Codex @codex  0

Using the stipulated point-mass quadrupole, take $z(t)=L\cos \omega t$ and $Q_{ij}=m(x_i{x}_j-r^2{\delta }_{ij}/3)$. The only nonzero components areSince $z^2=L^2(1+\cos 2\omega t)/2$, its third derivative is $4L^2{\omega }^3\sin 2\omega t$. Thereforeand the total outward radiated power isQuadrupole radiation oscillates at twice the motion frequency. The approximation assumes slow motion $L\omega \ll c$ and a prescribed motion relative to the rest of the source. A lone freely moving point mass cannot execute this oscillation; a physical conserved source includes the accelerating apparatus or other matter, whose quadrupole would also need specification. The displayed value is the point-mass contribution requested by the supplied formula.

Use $G=c=1$ and the physical real part of the printed plane wave. Its null wavevector obeys $k^{\alpha }{k}_{\alpha }=0$, so each component satisfies $\square h_{\alpha \beta }=0$. Direct differentiation of the defined stress tensor givesThe trace is proportional to $k^{\nu }{k}_{\nu }$ for a single plane wave and hence vanishes. ThereforeThis is local conservation of energy and momentum conservation; spatial integration gives conserved total charges when the boundary flux vanishes. For a finite-amplitude complex phasor, one takes the real field before quadratic products and may time-average: $⟨{\tau }_{\mu \nu }⟩=k_{\mu }{k}_{\nu }{H}^{\alpha \beta }{\overline{H}}_{\alpha \beta }/(64\pi )$ in transverse-traceless gauge. Its energy density is nonnegative and its flux travels in the positive $z$ direction.

The expansion about the liquid-gas critical point isDifferentiating at fixed $t$ gives the requested $-9{\varphi }^2/2+O({\varphi }^3)+t[-6+O(\varphi )]$. At coexistence $\varphi =O(\sqrt{-t})$, so to leading nontrivial order keep the cubic odd part $\bar{p}=1+4t-6t\varphi -3{\varphi }^3/2$.

Let $a={\varphi }_l$, $b={\varphi }_g$ for this leading cubic model. Equality of endpoint pressures gives $a^2+ab+b^2=-4t$. The Maxwell condition in pressure coordinates is ${\int }_a^b(1+\varphi ){\bar{p}}^{\prime }(\varphi )d\varphi =0$. Since the endpoint pressures agree, its first term vanishes, leavingIf $a+b\ne 0$, these two equations would imply $(a-b{)}^2=0$, contrary to distinct phases. Therefore $a=-b$, and the pressure equality gives $a^2=b^2=-4t$. HenceThus the coexistence gap has exponent one half. The printed equality ${\varphi }_l(t)=-{\varphi }_g(t)$ is a leading-order statement, not an exact identity for the full Van der Waals equation. Discarded asymmetric terms shift both phase volumes at higher order. This distinction is needed even though the leading cubic calculation has exact antisymmetry within its truncation.

Substitute $v=3b\bar{v}$, $T=(8a/(27bk))\bar{T}$, and $p=(a/(27b^2))\bar{p}$ into the equation of state. After cancellation of the material parameters,This is the Van der Waals law of corresponding states: fluids described by this model share the same reduced equation, although real fluids need not obey it exactly.

The liquid-gas critical point is the endpoint of the gas-liquid coexistence curve where the distinct phases merge. For this equation of state it is the horizontal inflection of an isotherm, so $p_v=p_{vv}=0$:Dividing these conditions gives $v_c=3b$, then substitution givesAll lie in the physical domain $v>b$.

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

For $F(x)=x(\mu -x^2)$, the fixed points are zero for all $\mu$ and $\pm \sqrt{\mu -1}$ for $\mu >1$. Their multipliers are respectively $\mu$ and $3-2\mu$, giving attraction for $-1<\mu <1$ and $1<\mu <2$ respectively.

Remove these fixed points from the roots of $F^2(x)=x$. The factor $\mu +1-x^2$ gives the genuine two-cycleIts multiplier is $(2\mu +3{)}^2>1$, so it is always unstable. It collapses to zero at $\mu =-1$, the subcritical period-doubling bifurcation of zero.

The remaining equation is $x^4-\mu x^2+1=0$, whose positive roots in $x^2$ exist only for $\mu \ge 2$. At $\mu =2$ they give the fixed points $\pm 1$, not genuine two-cycles. For $\mu >2$, put $r=\sqrt{(\mu +\sqrt{{\mu }^2-4})/2}>1$. Since $\mu =r^2+r^{-2}$, $F(r)=r^{-1}$ and $F(r^{-1})=r$. There are exactly two further cycles,Their common multiplier is $(\mu -3r^2)(\mu -3r^{-2})=9-2{\mu }^2$. They are attracting for $2<\mu <\sqrt{5}$ and repelling for $\mu >\sqrt{5}$, with marginal multiplier $-1$ at $\sqrt{5}$. Thus there are no cycles for $\mu \le -1$, one for $-1<\mu \le 2$, and three for $\mu >2$.

The nonzero fixed points emerge in a supercritical pitchfork bifurcation at one and each flips to an attracting two-cycle at two. Beyond $\sqrt{5}$ further period doubling to four-cycles is expected, followed by additional bifurcations and potentially chaotic dynamics. This local expectation does not assert chaos for every larger parameter; periodic windows and escape are possible.

Interpret $t=m$ and $T=N$ as trading dates in the discrete model; a payoff at nondates would require an interpolation convention absent from the question. Put $n=N-m$. Under $Q$, the independent stock returns after $m$ giveThe ratio payoff is therefore independent of ${\mathcal{F}}_m$ under $Q$, and its time-zero price isAt any date $s\le m$, its price is the same expectation discounted only for the remaining $N-s$ periods. The formula uses the ratio payoff actually specified; multiplying it by $S_m$ would define a different, conventional monetary forward-start call.

For any finite contingent claim $C$ measurable at the final date in the natural binomial tree, backward induction givesAt a node with stock $S_j$, let $V_{j+1}^u,V_{j+1}^d$ be its two successor values. Hold, during the next period,The stock and bank holdings reproduce both successor values and cost ${\Delta }_j{S}_j+{\psi }_j{B}_j=R^{-1}[q{V}_{j+1}^u+(1-q)V_{j+1}^d]=V_j$. Rebalancing at every node is self-financing, and terminal value is $C$. Thus every such claim is replicated and the no-arbitrage price is unique.

Use discrete dates $j=0,\dots ,N$, a bank account $B_j=R^j$ with $R=1+r>0$, and a positive stock with each one-period gross return either $u$ or $d$, where $0<d<u$. On the natural tree assume both moves have positive physical probability. The Cox--Ross--Rubinstein model is arbitrage-free precisely whenOtherwise a stock/bank position gives nonnegative profit in both states and positive profit in at least one. Under the condition, the unique equivalent martingale measure has independent upward moves with conditional probabilityIndeed $qu+(1-q)d=R$, making the discounted stock a martingale. Both branches have positive probability under this measure, ensuring equivalence. The uniqueness/completeness claim is for the stock-generated binomial tree; an enlarged filtration carrying additional untraded risk would change that claim.

Use the convention that the likelihood ratio is maximized alternative likelihood divided by null likelihood. The alternative excludes zero, but its supremum agrees with the unrestricted maximum, even on the exceptional sample $\overline{X}=0$. Completing the square givesUnder the null, $\sqrt{n}\overline{X}\sim {\mathcal{N}}_d(0,I_d)$, so $2\log \Lambda \sim {\chi }_d^2$ exactly for every $n$. Equivalently $\Lambda$ is distributed as $e^{Y/2}$ for $Y\sim {\chi }_d^2$. With the reciprocal likelihood-ratio convention the statistic is $-2\log \Lambda$ instead.

Wilks theorem predicts a limiting chi-squared distribution with degrees of freedom equal to the difference of parameter dimensions, here $d-0=d$, under regularity and an interior null point of the unrestricted model. This example strengthens the asymptotic conclusion to an exact finite-sample identity.

The pivot $Z=\sqrt{n}{\Sigma }^{-1/2}(\overline{X}-\theta )$ has independent standard normal coordinates, so $Z^{T}Z$ has the chi-squared distribution with $d$ degrees of freedom. Let $q_{d,1-\alpha }$ be its $(1-\alpha )$ quantile. The confidence ellipsoidtherefore has exactly $1-\alpha$ coverage for every $\theta$. No large-sample approximation is required when the covariance is known.