First assume $a-b\notin \mathbb{Z}$ and nonsingular parameters, so the two infinity solutions $w_1$ and $w_2$, obtained by exchanging $a,b$, are independent. Write $F=A{w}_1+B{w}_2$. Since the power series at $1/z=0$ starts at one, $w_1\sim (-z{)}^{-a}$ and $w_2\sim (-z{)}^{-b}$.

In the parameter region $\Re (b-a)>0$, let $z\to -\infty$ in the Pfaff transformation. The $w_2$ term vanishes after multiplication by $(-z{)}^a$, while $z/(z-1)\to 1$ and $(-z)/(1-z)\to 1$. The supplied value at one of the Gauss hypergeometric function therefore givesAnalytic continuation in the parameters extends this coefficient to the nonresonant parameter domain. Exchange $a,b$ to determine $B$. Hence the hypergeometric connection formula at infinity isThe branch is $|\arg (-z)|<\pi$. The existence of $F(a,b;c;1)$ alone does not exclude integer $a-b$. At such resonant values the two separate displayed terms can be singular; the correct statement takes their combined parameter limit, which can contain logarithms. This qualification is essential, for example when $a=b$.

For $F(a,1+a-c;1+a-b;w)$, the exponents at $w=0,1,\infty$ are $(0,b-a)$, $(0,c-a-b)$, and $(a,1+a-c)$. The substitution $w=1/z$ exchanges zero and infinity, giving $(a,1+a-c)$ at $z=0$, $(0,c-a-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(-z{)}^{-a}$ shifts the zero exponents by $-a$ and the infinity exponents by $+a$. The resulting Papperitz symbol is again $(0,1-c)$, $(0,c-a-b)$, $(a,b)$. Therefore $w_1$ satisfies the same hypergeometric equation, with the stated branch $|\arg (-z)|<\pi$.

Write the Papperitz symbol by listing pairs of exponents at $0,1,\infty$. For $F(a,c-b;c;w)$ they are $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. Under $w=z/(z-1)$ the points $0,1,\infty$ in the $z$-plane correspond respectively to $0,\infty ,1$ in the $w$-plane. Thus the pulled-back exponents are $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity.

Multiplication by $(1-z{)}^{-a}$ shifts the exponents at one by $-a$ and those at infinity by $+a$, yieldingThis is the original Gauss hypergeometric equation. Both functions are analytic and normalized to one at zero, so uniqueness of that normalized local solution proves the Pfaff transformationContinue the identity using the branch $|\arg (1-z)|<\pi$. As usual, the parameters must admit the normalized hypergeometric solution; excluded singular parameter values are treated only when their limiting solution exists.

At onset the determinant has a double root, soSquaring the neutral-boundary equation and canceling the positive factors gives $u_0=s/[d(2\sqrt{2s}-d)]$. Substitution yieldsFor a positive branch with $u_0>1/2$, the actual neutral curve lies strictly between $\sqrt{2s}$ and $2\sqrt{2s}$. If $d<\sqrt{2s}$, then $d<d_c(u_0)$ for every such branch: a homogeneously stable state cannot have this Turing instability. If $d>2\sqrt{2s}$, then $d>d_c(u_0)$ for every finite positive $u_0$, so every homogeneously stable positive state is spatially unstable. The negative value yielded by the displayed formula outside its boundary domain is not a physical critical wavenumber and does not imply renewed stability. One must also keep $u_0<s$, which can restrict the attainable neutral curve further.

A Fourier mode with $q=k^2\ge 0$ evolves under $J-q\mathrm{diag}({\kappa }_1,{\kappa }_2)$. Its trace remains negative for a homogeneously stable positive branch. Its determinant isIf $u_0{\kappa }_2\le s{\kappa }_1$, this is positive for every $q\ge 0$. Otherwise its minimum is attained at $q_{\ast }=(u_0{\kappa }_2-s{\kappa }_1)/(2{\kappa }_1{\kappa }_2)$. Stability requires $D(q_{\ast })>0$, orIn terms of $d=\sqrt{{\kappa }_2/{\kappa }_1}$, the boundary is the positive root of $u_0{d}^2-s=2d\sqrt{s(2u_0^2-u_0)}$:Within $1/2<u_0<s$, all modes decay for $d<d_c$; a finite band grows for $d>d_c$, giving a Turing instability. Equality is neutral at one nonzero wavenumber.

For $s=2$, the positive lower branch $0<u_0<1/2$ is unstable for every $d$. The upper branch is homogeneously stable for $1/2<u_0<2$: it is stable to all modes to the left of $d_c=2+\sqrt{4-2/u_0}$ and unstable to spatial modes to the right. For $u_0>2$ it is already homogeneously unstable. The zero branch is stable to every spatial mode when $r<0$, because its mode eigenvalues are $r-{\kappa }_1{k}^2$ and $-s-{\kappa }_2{k}^2$.

A homogeneous equilibrium has $v_0=u_0^2$. Besides $u_0=0$, a nonzero equilibrium satisfies $r=u_0^2-u_0$, giving $u_{\pm }=(1\pm \sqrt{1+4r})/2$. Concentrations are nonnegative, so only nonnegative roots are physically admissible. At a positive equilibrium, the reaction Jacobian matrix isNegative trace and positive determinant are equivalent toThus a positive homogeneous stable branch exists precisely when $s>1/2$ and $-1/4<r<s^2-s$, on the upper root $u_{+}$. The lower positive branch has negative determinant and is unstable. At $r=-1/4$ the branches meet and have a zero eigenvalue; at $u_0=s$ the strict linear stability condition also fails.

The trivial equilibrium $(0,0)$ has eigenvalues $r,-s$, so it is asymptotically stable for $r<0$. It exists for all $r$, independently of the nonzero branches. If negative $u_0$ were allowed mathematically, the negative branch for $r>0$ would also be homogeneously stable, but it does not represent a chemical concentration.

After its giant phases, the Sun ejects much of its envelope and leaves a carbon–oxygen white dwarf. The remnant mass lies below the Chandrasekhar limit. Its support is chiefly electron degeneracy pressure: the Pauli exclusion principle forces electrons into distinct momentum states even at low temperature, giving $P\propto {\rho }^{5/3}$ when they are nonrelativistic. Consequently pressure support need not cease when nuclear burning ends. The white dwarf subsequently cools and fades.

The process called inverse beta decay here is electron capture, $e^{-}+p\to n+{\nu }_e$, becomes possible for free particles when the electron's total energy exceeds approximately $(m_n-m_p)c^2\simeq 2.6m_e{c}^2$, corresponding to a kinetic energy about $1.6m_e{c}^2$. Electron capture on nuclei has thresholds modified by nuclear binding energies. In a solar-mass progenitor's ordinary white dwarf remnant, the Fermi energy is below the relevant capture thresholds, so the electron population providing support is stable. At much higher density, electron capture can reduce the electron density and pressure and promote collapse; this is why stability against electron capture matters in assessing degeneracy support. The Sun ends as a cooling white dwarf, rather than a neutron star.

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 $\mathcal{A}$.