For a single barotropic perfect fluid in general relativity, the perturbations obey $\delta P=w\delta \rho =w\bar{\rho }\delta$. This adiabatic closure is needed: constant background $P/\rho$ alone would not eliminate an independent entropy perturbation. The absence of scalar anisotropic stress allows the common potential used in Newtonian gauge in cosmology.

Substituting the density constraint into the pressure equation gives the gravitational potential evolution of a barotropic fluidSince $a\propto {\tau }^{2/(1+3w)}$, the conformal Hubble parameter is $\mathcal{H}=2/[(1+3w)\tau ]$. The bracket cancels identically, leavingFor a Fourier transform mode, ${\nabla }^2$ becomes $-k^2$.

During radiation domination, $w=1/3$ and ${\Phi }_k^{\prime \prime }+4{\Phi }_k^{\prime }/\tau +(k^2/3){\Phi }_k=0$. Set $x=k\tau /\sqrt{3}$ and write $\Phi =u/x$. The resulting equation is $u_{xx}+2u_x/x+(1-2/x^2)u=0$, so the two Spherical Bessel functions in the hint giveAt $x\ll 1$, the two solutions approach a constant and a mode proportional to ${\tau }^{-3}$. The regular adiabatic mode, normalized to its primordial potential, isAfter entry into the sound horizon, $x\gg 1$, the potential oscillates at cosmological sound speed $1/\sqrt{3}$ with envelope $x^{-2}\propto a^{-2}$. The Hubble radius and sound horizon differ by the sound-speed factor; outside the Hubble radius the regular potential is constant, while well inside it radiation supports acoustic oscillations.

During matter domination, $w=0$ gives ${\Phi }_k^{\prime \prime }+6{\Phi }_k^{\prime }/\tau =0$ at every wavenumber. HenceThe growing density mode has a constant potential both outside and inside the Hubble radius; the other potential mode decays. Pressureless matter has zero cosmological sound speed, so horizon entry does not produce the radiation acoustic decay. These formulas cover both independent solutions, while the subsequent sketches select the regular adiabatic growing mode.

The scaling hypothesis for critical phenomena says that the singular part of the equilibrium free energy is a generalized homogeneous function of its thermal and field-like controls. Smooth backgrounds must first be removed; the hypothesis does not claim that the entire free energy, including arbitrary regular terms, has a pure scaling form.

For the ordinary scalar quartic LG theory, set $r=r_{t}t$ with $r_t>0$ and $u>0$, and rescale the uniform order parameter asThe order-parameter-dependent free-energy density becomesDefine $f_{\pm }(H)$ as the minimum of the braces, with the sign of the quadratic term respectively positive or negative. ConsequentlyFor an extensive $A_s$, $a$ additionally contains the system volume. The printed double-inequality subscript labels the two temperature branches: the upper-temperature function is $f_{+}$ and the lower-temperature function is $f_{-}$. It does not classify positive and negative magnetic fields. In particular $f_{+}(0)=0$ and $f_{-}(0)=-1/4$; below the transition the field dependence has a cusp at zero, so derivatives are taken on a selected branch.

For the following derivatives, take $A_s$ to be a density, so $m$ is the magnetization density; for total free energy the derivative gives total magnetization instead. Differentiate this mean-field scalar free-energy scaling form. The magnetization is $m=-\partial A_s/\partial h=-ab|t{|}^{1/2}{f}_{\pm }^{\prime }(H)$, giving $m_0\propto (-t{)}^{1/2}$ and $\beta =1/2$. The magnetic susceptibility is $\chi =-{\partial }_h^2{A}_s=-a{b}^2|t{|}^{-1}{f}_{\pm }^{\prime \prime }(H)$, giving $\gamma =1$. At zero field the nonzero curvature amplitudes are $-f_{+}^{\prime \prime }(0)=1$ and $-f_{-}^{\prime \prime }(0^{+})=1/2$.

To allow nonclassical critical exponents, replace the fixed powers byThe heat-capacity critical exponent is defined by $C_{V,s}\sim |t{|}^{-\alpha }$; temperature differentiation gives the thermal exponent $2-\alpha$ in $A_s$. The order-parameter critical exponent has $m_0\sim (-t{)}^{\beta }$, and the magnetic-susceptibility critical exponent has $\chi \sim |t{|}^{-\gamma }$. Differentiating the scaling form givesEliminating $\Delta$ proves the Rushbrooke scaling relationThe critical-isotherm exponent is defined by $m(0,h)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$. At fixed small $h$, the $t\to 0$ limit requires $f_{\pm }(H)\sim c|H{|}^{(2-\alpha )/\Delta }$, so that the temperature factors cancel. Therefore $1/\delta =(2-\alpha -\Delta )/\Delta =\beta /\Delta$, yielding $\Delta =\beta \delta$. But the two differentiated identities also give $\Delta =\beta +\gamma$, and hence the Widom scaling relationThese are relations among the leading power indices. At marginal dimensions, multiplicative logarithms can accompany them, and an additive analytic background must not be mistaken for the singular scaling contribution.

For a group presentation $\mathcal{P}=⟨X\mid R⟩$ with $X$ finite, let $F(X)$ be its free group. For a nontrivial relator defineThe p-deficiency in the unshifted convention used here isIf the weighted sum diverges the value is $-\infty$; identity relators may be omitted or assigned weight zero. Roots are taken in the free group, not in the presented quotient. Some authors subtract one from this definition; here the requested threshold is $1$.

Two elementary bounds explain why p-deficiency detects infinitude. The p-rank of a group isHere $G^p$ denotes the subgroup generated by all $p$th powers. Each relator that is not a $p$th power imposes at most one linear relation in this vector space, and a $p$th-power relator imposes none. If there are $s$ relators of the first type, then

The second bound is the index-p rewriting bound for p-deficiency. Suppose $H⊴G$ has index $p$, and its preimage in $F(X)$ is $V$. The Nielsen–Schreier formula gives $V$ rank $1+p(|X|-1)$. For a relator $r=w^{p^k}$, there are two cases in the Reidemeister–Schreier theorem. If $w\in V$, its $p$ coset-conjugates are all $p^k$th powers in $V$, with total weight at most $p{p}^{-k}$. If $w\notin V$, then $k\ge 1$, since $r\in V$. Its cosets generate $F(X)/V$, so representatives $1,w,\dots ,w^{p-1}$ show that the $p$ rewritten conjugates of $r$ are redundant up to conjugation in $V$. One relator suffices, and $r=(w^p{)}^{p^{k-1}}$ has weight at most $p^{-(k-1)}$. In both cases the total weight is at most $p$ times the old weight. Thus the induced group presentation $\mathcal{Q}$ of $H$ satisfiesThe argument applies termwise to infinitely many relators whenever the weighted sum converges.

If ${\mathrm{def}}_p(\mathcal{P})\ge 1$, the p-rank of a group bound gives a surjection to $C_p$, hence a normal subgroup of index $p$. The rewriting bound gives that subgroup another presentation of p-deficiency at least one. Iterating produces subgroups of index $p^j$ for every $j$. p-deficiency at least one implies infinitude.

Now enumerate the nonidentity elements $w_1,w_2,\dots$ of $F(x,y)$ and choose the presentationIts p-deficiency obeysThe infinitude criterion shows that $T$ is infinite. It is generated by two elements, and every element is represented by some $w_i$ or is the identity; the imposed relation makes its order a power of $p$. Thus $\boxed{T\text{ is an infinite finitely generated torsion group}}$. This is a torsion group construction by p-power relators; the presentation intentionally has infinitely many relators.

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

A soluble group, also called a solvable group, has a terminating derived series:For a subgroup $S\le G$, induction gives $S^{(j)}\le G^{(j)}$, so subgroups of soluble groups are soluble. For a surjective group homomorphism $q:G\to Q$, $q(G^{(j)})=Q^{(j)}$, so quotients of soluble groups are soluble. Finally, in a group extension $1\to K\to E\to Q\to 1$, suppose $K^{(r)}=1$ and $Q^{(s)}=1$. Then $E^{(s)}\le K$ and $E^{(s+r)}=1$. Soluble groups are closed under subgroups, quotients and group extensions.

A virtually soluble group contains a soluble group as a finite-index subgroup. The finite-index facts proved in parts (i)–(iii) imply closure under subgroups and quotients: intersect a finite-index soluble subgroup with the chosen subgroup, or take its image under the quotient map.

For group extensions, no finite-generation hypothesis may be inserted. We first establish the finite-index characteristic soluble subgroup lemma. If $K$ is a virtually soluble group, the kernel of its action on the cosets of a finite-index soluble subgroup is a soluble normal subgroup $K_0$ of finite index. Among soluble normal subgroups containing $K_0$, choose $R$ with maximal $|R/K_0|$, possible because $K/K_0$ is finite. If $S$ is any soluble normal subgroup of $K$, then $RS$ is soluble: it is an extension of $R$ by $S/(R\cap S)$. Maximality forces $S\le R$. Thus $R$ is the unique largest soluble normal subgroup of $K$, making it a characteristic subgroup, and it has finite index.

Now suppose $1\to K\to G\stackrel{\pi }{\to }Q\to 1$ has both $K$ and $Q$ virtually soluble. Replace $G$ by the preimage $G_0$ of a finite-index soluble subgroup of $Q$. The subgroup $R$ just constructed is characteristic in $K$ and therefore normal in $G_0$. In $E=G_0/R$, the subgroup $A=K/R$ is a finite normal subgroup, and $E/A$ is a soluble group. The centralizer $C_E(A)$ has finite index in $E$, since conjugation gives a map $E\to \mathrm{Aut}(A)$ with finite image. Its intersection with $A$ is the center of a group $Z(A)$, an abelian group, while its quotient by $Z(A)$ embeds in the soluble group $E/A$. Thus $C_E(A)$ is a soluble group. Its preimage in $G_0$ is an extension by $R$, so it too is soluble and has finite index in $G$. Virtually soluble groups are closed under group extensions.

For the final example take the restricted direct sum of groupswhere $A_5$ is the nonabelian simple group of even permutations on five letters. Every finite collection of elements lies in a product of finitely many finite factors, so $D$ is a locally finite group and hence a torsion group. It cannot contain a nonabelian free group, which is a torsion-free group.

To show that $D$ is not a virtually soluble group, let $H$ be any finite-index subgroup and let $N\le H$ be the kernel of the finite coset action. Each coordinate $A_5$ maps either injectively or trivially into the finite quotient $D/N$, by Simplicity of the alternating group A5. The nontrivial images of distinct factors commute, and each has trivial centre, so any $k$ of them generate a direct product of groups of order $60^k$. Only finitely many such images can occur in a finite quotient. Therefore $N$, and hence $H$, contains a whole coordinate copy of $A_5$, which is not soluble: its nontrivial commutator subgroup is normal and therefore equals $A_5$. No finite-index subgroup of $D$ is soluble.

For mutually exclusive event types, let $T$ be the first event time and $J$ its type. The cause-specific hazard for cause $j$ isIt is a rate conditional on having had no event of any cause. The cumulative incidence function, also called the cumulative risk function, is the actual probability $F_j(t)=P(T\le t,J=j)$.

The total hazard function is $h(t)={\sum }_k{h}_k(t)$, giving $S(t)=\exp \{-{\int }_0^t{\sum }_k{h}_k(u)du\}$. Surviving every cause to time $u$ and then experiencing cause $j$ givesIn competing risks, $F_j$ is generally not $1-e^{-{\int }_0^t{h}_j}$, because competing events remove individuals before they can experience cause $j$. This formula does not require assuming independent latent failure times for the different causes.

For an event at $t_j$ from subject $i_j$, with covariate vector $z_i$, define the Schoenfeld functionThe second term is the hazard-weighted mean covariate in the risk set just before the event. The Schoenfeld residual is this function evaluated at the fitted coefficient, $r_j=s_j(\hat{\beta })$. Calculate one residual vector per event, using every at-risk subject, including those who will subsequently be censored. There is no ordinary event residual assigned at a right censoring time.

The Cox partial likelihood score function is ${\sum }_j{s}_j(\beta )$. At the true constant coefficient in a Cox proportional-hazards model, the conditional event subject is selected with weights proportional to $e^{{\beta }^T{z}_i}$, so each Schoenfeld function has conditional mean zero. If the coefficient varies with time, that centering changes. Plot residuals against event time or a transformation of it, smooth them, and investigate departures from zero. Scaled Schoenfeld residuals account for the risk-set covariate variance and can display departures in coefficient units; score function tests based on residual-time association provide a formal check. Risk-set composition affects unscaled residual variance, and the total residual score function can be zero by fitting even when a time trend is present.

Write the year-$j$ count as a sum of its $m_j$ individual counts, with $m_j>0$. The specified conditional independence within that year givesDividing by $m_j$ and scaling the conditional variance by $m_j^{-2}$ therefore givesThis is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.

Define the population parameters and total observed exposure byHere $v$ is the expected process variance and $a$ the variance of hypothetical means. Assume finite second moments. The law of total variance gives $\mathrm{Var}(X_j)=a+v/m_j$, and conditional independence between years gives $\mathrm{Cov}(X_j,X_k)=a$ for $j\ne k$: their only shared variation is the latent conditional mean.

To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of $\mu (\Theta )$. For coefficients $b_j$ and $B={\sum }_j{b}_j$, the optimal intercept is $m_0(1-B)$. Centering at $m_0$ and conditioning on $\Theta$ shows that the resulting error isThe cross term vanishes because $X_j-\mu (\Theta )$ has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.

For fixed $B$, the Cauchy-Schwarz inequality giveswith equality when $b_j=B{m}_j/W$. Minimize the remaining quadratic $a(1-B{)}^2+(v/W)B^2$. Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is $m_{n+1}\mu (\Theta )$, multiplying the optimal estimate by the known future exposure givesIt is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If $a=0$, the conditional mean is a known constant almost surely and one takes $Z=0$; if $v=0<a$, the observations reveal it without process noise and $Z=1$. The ordinary case has $a,v>0$.