi ⟩$, soThe global minus sign has no observable effect. Prepare $|\psi ⟩=H^{\otimes 2}|00⟩$, query $I_{x_0}$ once, apply the fixed operator $A$, and perform a quantum measurement in the computational basis. The outcome is $x_0$ with probability one. This is exact Grover search on four entries.

Let $P=|\psi ⟩⟨\psi |$. Since $|\psi ⟩$ is normalized, $P^2=P=P^{†}$. Hence the Householder reflectionis Hermitian andTherefore $A$ is unitary.

For a bosonic degree of freedom, the supplied integral givesFor a fermionic degree of freedom, useChanging variable $u=2y$ in the second integral givesThis is the energy-density case of the fermion-to-boson thermal integral ratio.

Summing the result of part (a) over all particle species therefore giveswhere the effective number of relativistic energy degrees of freedom is

In the very early universe, assume species $i$ is in thermal equilibrium, is ultrarelativistic so that $kT\gg m_i{c}^2$ and $E(p)\simeq pc$, and has negligible chemical potential, ${\mu }_i\simeq 0$. Its occupation factor is then the Bose-Einstein distribution for a boson or the Fermi-Dirac distribution for a fermion.

Substitute $E=pc$ and set $y=pc/(kT)$. Since $p=(kT/c)y$ and $dp=(kT/c)dy$,

The original grammar generatesbecause each use of $S\to ASc$ adds one $a$ on the left and one $c$ on the right, and $S\to B\to b$ terminates the derivation.

Introduce nonterminals $S_0,X,C$ and useEvery production is in Chomsky normal form. The production $X\to SC$ completes one matched $a,c$ pair, while either start-like symbol terminates with $b$, so the grammar generates exactly the same language. These productions are a valid $G_{\mathrm{Chom}}$.

The standard conversion of a context-free grammar to Chomsky normal form proceeds as follows:

Add a fresh start symbol $S_0$ with $S_0\to S$.
Find the nullable nonterminals, add the variants obtained by omitting nullable occurrences, and remove all epsilon productions. Omit $S_0\to ϵ$ as well because the requested language excludes $ϵ$.
Remove every unit production $A\to B$ by copying the non-unit productions reachable from $B$ to $A$.
Remove nonterminals that are unreachable from $S_0$ or cannot derive a terminal word.
In every right-hand side of length at least two, replace a terminal $a$ by a fresh nonterminal $T_a$ with $T_a\to a$.
Replace each right-hand side of length greater than two by a chain of binary productions using fresh nonterminals.

Each remaining production is of the required binary or terminal form, and these transformations preserve every nonempty generated word. Thus $\mathcal{L}(G_{\mathrm{Chom}})=\mathcal{L}(G)\setminus \{ϵ\}$.

A context-free grammar is in Chomsky normal form when every production rule has one of the formswhere $A,B,C$ are nonterminal symbols and $a$ is a terminal symbol. If the language contains the empty word, one may additionally permit $S_0\to ϵ$, with the start symbol $S_0$ absent from all right-hand sides.

For the same Fourier mode, the leapfrog finite-difference scheme for the diffusion equation givesThe two amplification roots areFor every nonzero mode $q>0$, one hasThus every $\mu >0$ admits a growing Fourier mode, and

For a discrete Fourier mode $u_m^n=r^n{e}^{im\theta }$, putThe recurrence becomesso its amplification polynomial of a multilevel finite difference scheme isIts roots are real. Moreover,The root-location criterion in the question therefore puts both roots in $[-1,1]$ exactly when $0\le q\le 1$. Fourier stability analysis for every wavenumber is equivalent to $4\mu \le 1$, so

Let ${\delta }_x^2{u}_m=u_{m-1}-2u_m+u_{m+1}$. For an exact smooth solution of the diffusion equation, the second-order central difference givesSince the Courant number $\mu =k/h^2$ is fixed, $h^2=O(k)$. Taylor expansion in time givesUsing $u_t=u_{xx}$ and $u_{tt}=u_{xxt}$ cancels the leading terms. Because $k{h}^2=O(k^2)$, the residual, and hence the local truncation error in the convention of the question, is