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Hold $p=m\gamma$ fixed. At a point with $x\ne t$,butThe small-$A$ expansions areConsequently, with $H=4p/|x-t|$, the Ultrarelativistic limit of the boosted Schwarzschild metric away from the null plane isEquivalently,with all other components zero.

PutThe rest-frame metric can be written as $Cd{\bar{s}}_{\mathrm{Minkowski}}^2+(C-B^2)d{\bar{t}}^2$. The Minkowski term is invariant under the Lorentz transformation, while $d\bar{t}=\gamma (dt-vdx)$. ThereforeandThese substitutions give precisely

The Lorentz transformation isIt is a boost along the $x$-axis from the frame in which the black hole moves to its rest frame. Direct matrix multiplication shows that replacing $v$ by $-v$ gives its inverse.

For each one-particle state $i$, a boson occupation number can be any $n_i\ge 0$. Its contribution to the grand canonical partition function is the geometric seriesThe noninteracting modes factorize, soConvergence requires $\mu <{\min }_i{ϵ}_i$. Since the lowest energy is zero, the limiting maximum chemical potential is

Put $q=u/x^2$, so $F=xG(q)$ andSubstitution into the Lie-symmetry determining equation for a first-order ordinary differential equation and separation with respect to the arbitrary function $G$ givesThese equations imply $\xi =Cx$ and $\eta =2Cu$. Apart from the zero generator, the common symmetry for arbitrary $G$ is therefore generated byIts one-parameter group is the scaling symmetrywhich leaves $u/x^2$ invariant. Special choices of $G$ can have additional symmetries, but this is the complete symmetry valid for an arbitrary smooth $G$.

Let $\Delta (x,u,u^{\prime })=u^{\prime }-F(x,u)$. Applying the first prolongation givesOn the solution manifold $\Delta =0$, one has $u^{\prime }=F$. The infinitesimal symmetry criterion is therefore exactly

On the jet space of a scalar ordinary differential equation, the $n$th prolongation of a vector field iswhere $D_x$ is the total derivative operator. In particular,

Writing ${\psi }^s(x,u)=(\tilde{x}(s;x,u),\tilde{u}(s;x,u))$, the infinitesimal generator of a Lie point symmetry is

Use the real integral representation of the modified Bessel function of the second kindThe unique stationary point is the strict maximumThe Laplace approximation is consequentlyFor fixed $x>0$, as $\nu \to +\infty$,Substitution gives the required large-order asymptotic of the modified Bessel function of the second kind:

Put $t=u+iv$ and $\Phi (t)=\sinh t-t$. Sincea steepest-descent curve through the saddle point $t=0$ lies inThe saddle is cubic rather than simple:Thus the curves meet the origin in the three descent directions for which $t^3$ is negative real,One curve is the negative real axis $v=0$, $u<0$. The other two are conjugate curves $v=\pm v_{+}(u)$ with $u>0$ and $0<v_{+}(u)<\pi$. Near the origin, $v_{+}(u)\sim \sqrt{3}u$, while for $u\to +\infty$ the defining equation gives $\sin v_{+}\sim 2v_{+}{e}^{-u}$, so $v_{+}\to \pi$. Their three asymptotes are therefore

The minimizer in the quadratic expression from part (b) isRelabelling the time index,

If $\mathrm{Var}(x_0)$ or the observation-noise variances change, the Kalman filter gains and the sequence $V_t$ change. The control gain $K_t$ does not change, because the state dynamics and quadratic cost are unchanged. This is the separation principle: estimation uncertainty changes the additive constants $d_t$, while the optimal linear feedback gain is the full-information gain.

At the terminal time,so $P_T=1$ and $d_T=V_T=1/(T+1)$.

Assume $F(W_t)=P_t{\hat{x}}_t^2+d_t$. Conditional on $W_{t-1}$, the random part of the next Kalman filter estimate is $h_t$ times the innovation. Its variance isTherefore the Bellman equation givesCompleting the square givesStarting from $P_T=1$ yieldsand hence

Suppose inductively thatThe observation innovation isso its conditional variance is $V_{t-1}+1$ and its conditional covariance with $x_{t-1}$ is $V_{t-1}$. The scalar static-state Kalman update therefore givesAdding the known control $u_{t-1}$ shifts the conditional mean but not the variance, giving exactlySince $V_0=1$, induction yieldsEvery update is an affine conditioning operation on jointly Gaussian random variables, so

Suppose an arbitrage existed. Its discounted terminal wealth would satisfy $V_T\ge 0$ $\mathbb{P}$-almost surely and $\mathbb{P}(V_T>0)>0$. Equivalence of $\mathbb{Q}$ and $\mathbb{P}$ gives the same two statements under $\mathbb{Q}$, soBut part (c), after the usual admissibility or localization needed for integrability, givesa contradiction. Hence the existence of the stated martingale measure rules out arbitrage.