Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2

Let $g$ generate the cyclic group of nonzero residues, whose order $p-1=2^{2^n}$ is a power of two. Write $a=g^j$. Its multiplicative order is $(p-1)/\gcd (j,p-1)$, so it is a primitive root precisely when $j$ is odd. The quadratic residues are exactly the even powers of $g$. Thus the primitive roots are precisely the quadratic nonresidues, with nonzero residue classes understood.

For $n\ge 1$, $p\equiv 1(\mathrm{mod}4)$, so quadratic reciprocity gives $(7/p)=(p/7)$. Since $2^3\equiv 1(\mathrm{mod}7)$ and $2^n$ alternates between $1$ and $2$ modulo three, $p$ is respectively $3$ or $5$ modulo seven. The nonzero squares modulo seven are $1,2,4$. Consequently $(7/p)=-1$, andThere is a genuine endpoint exception if the printed form permits $n=0$: then $p=3$ and $7\equiv 1(\mathrm{mod}3)$, which has order one and is not primitive. The first equivalence remains true for $p=3$, but the final assertion requires excluding that prime.

The Liouville approximation theorem states that if $\alpha$ is a real irrational algebraic number of degree $d$, there is a constant $C(\alpha )>0$ such that, for all integers $p,q$ with $q\ne 0$,Reduction of the fraction is not necessary: an unreduced denominator only weakens this bound. More generally the same type of bound holds for an algebraic number of degree $d$ whenever $p/q\ne \alpha$. The exception is essential for a rational $\alpha$, since an exact rational representation gives zero error. A rational number of degree one obeys the bound directly: if $\alpha =a/b$ in lowest terms and $p/q\ne \alpha$, then $|\alpha -p/q|=|aq-bp|/(|bq|)\ge 1/(|bq|)$.

Set $r=p/q$. If $r\notin [1,2]$, then $|\sqrt{2}-r|\ge \sqrt{2}-1>1/4\ge 1/(4q^2)$. If $1\le r\le 2$, irrationality of the square root of two implies that the integer $2q^2-p^2$ is nonzero. Factoring a difference of squares therefore givesThus the requested bound holds for every integer $p$ and every nonzero integer $q$, without using the Liouville approximation theorem. For completeness, irrationality follows because $p^2=2q^2$ in lowest terms would make $p$ even and then $q$ even, contradicting coprimality.

Use the curvature-minus-one Poincare disc model, whose Riemannian metric isIts geodesics are Euclidean diameters and arcs of circles meeting the boundary circle orthogonally. The hyperbolic area of a measurable region $D$ isAn isometry can move the center of a circle to the origin. Its hyperbolic radius $r$ and Euclidean radius $\rho$ then satisfy $r={\int }_0^{\rho }2du/(1-u^2)=2\mathrm{artanh}\rho$, so $\rho =\tanh (r/2)$. Integrating the area element of a surface in polar coordinates yieldsIntegrating the line element around the circle givesA geometric definition of Pi valid in this geometry isIndeed $\sinh r=r+O(r^3)$ and $\cosh r=1+r^2/2+O(r^4)$. The circumference-to-diameter ratio at a fixed positive radius is not constant, so the limiting qualification is necessary.

A binary linear code of length $n$ is a vector subspace $C\subseteq {\mathbb{F}}_2^n$. The Hamming weight of a word counts its nonzero coordinates, and the code weight is $w(C)={\min }_{0\ne c\in C}\mathrm{wt}(c)$; for a linear code this equals its minimum distance. One may assign infinite minimum weight to the zero code when needed.

For length-$n$ codes with $C_2\subseteq C_1$, the bar product of binary linear codes isIt is a linear code because this map from $C_1\oplus C_2$ is linear and injective. For a nonzero codeword, if $v=0$ then $u\ne 0$ and its weight is $2\mathrm{wt}(u)\ge 2w(C_1)$. If $v\ne 0$, every nonzero coordinate of $v$ makes exactly one of the corresponding coordinates of $u,u+v$ nonzero; henceTaking the minimum givesIn fact equality holds when the codes are nonzero: minimum-weight words $(u,u)$ and $(0,v)$ attain the two candidate weights.

Each scan() command reads a numeric vector from the entered data: the first creates the nine counts, and the second their nine predictor values. The last command fits a generalized linear model with the Poisson distribution and its default log link, including an intercept, then prints a summary. Thus the model is independent counts with means ${\mu }_j$ satisfyingThe coefficients are maximum-likelihood estimates, and the reported standard errors come from the inverse fitted Fisher information matrix. Increasing the predictor by one multiplies the fitted mean by $e^{0.3106}\simeq 1.364$, about a $36.4\%$ increase. The intercept corresponds to a mean $e^{1.363}\simeq 3.91$ at predictor zero, outside the observed range. Approximate coefficient-to-standard-error ratios are $6.17$ and $8.13$; the slope is strongly different from zero under the assumed model. A $95\%$ Wald confidence interval for the slope is about $[0.236,0.385]$, or $[1.266,1.470]$ for the multiplicative change.

The deviance is twice the log-likelihood advantage of the saturated model over the fitted model. For these counts it iswith the logarithmic term interpreted as zero when a count is zero. The saturated model has nine freely fitted means; the fitted model has two independent regression coefficients, so the residual degrees of freedom are $9-2=7$.

Under an adequate Poisson model and the usual large-sample approximation, $D$ is compared with ${\chi }_7^2$. The reported $13.218$ is rather large relative to seven, but below the $5\%$ upper critical value about $14.07$; its approximate upper-tail probability is $0.067$. Thus there is some indication of lack of fit or overdispersion, but not a rejection at the $5\%$ level on this diagnostic. The deviance-to-d.f. ratio is about $1.89$; it does not by itself prove an overdispersed model, and the chi-square approximation is imperfect for small counts.

There is one inconsistency in the supplied numerical output. Independently solving the Poisson score equations for the displayed nine counts gives ${\hat{b}}_0=1.363454$, ${\hat{b}}_1=0.310563$ and deviance $13.217971$, agreeing with the printed estimates and deviance. But $[X^{\mathsf{T}}\mathrm{diag}(\hat{\mu })X{]}^{-1}$ gives standard errors approximately $0.221295$ and $0.030850$. In particular the printed slope standard error $0.0382$ is not the standard Poisson-model value for these data. The interpretations above describe the supplied numbers; the directly recomputed slope ratio is $10.07$ and its normal-approximation $95\%$ interval is about $[0.2501,0.3710]$. The conclusion of a strong increasing trend is unchanged.

Apply the formal infinitesimal generator to the molecule count: its increment is $+1$ at rate $\lambda$ and $-2$ at rate $\beta x^2$. Thus the exact first-moment identity for the specified generator isNeglecting fluctuations gives the mean-field approximation $\dot{m}=\lambda -2\beta m^2$, so the positive steady mean isThere is a physical boundary qualification: the printed loss propensity is positive at $x=1$ and would produce a negative molecule count. Consequently it does not by itself define a molecule-count chain on all nonnegative integers. If losses are forbidden below two, the exact loss moment becomes $2\beta E[x^2{1}_{\{x\ge 2\}}]$; a mass-action two-molecule propensity instead uses $x(x-1)$. The calculation requested here uses the printed formal generator and its large-copy-number approximation, where such boundary corrections are negligible.

Let $J_{+}=\lambda$ and $J_{-}=2\beta m^2$ be the production and loss fluxes, counted in molecules per unit time. At the steady state, both equal $J=\lambda$. The restoring elasticity and turnover lifetime areFlux-weighting the magnitudes of the reaction steps, one for production and two for loss, givesThis is an average over molecular fluxes, as requested, not an average over reaction-event counts; those are different when a reaction removes two molecules at once.

For a stable one-species linear noise approximation, the normalized stationary fluctuation-dissipation relation for a reaction network becomesTo see the factors explicitly, the linearized restoring rate is $a=H/\tau =4\beta m$, while the chemical noise intensity is the rate-weighted squared step sizeStationarity of the variance gives $2a\mathrm{Var}x=B$. Using $\lambda =2\beta m^2$ therefore yieldsThis is the stationary fluctuation result within the stated negligible-fluctuation/linear-noise approximation, not an exact identity for the boundary-corrected discrete chemistry.

A fixed point $x_0$ is Lyapunov stable if for every $\epsilon >0$ there is $\delta >0$ such that solutions starting within $\delta$ of $x_0$ stay within $\epsilon$ for all forward times. It has quasi-asymptotic stability if some neighborhood of $x_0$ consists of initial points whose solutions converge to $x_0$. Asymptotic stability requires both properties.

Choose the Lyapunov functionIt vanishes only at the origin, is positive elsewhere, and has compact sublevel sets. The vector field is polynomial, hence locally Lipschitz. Along a solution,A small sublevel set is forward invariant and can be fitted inside any prescribed neighborhood, proving Lyapunov stability; boundedness also prevents finite-time escape.

For attraction, a bounded forward solution has a nonempty compact limit set, and $V$ decreases to a limit. Every point in that limit set has that same $V$ value. Continuous dependence of solutions makes the limit set forward invariant, so $\dot{V}=0$ along every trajectory in it. Such a trajectory must have $x=0$ at all times. But then $\dot{x}=-y$, forcing $y=0$. Thus the only possible limit point is the origin, and every solution converges there. This verifies the invariant-set condition behind the LaSalle invariance principle, rather than incorrectly treating $\dot{V}$ as strictly negative at every nonzero point. The origin is therefore asymptotically stable, indeed globally attracting.

Fix a Hankel contour with a positive-radius circle around zero and a branch $-\pi <\arg t<\pi$, traversing the lower bank inwards and the upper bank outwards. Write its numerator as $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$. For fixed circle radius, $I$ is an entire function of $z$: the finite contour pieces avoid zero, and the tails converge uniformly on compact $z$ sets by exponential decay, also after differentiation with respect to $z$.

At $z=-n$, the two straight-bank contributions cancel because $t^{-n-1}{e}^t$ is single-valued. The remaining circle is positively oriented, so the residue theorem and the exponential series giveSince $\sin (\pi z)=(-1{)}^n\pi (z+n)+O((z+n{)}^3)$, the quotient has the simple poleFor cancellation of positive-integer gamma singularities on a Hankel contour, at a positive integer $n$, the integrand $t^{n-1}{e}^t$ is entire and its bank contributions again cancel; its circular integral is zero. Hence $I(n)=0$. The numerator is analytic near $n$, while the denominator has a simple zero with nonzero derivative. Factoring $I(z)=(z-n)J(z)$ therefore cancels that zero, proving there is no pole at a positive integer using only this representation. Cancellation of the numerator, not merely the zero of the sine, decides the issue. One can also shrink the contour for $\Re z>0$, obtaining $I(z)=2i\sin (\pi z){\int }_0^{\infty }{u}^{z-1}{e}^{-u}du$ and hence the finite value $(n-1)!$.

The components in Euler equations for a torque-free rigid body are evaluated in the body-fixed principal-axis frame, with $L_i=I_i{\omega }_i$. In an inertial space frame the angular momentum vector is constant; its components in the rotating body frame need not be.

Assume $\Omega \ne 0$ and distinct positive principal moments of inertia. Write $\omega =(\Omega +{\delta }_1,{\delta }_2,{\delta }_3)$. To first order,soThe coefficient is positive when $I_1$ lies between the other moments, giving an exponentially growing mode and instability. When $I_1$ is an extreme moment, it is negative and the transverse perturbations oscillate.

To establish actual nonlinear Lyapunov stability in the latter case, use the conserved kinetic energy $E=\frac{1}{2}\sum I_j{\omega }_j^2$ and squared angular momentum $L^2=\sum I_j^2{\omega }_j^2$. Their combination isFor an extreme $I_1$ its two coefficients have the same sign. Conservation controls both transverse components by their initially small values, while conservation of $E$ controls ${\omega }_1^2$ near ${\Omega }^2$. Continuity preserves its initial sign for sufficiently small perturbations. Thus nonzero steady rotation is stable exactly about the largest or smallest principal moment, the intermediate axis theorem. It is not asymptotically stable, since energy is conserved. If $\Omega =0$, energy makes the zero angular-velocity state stable for every ordering; the nonzero-rotation assumption is essential to the stated equivalence.

Let the surface be at $R$ with vanishing external pressure. Multiply hydrostatic equilibrium by $4\pi r^3$ and integrate:The other side is $-{\int }_0^{R}4\pi Gm(r)\rho (r)rdr=-{\int }_0^{M}Gm/r(m)dm$. Building the star one shell at a time identifies this as its negative gravitational potential energy $E_{\mathrm{grav}}$. ThereforeIf the star has nonzero surface pressure $P_s$, the integration instead gives $3⟨P⟩V-3P_{s}V=-E_{\mathrm{grav}}$; the requested form assumes $P_s=0$.

Within the prescribed uniform-density approximation, $n=3M/(4\pi m_p{R}^3)$. Treating the given approximate pressure coefficient as one, the degeneracy pressure and specified kinetic-energy relation giveFor a uniform sphere, $m(r)=M(r/R{)}^3$, so shell assembly givesThusAt fixed mass, this energy has its minimum at $R_{\ast }=2\alpha /\beta$, with $E^{\prime \prime }(R_{\ast })=2\alpha /R_{\ast }^4>0$. Expansion beyond this radius raises the energy, while strong compression costs positive energy growing as $R^{-2}$, faster than the attractive $R^{-1}$ term. This explains radial stability of the model white dwarf. The numerical coefficient in $\alpha$ inherits the approximate pressure law; a relativistic degeneracy law would change this stability argument.

The Baire category theorem says that in a nonempty complete metric space, a countable intersection of open dense subsets is dense. Equivalently, such a space cannot be a countable union of nowhere dense sets: a set is nowhere dense when the interior of its closure is empty. Indeed the complements of those closures are open dense; their intersection is dense and in particular nonempty, so a countable union of the original sets cannot exhaust the space. The same argument shows that the complement of such a union is dense.

First interpret the decimal expansions as fractional expansions $0.d_1{d}_2\cdots$, with every $d_j\in \{4,5\}$. The resulting set $X_0$ is compact: a diagonal subsequence argument selects successively stabilized digits from any sequence in $X_0$, and the tail bound ${\sum }_{j>m}9\cdot 10^{-j}=10^{-m}$ proves convergence to the corresponding expansion.

It has empty interior. At level $m$ its points lie in $2^m$ decimal intervals, each of length $10^{-m}$, whose total length is $(1/5{)}^m$. An interval of positive length could not lie in all these covers. Thus $X_0$ is closed and nowhere dense. Each translate $X_0+q$ is also nowhere dense, and the rational numbers are countable. The Baire category theorem applied to $\mathbb{R}$ yieldsAny point in this complement has the required property. If the printed description also permits integer digits before the decimal point, there are only countably many such integer parts, and negative expansions merely add reflected translates. This makes $X$ a countable union of translates or reflections of $X_0$. The same countable-union argument therefore still proves that some real $t$ cannot be written as $x+q$.

A linear-feedback shift register of length $d$ has binary state $(x_r,\dots ,x_{r+d-1})$ and updateIts output obeys $x_{r+d}={\sum }_{j=0}^{d-1}{a}_j{x}_{r+j}$ over ${\mathbb{F}}_2$. Since the state space is finite, its trajectory consists of a transient of length $M$ followed by a cycle of length $N\ge 1$, and the output satisfies $x_{r+N}=x_r$ for $r\ge M$.

If the trajectory never reaches zero, all its $M+N$ distinct transient and cyclic states are nonzero; hence $M+N\le 2^d-1$. If it reaches zero, let $M$ be its first zero-state time. The vectors $v,fv,\dots ,f^{M-1}v$ are linearly independent: in a dependence, apply $f^{M-1-j}$ where $j$ is the smallest index with a nonzero coefficient; every later term vanishes and the remaining term is $f^{M-1}v\ne 0$. Therefore $M\le d$. The zero cycle has $N=1$, and $d+1\le 2^d-1$ for $d\ge 2$. This proves the requested bound for all $d\ge 2$, and for every invertible register.

The unrestricted printed assertion has one exception: at $d=1$, the zero-feedback register with initial state one outputs $1,0,0,\dots$. It requires $M\ge 1,N\ge 1$, so $M+N\ge 2>2^1-1$. The universally valid finite-state bound is $M+N\le 2^d$. A register with feedback from its oldest bit is invertible, in which case a nonzero state never reaches zero and the sharper bound holds already with $M=0$.

To recover an unknown register from a known plaintext/ciphertext keystream, the Berlekamp-Massey algorithm finds a shortest linear recurrence consistent with the observed bits. Use a connection polynomial $C(z)=1+{\sum }_{j=1}^L{C}_j{z}^j$, so the recurrence discrepancy at index $r$ is $\delta =x_r+{\sum }_{j=1}^L{C}_j{x}_{r-j}$. Initialize $C=B=1,L=0,b=1,m=1$. Process the bits in order. If $\delta =0$, increment $m$. If $\delta \ne 0$, save $T=C$ and replaceIf $2L\le r$, also replace $L$ by $r+1-L$, set $B=T,b=\delta ,m=1$; otherwise retain $L,B,b$ and increment $m$. Arithmetic is binary, so every nonzero discrepancy equals one. The stored polynomial $B$ last failed at an earlier index with discrepancy $b$; its shifted contribution cancels the new failure without changing previous successful discrepancies.

Here is the minimum recurrence length after a new discrepancy argument behind the update, not just its name. Suppose a span-$L$ recurrence $C$ succeeds through index $r-1$ but fails at $r$. If another polynomial $D$ of span $L^{\prime }$ succeeded through $r$ and $L+L^{\prime }\le r$, compute the convolution $(CD)x$ at $r$ in two ways. Applying $D$ first gives zero, since all indices $r-j$ used by $C$ lie at least at $L^{\prime }$. Applying $C$ first gives its nonzero discrepancy at $r$, since its earlier discrepancies used by $D$ vanish and $D_0=1$. This contradiction proves $L^{\prime }\ge r+1-L$. The shifted-polynomial update attains the larger of the existing minimum length and this new lower bound; induction therefore gives the shortest recurrence for every prefix. Once at least twice the true linear complexity has been observed, this recurrence determines the whole stream: the difference from a true degree-$d$ recurrent stream satisfies a recurrence of degree at most $d+L$, and a prefix of that many zeros forces every later value to be zero. Without any bound on complexity, a finite observed prefix cannot exclude a longer register that diverges later; the breaking claim uses the finite-register hypothesis and adequate known data.

Finally each of the three input streams is eventually periodic. Let $M$ be the maximum of their transient lengths and $N$ a common multiple of their periods. Their joint triple, and hence the bit chosen by the specified rule, repeats after $N$ once $r\ge M$. ThereforeThis itself is a linear recurrence of length $M+N$, so it constructs a linear-feedback shift register producing the selected stream from its first $M+N$ bits. No closure theorem is being assumed: eventual periodicity and this explicit recurrence prove the combined stream is also an LFSR stream, even though the combining rule is nonlinear in its three inputs.

Choose the nondimensionalizationThe chain rule gives ${\partial }_{\tau }={\beta }_u{\partial }_t$ and ${\partial }_x^2=({\beta }_u/D_u){\partial }_s^2$. Dividing the first equation by ${\beta }_u$ and the second by ${\beta }_v$ gives the required scaled reaction–diffusion system. Positive degradation and diffusion constants are assumed.

A nonzero spatially constant steady state satisfies $v=u$ and $v=u^2$, hence $(u,v)=(1,1)$. The first derivatives of $u^2/v-u$ there are $1,-1$, while those of $Q(u^2-v)$ are $2Q,-Q$. ThusSince $Q>0$, both eigenvalues have strictly negative real part exactly when $Q>1$. For $0<Q<1$ the equilibrium is unstable. At $Q=1$ its linear eigenvalues are $\pm i$; this is the marginal boundary and does not give asymptotic stability from linearization. The positive-trace criterion in the hint tests positive real parts; stability requires applying it to $-J$.

Set $u=1+\hat{u}(t)\cos (s/L)$ and $v=1+\hat{v}(t)\cos (s/L)$ and retain linear terms. With $k=1/L$, the Fourier mode has second derivative $-k^2\cos (ks)$. HenceThe trace and determinant of this linearization are

Assume $Q>1$, so $T(k)<0$ for every wavenumber. A Turing instability therefore occurs exactly when $D(k)<0$ for some $k$. Put $z=k^2\ge 0$. The upward quadratic $1+(1-P)z+P{z}^2$ has a negative minimum on this half-line precisely whenCombining these givesThe unstable band is $z_{-}<k^2<z_{+}$, whereThe degradation-limited diffusion distances are proportional to ${\ell }_u=\sqrt{D_u/{\beta }_u}$ and ${\ell }_v=\sqrt{D_v/{\beta }_v}$. Thus $P=({\ell }_v/{\ell }_u{)}^2$ and the criterion says the inhibitor must travel more than $(1+\sqrt{2})$ times as far as the activator before degradation. Short-range activation and longer-range inhibition permit a local peak to reinforce itself. On a finite domain the allowed wavenumbers must also intersect the unstable band; the displayed condition assumes continuously available modes.

At the threshold $P_{\ast }=3+2\sqrt{2}$, the determinant quadratic has a double zero atConsequently the scaled disturbance length in $\cos (s/L)$ isThe physical length is ${\ell }_u{L}_{\mathrm{crit}}=\sqrt{{\ell }_u{\ell }_v}$ at threshold; its full spatial wavelength is $2\pi$ times this length, not this length itself. For much shorter lengths, the large negative diffusion terms smooth both components. For much longer lengths, diffusion contributes little and the stable homogeneous reaction feedback, especially the inhibitor response, suppresses disturbances. The selected length lies between those two stabilization mechanisms.

Let the three orbit points be $a<b<c$. Up to reflection of the interval, their images are $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b]$ and $J=[b,c]$. Continuity and the intermediate value theorem imply the covering relationsWe first prove the needed interval-covering principle using the permitted pullback fact. If a cyclic list of closed intervals $I_0,\dots ,I_{n-1},I_n=I_0$ obeys $F(I_j)\supseteq I_{j+1}$, pull back successively from the end to obtain a closed interval $K\subseteq I_0$ with $F^n(K)=I_0$ and $F^j(K)\subseteq I_j$. Since $F^n(K)\supseteq K$, the intermediate value theorem gives a fixed point of $F^n$ in $K$: choose preimages of the two endpoints of $K$, which give opposite weak signs of $F^n(x)-x$, irrespective of their order. This point realizes the cyclic itinerary.

For $n\ge 2$, use the word $I{J}^{n-1}$. It obeys the covering relations and therefore yields a point with period dividing $n$. A point away from the shared endpoint $b$ has an unambiguous itinerary, and this word has exactly one $I$ per cycle; it cannot be a repetition of a shorter word. Its least period is consequently $n$. If an orbit touches $b$, it is the original three-cycle. That realizes the case $n=3$ itself. It cannot realize any of the other words: for $n$ not divisible by three it is not fixed by $F^n$, and for a multiple of three larger than three it visits $a\in I\setminus J$ more than once, contradicting the itinerary. Thus the construction works for every $n\ge 2$. For $n=1$, $F(J)\supseteq J$ and the same fixed-point argument gives a fixed point. This proves period three implies all periods: there is an orbit of every positive least period.

For the two five-cycles forced by a three-cycle, use the two cyclic words $IJJJJ$ and $IJIJJ$. Both are admissible. Their points cannot be endpoints from the three-cycle, and because five is prime and the words use both intervals, each has least period five. Their itineraries have respectively one and two visits to the interior of $I$; cyclically shifting the starting point cannot change this count. Hence they give at least two distinct period-five orbits.

In the Newtonian dust calculation, use a comoving spherical shell of physical radius $R(t)=a(t)\chi$. Homogeneity gives its enclosed mass $M=4\pi \rho R^3/3$, and the shell theorem removes the gravitational force of exterior shells. Assume pressureless matter, conserved enclosed mass, no cosmological constant, and no shell crossing. Its specific conserved mechanical energy isHomogeneity permits the energy per ${\chi }^2$ to be a constant, written $-k{c}^2/2$. Dividing by $a^2$ givesThis is a Newtonian derivation for dust. For radiation the enclosed gravitating mass is not conserved, so this same mechanical-energy argument cannot simply be reused; general relativity supplies the same Friedmann constraint with the total energy density divided by $c^2$, together with pressure-dependent fluid conservation.

The fluid conservation equation is $\dot{\rho }+3H(\rho +P/c^2)=0$. With the radiation equation of state, it becomes $\dot{\rho }+4H\rho =0$, giving $\rho a^4={\rho }_0$. Define the critical density ${\rho }_{\mathrm{crit}}=3H^2/(8\pi G)$ and cosmological density parameter $\Omega =\rho /{\rho }_{\mathrm{crit}}$. Since $a_0=1$,Evaluating the Friedmann equation today gives $H_0^2+k{c}^2=H_0^2{\Omega }_0$, hence

For ${\Omega }_0>1$, introduce conformal time by $dt/d\tau =a$. Since $a^{\prime }=a\dot{a}$, the constraint becomesOn the initial expanding branch, separate variables and choose $a=t=\tau =0$ at the Big Bang. With $\alpha =H_0\sqrt{{\Omega }_0-1}$,The second formula follows by integrating $dt/d\tau =a$. The model reaches maximum size at $\alpha \tau =\pi /2$ and recollapses at $\alpha \tau =\pi$.

Today $a=1$ and $H_0>0$ selects the expanding branch, soSubstitution yields the requested ageIn particular the numerator is $\sqrt{{\Omega }_0}-1$, not $\sqrt{{\Omega }_0-1}$; the PDF displays the former. The flat-model limit is $1/(2H_0)$.