A thin spherical shell has mass $4\pi r^2\rho dr$ and inward gravitational force $Gm(r)(4\pi r^2\rho dr)/r^2$. The net outward pressure force is $-4\pi r^2{P}^{\prime }(r)dr$. Their balance gives the hydrostatic equilibrium equationSince $m^{\prime }=4\pi r^2\rho$, multiplying by $r^2/\rho$ and differentiating yields $\frac{d}{dr}(r^2{P}^{\prime }/\rho )=-4\pi G{r}^2\rho$.

At the centre, require $m(0)=0$ and finite $P(0)$ and $\rho (0)$. Spherical symmetry gives $P^{\prime }(0)=0$; indeed $m(r)=O(r^3)$ makes the pressure gradient $O(r)$ for regular density. At the surface require $P(R)=P_{\mathrm{ext}}$, usually approximated by zero, because pressure matches the surrounding medium. If the stellar mass is prescribed, also impose $m(R)=M$. An equation of state is needed to close the system. A vanishing surface density is appropriate for some stellar models but is not a universal boundary condition.

Write $c=\cosh \varphi$. The proposed function is $u=(c-2)/(3M(c+1))$. Differentiating using $c^{\prime }=\sinh \varphi$ and $(c^{\prime }{)}^2=c^2-1$ verifies $u^{\prime \prime }+u-3M{u}^2=0$. Its physical branches have $u>0$, namely $|\varphi |>{\varphi }_0=\mathrm{arcosh}2$, andOn the positive branch it comes from infinity at ${\varphi }_0$ and spirals toward $r=3M$ from outside as $\varphi \to \infty$. The negative branch gives the reversed orbit after reflection. The interval where $u<0$ does not describe positive physical radius.

Near ${\varphi }_0$, $\cosh \varphi -2=\sqrt{3}(\varphi -{\varphi }_0)+O((\varphi -{\varphi }_0{)}^2)$. Thus $r\sim 3\sqrt{3}M/(\varphi -{\varphi }_0)$, while $\sin (\varphi -{\varphi }_0)\sim \varphi -{\varphi }_0$. HenceThis is the perpendicular offset of the incoming asymptote and agrees with $|L/E|$ from the circular orbit. The orbit is the critical separatrix between scattering and capture.

The constant $u=1/(3M)$ satisfies $u=3M{u}^2$. It is the circular photon orbit at $r=3M$, the photon sphere. A perturbation $u=u_c+\eta$ satisfies ${\eta }^{\prime \prime }=\eta$ to first order, so perturbations grow exponentially with orbital angle: the circular null orbit is unstable. Substitution into the first integral gives $E^2/L^2=1/(27M^2)$.

Take $x_j=-1+j/4$, $0\le j\le 8$. These nine points satisfy $\cos (4\pi x_j)=(-1{)}^j$. Thus the zero polynomial has nine alternating error extrema of magnitude one and is minimax among polynomials of degree at most seven by the Chebyshev alternation theorem. Hence every such polynomial has uniform error at least one.

One can also prove the bound directly. If the error were strictly below one everywhere, $g(x_j)$ would have sign $(-1{)}^j$. The intermediate value theorem would give a distinct zero in each of the eight intervening intervals, contradicting the degree bound. Since the continuous error attains its maximum on the compact interval, there is an $x$ with $|f(x)-g(x)|\ge 1$.

The Chebyshev alternation theorem says that a polynomial $p$ of degree at most $n$ is a best uniform approximation to a real continuous function $f$ on a compact interval if and only if there are $n+2$ ordered points at which $f-p$ attains its uniform norm with alternating signs. For nonzero error these are equal-magnitude alternating extrema; the zero-error case is understood separately. The best approximating polynomial is unique.

A zero-cost stock holding $\theta$ has terminal payoff $\theta (S_1-1)$. If $a\le 1$, shorting the share and holding a bond gives $1-S_1\ge 0$ and a strictly positive payoff almost surely. If $a>1$, both signs of $S_1-1$ occur with positive probability, so no nonzero holding is an arbitrage. Hence arbitrage freedom is equivalent to $a>1$.

The equivalent martingale measures are exactly the densities $Z(s)>0$ almost everywhere on $(0,a)$ satisfyingThey exist for $a>1$: exponential tilts of the uniform law have means ranging continuously from zero to $a$, so one has mean one.

Under the usual utility assumptions $U^{\prime }>0$, $U^{\prime \prime }<0$, and existence of an admissible optimum, $J(\theta )=E[U(w+\theta (S_1-1))]$ is strictly concave and $J^{\prime }(0)=U^{\prime }(w)(a/2-1)$. Its decreasing derivative implies that the optimizer is positive exactly when $a>2$, negative when $a<2$, and zero when $a=2$, provided the feasible interval contains zero in its interior. The hypothesis $C^2$ alone is insufficient: a linear increasing utility can give an unbounded objective rather than a finite optimizer. Concavity and existence are implicit in the standard utility formulation.

For $a=2$, $\theta =0$ makes marginal utility constant, so the pricing measure is the original uniform law andFor any equivalent pricing measure, the call expectation is positive and strictly less than $E_Q{S}_1/2=1/2$, since $(s-1{)}^{+}<s/2$ for $0<s<2$. Both endpoints can be approached. Symmetric densities concentrated in $[1-\delta ,1+\delta ]$ give prices tending to zero; symmetric densities concentrated near zero and two give prices tending to one half. Mix either density with an arbitrarily small amount of the uniform law to keep it everywhere positive; symmetry keeps its mean one. The set of positive mean-one densities is convex and the call price is linear, so every intermediate value is attained. Thus the exact price range is $(0,1/2)$, with neither endpoint attained.

An arbitrage is a portfolio with nonpositive initial cost, nonnegative terminal payoff almost surely and a positive payoff with positive probability, allowing unused initial cash to be invested in the riskless asset. Equivalently it is a zero-cost portfolio with those terminal properties. An equivalent martingale measure $Q$ has the same null sets as the physical measure and satisfies $E_Q[S_1^i]=(1+r)S_0^i$ for every traded asset, with appropriate integrability.

For an agent with differentiable increasing concave utility and an optimal portfolio, let $W^{\ast }$ be its terminal wealth. The first-order conditions for variations in risky holdings are $E[U^{\prime }(W^{\ast })(S_1^i-(1+r)S_0^i)]=0$. If $U^{\prime }(W^{\ast })>0$ and integrable, normalize it to the probability density $Z=U^{\prime }(W^{\ast })/E[U^{\prime }(W^{\ast })]$. The first-order conditions make $Q$ with $dQ=ZdP$ an equivalent martingale measure. A small additional payoff $H$ then has marginal priceThis follows by setting the first variation of expected utility for buying a small claim at price $p$ equal to zero.

One complete cycle consists of ten stops of mean one minute and ten journeys of mean four minutes, so its mean duration is fifty minutes. Over the eight-hour shift, the elementary renewal approximation gives $480/50=9.6$ completed circuits on average.

For the end-of-shift calculation use the long-run occupation probabilities of the twenty successive phases. Each stop phase has probability $1/50$, and each travel phase probability $4/50$. Alighting stop $j$ is therefore selected with probability $1/50+4/50=1/10$, from either stopping there or travelling toward it. If that stop is the station, no further stops are passed. If it is stop $j\ge 2$, the intermediate stops are $j+1,\dots ,10$, numbering $10-j$. Thus the requested approximate mean isThe approximation treats eight hours as long compared with a cycle and ignores the initial-phase transient. Exponential holding times justify the twenty-phase Markov description; independent cycle times justify the renewal rate.

For independent identically distributed positive holding times $S_i$, define $T_0=0$, $T_n={\sum }_{i=1}^n{S}_i$ and the renewal count $X_t=\max \{n:T_n\le t\}$. With $0<\mu =E{S}_1<\infty$, the renewal strong law is $X_t/t\to 1/\mu$ almost surely. The elementary renewal theorem is $m(t)/t\to 1/\mu$, where $m(t)=E{X}_t$. More generally the rate is zero when the mean holding time is infinite, under the usual nonexplosion assumptions.

Set $Y_n={\sup }_{m\ge n}|f_m|$. It is an extended-real measurable function and decreases to $Y={\mathrm{limsup}}_m|f_m|$. If $f_n\to 0$ almost surely, then $Y_n\to 0$ almost surely. For any $ϵ>0$, the decreasing sets $\{Y_n>ϵ\}$ have intersection contained, up to a null set, in the exceptional set for convergence. Continuity from above of the probability measure gives $P(Y_n>ϵ)\to 0$.

Conversely, if $Y_n\to 0$ in probability, then $P(Y>ϵ)\le P(Y_n>ϵ)\to 0$ for every positive $ϵ$. Taking rational $ϵ>0$ gives $Y=0$ almost surely, so $|f_n|\to 0$ almost surely. This proves both directions, even when some individual $Y_n$ take the value infinity.

Convergence to a finite real limit is equivalent to the Cauchy property. The convergence set is thereforeEach set in braces is measurable, and sigma-algebras are closed under all the displayed countable operations. Completeness of $\mathbb{R}$ ensures that the Cauchy property gives a finite limit, proving the claim.

The original PDF has height $1/n$ on the interval $[0,e^n]$. For finite $p$,so the sequence cannot converge in $L^p$, since a norm-convergent sequence is bounded. For $p=\infty$, $\Vert f_n{\Vert }_{\infty }=1/n\to 0$, so it converges to zero in $L^{\infty }$. It also converges pointwise, and hence almost everywhere, to zero. Finally for each $ϵ>0$, once $n>1/ϵ$ the set $\{|f_n|>ϵ\}$ is empty. Thus it converges in measure to zero, even on this infinite-measure space.

A measure space consists of a set $\Omega$, a sigma-algebra $\mathcal{A}$ of subsets, and a countably additive map $\mu :\mathcal{A}\to [0,\infty ]$ with $\mu (\varnothing )=0$. A sigma-algebra contains $\Omega$ and is closed under complements and countable unions. Countable additivity means $\mu ({\bigcup }_n{A}_n)={\sum }_n\mu (A_n)$ for pairwise disjoint measurable sets. Such sets are called measurable; a real function is measurable when the inverse image of every Borel set belongs to $\mathcal{A}$.

In normal coordinates, each radial line $u(t)=tv$ is an affinely parametrized geodesic. Its coordinate geodesic equation gives ${\Gamma }_{ij}^k(tv)v^i{v}^j=0$. At $t=0$ this holds for every tangent vector $v$. Since ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$, polarization of this quadratic form gives every coefficient zero. Thus all Christoffel symbols vanish at the centre $p$; they need not vanish elsewhere.

For a parametrization $X(u^1,u^2)$, put $g_{ij}=X_i\cdot X_j$. Decompose the second derivatives into tangent and normal components:These tangent coefficients are the Christoffel symbols. Differentiate $g_{ij}$ and combine the resulting three identities to obtainHere $(g^{kl})$ is the inverse matrix. The formula proves that these symbols depend only on the first fundamental form and its first derivatives, not on the second fundamental form.

For $v\in T_{p}S$ sufficiently small, let ${\gamma }_v$ be the geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. Define ${\exp }_p(v)={\gamma }_v(1)$. Geodesic uniqueness and rescaling give ${\gamma }_v(t)={\exp }_p(tv)$, soThus its derivative at zero is the identity, an isomorphism. Assuming the stated smoothness, the inverse function theorem proves ${\exp }_p$ is a diffeomorphism between neighbourhoods of zero and $p$.

Choose centred charts so the coordinate expression $F$ satisfies $F(0)=0$. On a component where the map is nonconstant, the identity theorem makes $F$ nonconstant. Its Taylor expansion factors as $F(z)=z^{n}h(z)$, with $n\ge 1$ and $h(0)\ne 0$. In a sufficiently small disc, choose a holomorphic logarithm of $h$, and hence a holomorphic $n$th root $q$ with $q^n=h$. The function $zq(z)$ has nonzero derivative at zero. The holomorphic inverse function theorem, which says that a holomorphic function with nonzero derivative has a holomorphic local inverse, makes it a valid new source coordinate. In that coordinate,If the map is bijective, local injectivity forces $n=1$: for $n>1$, a small nonzero value has $n$ distinct preimages in this local disc. The inverse is therefore holomorphic in a neighbourhood of every point, and the local inverses agree, proving the map is biholomorphic. The nonconstant hypothesis must apply to the component containing $p$ if disconnected Riemann surfaces are allowed; otherwise a constant component is an exception to the proposed local form.

Let $A$ be the set of points with a neighbourhood on which $f=g$. It is nonempty and open. If $p$ is in its closure, continuity and Hausdorffness give $f(p)=g(p)$. Choose a common target chart and a source disc around $p$ on which both images lie in that chart. Their coordinate difference is holomorphic and vanishes at accumulating points from $A$, so the identity theorem makes it vanish on the disc. Thus $p\in A$. Therefore $A$ is also closed, and connectedness implies $f=g$ on all of $R$.

The classes of $A_4$ are the identity, the three double transpositions, and two classes of four three-cycles. Its quotient by $V_4$ is cyclic of order three, giving characters $1,\chi ,{\chi }^2$, with $\chi$ taking values $1,1,\omega ,{\omega }^2$ on these classes. The restriction of $V$ has character $(3,-1,0,0)$ and norm $(9+3)/12=1$, so it is the three-dimensional irreducible representation $T$. These four representations are complete since $1+1+1+9=12$.

The sign is trivial on $A_4$, soFor the last equality the restricted character is $(2,2,-1,-1)$, the sum of the two indicated one-dimensional characters. Equality of characters proves the decomposition.

The conjugacy classes are specified by cycle type, with sizes $1,6,3,8,6$ for $1,(12),(12)(34),(123),(1234)$. The trivial and sign representations give the first two rows. The permutation representation on four letters contains a trivial line; its orthogonal complement has character equal to the number of fixed letters minus one. Tensoring that representation by the sign gives another row. Finally the quotient $S_4/V_4\cong S_3$ supplies the two-dimensional standard representation. The resulting table isWith the class-size-weighted inner product, these characters have norm one and are pairwise orthogonal, so all are irreducible and inequivalent. Their squared dimensions sum to $1+1+9+9+4=24=|S_4|$, proving completeness.