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Write $r={\varphi }_0^2/M_{\mathrm{Pl}}^2$. As $t\to 0$,so $\varphi \to -\infty$ andBecause $2/3<r<2$, the power $r/2$ lies in $(1/3,1)$; thus $\ddot{a}<0$ and the early expansion decelerates. As $t\to \infty$, $\varphi \to 0$ and $\sinh (2\lambda t)\sim e^{2\lambda t}/2$, givingso the late universe has accelerating expansion.

The transition satisfies $\ddot{a}/a=H^2+\dot{H}=0$. Substitution givesor $r{\cosh }^2(2\lambda t)=2$. ThereforeThis is the transition described by the exact hyperbolic-sine scalar-field cosmology.

Put $u=\tanh (\lambda t)=e^{\varphi /{\varphi }_0}$. The hyperbolic cotangent double-angle identity givessoAlsoSolving the Friedmann equation for the scalar potential,therefore gives

Differentiating the ansatz givesThe Raychaudhuri equation therefore becomesSince $d[\coth (2\lambda t)]/dt=-2\lambda {\mathrm{csch}}^2(2\lambda t)$, the requested specific solution takes the integration constant to be zero:Now $\dot{a}/a=H$, soand

Differentiate the Friedmann equation:The scalar-field equation gives $\ddot{\varphi }+c^2{V}^{\prime }=-3H\dot{\varphi }$, henceBy continuity this also holds through isolated zeros of $H$, so the Raychaudhuri equation is

For real $z>0$, the exponential decays as $\mathrm{Re}t\to -\infty$, so two allowed contours run from $-\infty$ to $i$ and from $-\infty$ to $-i$. For real $z<0$, it decays as $\mathrm{Re}t\to +\infty$, so reverse the construction: run from $i$ or $-i$ to $+\infty$. These are the infinite contours in the diagram.

For $z>0$ their integrals arefor $z<0$ the corresponding orientation changes only an overall sign. Taking the sum of the two exponentials produces a solution independent of $\sin z/z$:Its nonzero Wronskian with $w_1$ on $z\ne 0$ proves linear independence.

Because $\mathrm{Re}a>0$, the endpoint factor $(1+t^2{)}^a$ vanishes at the branch points $t=\pm i$. A finite contour may therefore run from $-i$ to $i$ on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,Hence a nonzero constant multiple gives the solution

Insertinto the equation. Since $w^{\prime }={\int }_{\gamma }t{e}^{zt}fdt$ and $w^{\prime \prime }={\int }_{\gamma }{t}^2{e}^{zt}fdt$, use $z{e}^{zt}={\partial }_t{e}^{zt}$ and integration by parts:The integral vanishes whenso, up to a constant and on a consistently chosen branch,The remaining contour condition isThis is the Laplace contour-integral method for a linear differential equation.

Fit a Poisson regression to artificial responses $Z_i=1$ with meansWith the logarithmic link, this is the generalized linear modelwhere $\log \Lambda (Y_i)$ is a known offset. Its log likelihood, apart from constants, isThe differencedoes not depend on $\beta$. The Poisson surrogate therefore has exactly the same maximum-likelihood estimate of $\beta$. This is the Poisson surrogate for an uncensored proportional-hazards likelihood.

Letbe the baseline cumulative hazard. Under the proportional hazards model,ThusIndependence makes the log likelihood

Writing $S_i(t)=1-F_i(t)$ for the survival function, the definition of the hazard function givesSince $S_i(0)=1$, integration yieldsThereforeThe exponent is the negative cumulative hazard.

Let $S(n)=n+1$ be the successor function. Define $f$ by primitive recursionInduction gives $f(n)=2n$. Then defineby composition. Initial functions, composition, and primitive recursion produce primitive recursive functions, so $f$ and $g$ are partial recursive functions and are visibly defined for every $n$. Both functions are total.

A many-one reduction $A{\le }_{m}B$ is a total computable function $f$ such thatIf $B$ is a computably enumerable set, compute $f(n)$ and run a semidecision procedure for $B$ on that value. It halts exactly when $f(n)\in B$, equivalently exactly when $n\in A$. Therefore $A$ is computably enumerable.

Fix an effective enumeration ${\phi }_0,{\phi }_1,\dots$ of unary partial computable functions. The diagonal halting set isIt is a computably enumerable set: on input $e$, simulate ${\phi }_e(e)$ and accept if that computation halts.

Suppose $K$ were a computable set. A program could then halt on input $e$ exactly when the decider says $e\notin K$. Let $d$ be its own index. If $d\in K$, the program does not halt, while if $d\notin K$, it halts. Both alternatives contradict the definition of $K$. Thus $K$ is computably enumerable but not computable.

This is superdense coding. To encode $00,01,10,11$, Alice applies respectively $I,X,Z,XZ$ to her half of $|{\Phi }^{+}⟩$. These local Pauli operators transform the shared state into the four mutually orthogonal Bell states $|{\Phi }^{+}⟩,|{\Psi }^{+}⟩,|{\Phi }^{-}⟩,|{\Psi }^{-}⟩$, up to irrelevant global phases. Alice sends her one qubit to Bob. Bob now holds both qubits and performs a Bell-basis measurement, whose outcome identifies Alice's two-bit message with certainty. One transmitted qubit communicates two classical bits because one shared Bell pair was available beforehand.

A pure bipartite state is entangled when it is not a product state $|u{⟩}_A\otimes |v{⟩}_B$.

Sincewe haveIts coefficient matrix is proportional towhose determinant is $-4\ne 0$. By the two-qubit product-state determinant criterion, a product state would have determinant zero. Therefore $|\alpha ⟩$ is entangled.

Let $D_i$ be the degree of a vertex and putIf $i$ and $j$ are adjacent, then ${\pi }_i{p}_{ij}=1/{\sum }_k{D}_k={\pi }_j{p}_{ji}$; if they are not adjacent, both sides vanish. Thus the walk is in detailed balance, and its invariant distribution is proportional to vertex degree.

Detailed balance means ${\pi }_i{p}_{ij}={\pi }_j{p}_{ji}$ for all $i,j$. Summing over $i$ givesso detailed balance implies invariance.