Hamilton's equations give $\dot{H}=H_q\cdot H_p-H_p\cdot H_q=0$. Also $\mathrm{div}{V}_H={\sum }_a({\partial }_{q_a}{\partial }_{p_a}H-{\partial }_{p_a}{\partial }_{q_a}H)=0$.

No. Context-free languages are not closed under symmetric difference (a standard counterexample is obtained from the two CFLs enforcing respectively equality of the first two and last two blocks in $a^{\ast }{b}^{\ast }{c}^{\ast }$). Transport such a pair through the bijection $\beta$; pointwise addition corresponds exactly to symmetric difference.

No. Let $X_f=\mathbb{N}$ and let $X_g=K$ be an r.e. undecidable set. Both are r.e., but $X_{f+g}=\mathbb{N}\setminus K$ is not r.e.; otherwise $K$ would be decidable.

Run the algorithms for $f(n)$ and $g(n)$ and output their XOR. Closure of computable functions under composition makes $f+g$ computable.

Put $G=M^{-1}N\ge 0$ and note $A=M(I-G)$. If $\rho (G)<1$, then $A^{-1}=(I-G{)}^{-1}{M}^{-1}\ge 0$. Conversely, if $A^{-1}\ge 0$, take $x\ge 0$, $x\ne 0$, with $Gx=\rho x$. Since $M^{-1}=A^{-1}(I-G)$, $M^{-1}x=(1-\rho )A^{-1}x$. The left side is nonnegative and nonzero, forcing $1-\rho >0$. Finally the iteration error is $e_{k+1}=G{e}_k$, so part (a) gives convergence for every start exactly under this condition.

If $B\ge 0$ and $\rho (B)<1$, then $(I-B{)}^{-1}={\sum }_{k\ge 0}{B}^k$, an entrywise nonnegative limit.

The same Jordan calculation gives convergence of $\sum A^k$ exactly for $\rho (A)<1$. Finite sums satisfy $(I-A){\sum }_{k=0}^m{A}^k=I-A^{m+1}$; taking limits gives $(I-A{)}^{-1}$. Conversely convergence forces $A^k\to 0$, so part (a) applies.

Jordan form shows each block of $A^k$ is a polynomial in $k$ times ${\lambda }^k$. These tend to zero exactly when every $|\lambda |<1$, i.e. $\rho (A)<1$.

Apply the normal-shock relations to $M_{1n}=M_1\sin {\theta }_1$. Tangential velocity continuity and mass conservation giveAs $M_1\to \infty$, maximize ${\theta }_1-{\theta }_2$; with $k=(\gamma -1)/(\gamma +1)$ the optimum has ${\tan }^2{\theta }_1=1/k$, and the maximum deflection is $\arcsin (1/\gamma )$.

Using $T\propto p/\rho$ together with the supplied density ratio and $p_2/p_1=[2\gamma M_1^2-(\gamma -1)]/(\gamma +1)$ gives

Mass, momentum, energy and $c^2=\gamma p/\rho$ eliminate the density and pressure ratios. The result isso a supersonic upstream flow becomes subsonic.

The three jumps express conservation of mass flux, normal and tangential momentum flux, and total enthalpy (energy) flux across the stationary shock.

A radial timelike geodesic has conserved $E=f\dot{t}$ and normalization ${\dot{r}}^2=E^2-f$. Choose $E>1$ and the outward sign. Thenso the observer reaches $r=b$ in finite proper time even though static coordinate $t$ diverges.

Radial nullness factors as $du(f,du+2dr)=0$, giving the two families. For $r<b$, $u=$ constant has $dr/dt=+f$ and is outgoing; $du/dr=-2/f$ has $dr/dt=-f$ and is ingoing. In the $r,t^{\ast }=r+u$ plane they cross the horizon according to these slopes. The cosmological horizon prevents a future-directed inward light signal sent from $r>b$ from reaching the static observer inside.

Let $f=1-r^2/b^2$. Then $U_{\mu }=(1,-1/f,0,0)$ and $g^{\mu \nu }{U}_{\mu }{U}_{\nu }=-1/f+f/f^2=0$. Since $dt=du+dr/f$, the metric becomeswhich is regular at $r=b$.

Integrating the normal Maxwell traction (the radial parts cancel) givesIt points toward the dielectric when ${ϵ}_r>1$. This equals the Coulomb force from the image charge, as it must.

With no free surface charge, tangential $E$ and normal $D$ are continuous. The image ansatz givesAt $z=0^{-}$, with $R=(r^2+d^2{)}^{1/2}$,

Insert $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{E}_t$ into the Lorentz force and use vector identities plus Faraday's law. Rearrangement gives ${\partial }_t{g}_i+{\partial }_j{\sigma }_{ij}=-f_i$. Here $g={ϵ}_{0}E\times B$ is field momentum density and $\sigma$ is the Maxwell stress tensor, whose surface traction transports momentum.