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Apply a Pauli X gate and then a Hadamard gate to the third qubit, and apply Hadamard gates to the first two. This preparesOne application of the Boolean quantum oracle now gives, by quantum phase kickback,The answer qubit is unentangled and may be ignored. If all three qubits must be restored to the displayed register form, applying $H$ and then $X$ to the answer qubit maps $|-⟩$ back to $|0⟩$. Thus the first register is exactly $|f⟩$ after one use of $U_f$.

The physical particle horizon isFor an Einstein-de Sitter universe, $a(t)\propto t^{2/3}$, soSince $H_0=2/(3t_0)$,The horizon problem is that widely separated regions of the observed universe have nearly identical conditions even though, in a decelerating standard Big Bang history, their past light cones do not overlap early enough to establish causal thermal equilibrium.

Spatial homogeneity says that the velocity of $B$ relative to $A$ depends only on their separation. Comparing both galaxies with $C$ givesThus $\mathbf{v}(\mathbf{x}-\mathbf{y})=\mathbf{v}(\mathbf{x})-\mathbf{v}(\mathbf{y})$, so continuity of the macroscopic velocity field turns this Cauchy functional equation into a linear map:Isotropy requires $RH=HR$ for every rotation matrix $R$. The only such rank-two tensor is a scalar multiple of the identity, and henceThis is the Hubble law. A comoving galaxy has physical position $\mathbf{r}(t)=a(t)\mathbf{x}$, so

A word $w$ is accepted by the Epsilon-NFA precisely whenBy part (b), this reached subset is exactly ${\hat{\delta }}_D(q_D,w)$, and by the definition of $F_D$ the intersection condition is equivalent to ${\hat{\delta }}_D(q_D,w)\in F_D$. Hence

Let ${\hat{\delta }}_E(q_0,w)$ mean all states reachable after reading $w$, allowing arbitrary epsilon transitions before, between, and after its symbols. We prove by mathematical induction on $|w|$ thatFor $w=ϵ$, both sides equal $E(\{q_0\})=q_D$. Suppose the claim holds for $w$, and append $a\in \Sigma$. By the definitions of the extended transition functions and the subset construction,This completes the induction.

For a set $S\subseteq Q_E$, let $E(S)$ be its epsilon closure, and putThe subset construction with epsilon transitions gives the deterministic finite automatonUnreachable subsets may be deleted from $Q_D$ without changing the recognized language.

For the high-frequency indices $(n+1)/2\le k\le n$, the angle ${\theta }_k=k\pi /(n+1)$ ranges from $\pi /2$ to $\pi -\pi /(n+1)$. Sincethe large-$n$ high-frequency smoothing factor of weighted Jacobi isFor $0<\omega <1$, this is strictly less than one independently of $n$, so high-frequency error decays geometrically much faster than the low-frequency factor $1-O(n^{-2})$.

To minimize ${\mu }_{\omega }$, balance the endpoint magnitudes:Thus the optimal relaxation and attenuation factors areFor finite $n$, retaining the endpoint $\cos (\pi -\pi /(n+1))=-\cos (\pi /(n+1))$ giveswhich tend to $2/3$ and $1/3$.

A stationary iterative method for a linear system converges from every initial vector exactly when the spectral radius of its iteration matrix is less than one. Here $\cos (kx)$ ranges between $\cos x$ and $-\cos x$, so all modes decay for a fixed $n$ exactly whenRequiring this for every grid size givesIndeed, $\omega \le 0$ leaves or amplifies the lowest mode, while any $\omega >1$ makes the highest-mode eigenvalue tend to $1-2\omega <-1$ as $n\to \infty$.

The slowest mode is $k=1$, for which the Taylor expansion isIf the initial error has a nonzero ${\mathbf{v}}_1$ component, that component is multiplied by ${\lambda }_1(\omega {)}^{\nu }$. Hence for a constant $c>0$ independent of $n$,This is the low-frequency obstruction to grid-independent convergence.

Subtracting the fixed point from the weighted Jacobi method givesPut $x=\pi /(n+1)$. Acting on the supplied sine eigenvector givesThereforeand the corresponding eigenvalues of the iteration matrix are

With $\mathbf{u}=(u_1,\dots ,u_n{)}^T$ and $\mathbf{b}=h^2(f_1,\dots ,f_n{)}^T$, the system isIts diagonal part is $D=-2I$. The weighted Jacobi method, also called the relaxed Jacobi method, isComponentwise,with $u_0^{(\nu )}=u_{n+1}^{(\nu )}=0$.

For pressureless matter, $w=0$ and part (a) gives $\rho =3\mu /(8\pi a^3)$. With $\Lambda =0$, the Friedmann equation becomesorThis is the $\Lambda =0$ case of the Friedmann effective potential for pressureless matter. The potential rises monotonically from $-\infty$ at $a=0$ to zero from below as $a\to \infty$.

For $k=+1$, $C=-1$ intersects the potential at $a_{\max }=\mu$. An initially expanding closed matter-dominated Friedmann solution reaches this turning point and recollapses to a Big Crunch. For $k=0$, $C=0$ gives expansion for all time with $\dot{a}\to 0$ and $a\propto t^{2/3}$. For $k=-1$, $C=1$ gives eternal expansion with $\dot{a}\to 1$ in these units. Thus positive curvature recollapses, whereas zero and negative curvature expand forever.

The cosmological perfect-fluid continuity equation and the constant barotropic equation of state $P=w\rho$ giveOn every interval where $a$ is monotone, separation of variables yieldsHence the constant-equation-of-state density scaling iswhere $\mu$ is a constant. The same expression extends through a regular turning point by continuity.

In the canonical ensemble, a one-particle velocity state has kinetic energy $E(\mathbf{v})=m|\mathbf{v}{|}^2/2$ and hence probability density proportional to $e^{-\beta E}$, where $\beta =(k_{B}T{)}^{-1}$. The three Cartesian Gaussian integrals normalize the density to the Maxwell-Boltzmann velocity distributionIntegrating over a spherical shell of radius $v$ and area $4\pi v^2$ gives the Maxwell speed distributionSinceits unique positive maximum is the most probable speed in the Maxwell distribution