Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3

Apply the reciprocal-and-integer-part algorithm for the continued fraction. The successive complete quotients after the integer part two are $(\sqrt{7}+2)/3$, $(\sqrt{7}+1)/2$, $(\sqrt{7}+1)/3$, and $\sqrt{7}+2$. Their integer parts are $1,1,1,4$, and taking the next reciprocal repeats the first complete quotient. ThereforeThe first four continued fraction convergents are $2,3,5/2,8/3$. The last gives $8^2-7\cdot 3^2=1$. Squaring its unit $8+3\sqrt{7}$ gives $127+48\sqrt{7}$, whose norm is again one. Thus two positive solutions of the Pell equation areDirect verification gives $127^2-7\cdot 48^2=16129-16128=1$.

The polynomial form of Runge theorem says that a function holomorphic on a neighbourhood of a compact set $K\subset \mathbb{C}$ with connected complement can be approximated uniformly on $K$ by polynomials.

For $n\ge 3$, takeThe complement is connected: the missing angular sector connects the inner disk to the exterior, and removing the single point zero from the inner disk does not disconnect it. On a neighbourhood of the annular sector choose an analytic logarithm with arguments between $-1/(2n)$ and $2\pi -1/(2n)$. Its exponential $\exp (3\log z/2)$ agrees with the prescribed function on the sector. On a disjoint small neighbourhood of zero prescribe the analytic function zero. These two neighbourhoods define a holomorphic function $g_n$ near $K_n$ with $g_n=f$ on $K_n$.

By Runge theorem, choose a polynomial $p_n$ with ${\sup }_{K_n}|p_n-g_n|<1/n$. Every fixed nonzero point of the disk eventually belongs to all $K_n$, including points on the positive real axis; zero belongs to every $K_n$. ConsequentlyEach polynomial restricts to an analytic function on $D$. The excluded sector lies immediately below the positive axis and shrinks with $n$, so the construction correctly preserves the chosen value on the branch cut. It gives pointwise convergence, not local uniform convergence across that cut. This is pointwise approximation by a Runge exhaustion.

For a plane crystallographic group, the translation subgroup consists of translations by vectors in a rank-two lattice $\Lambda$; the lattice is this additive group of translation vectors. The point group is the image under the linear-part homomorphism $G\to O(2)$, equivalently $G/\Lambda$. Conjugating a translation by an isometry with linear part $A$ gives translation by $Av$, so every point-group element preserves $\Lambda$.

Choose a lattice basis. A point-group rotation is represented in that basis by an integral matrix, because it maps lattice vectors to lattice vectors, and its inverse does likewise. The trace is therefore an integer. In an orthonormal basis the same rotation has trace $2\cos \theta$, which lies between minus two and two. Hence $2\cos \theta \in \{-2,-1,0,1,2\}$. The value two gives the identity, while the other values give angles $\pi$, $\pm 2\pi /3$, $\pm \pi /2$, or $\pm \pi /3$. Thus every nonidentity rotation has orderThis proves the crystallographic restriction theorem from the lattice action rather than assuming the classification of wallpaper groups.

The polynomials are over ${\mathbb{F}}_2$, and multiplication gives $gh=X^7+1=X^7-1$. A cyclic code of length seven generated by a degree-three polynomial has dimension $7-3=4$: the multiples $ag$ with $\mathrm{deg}a<4$ are distinct representatives, so its rank is four.

The root $\alpha$ has order seven in ${\mathbb{F}}_8^{\times }$: it is neither zero nor one, and the multiplicative group has prime order seven. A weight-one word cannot vanish at $\alpha$. A weight-two word $X^i+X^j$, with distinct indices between zero and six, cannot vanish either, because that would give ${\alpha }^{i-j}=1$. Every codeword vanishes at $\alpha$, so its nonzero weight is at least three. Since $g$ itself has weight three,From ${\alpha }^3=\alpha +1$ obtain ${\alpha }^4={\alpha }^2+\alpha$ and ${\alpha }^5={\alpha }^2+\alpha +1$. Therefore $r(\alpha )={\alpha }^5+{\alpha }^3+\alpha ={\alpha }^2+\alpha ={\alpha }^4$. The one-bit error $X^4$ has precisely this syndrome, so correcting it givesIts distance from the received word is one. The minimum distance three makes this nearest codeword unique: two codewords each at distance one from the received word would be at distance at most two from each other.

Write $S(\beta )={\sum }_{ij}{\mu }_{ij}(\beta )$. Ignoring only terms independent of $\beta$, the Poisson log-likelihood is ${\ell }_P=\sum y_{ij}\log {\mu }_{ij}-S(\beta )$. The multinomial model is defined on the constraint $S(\beta )=n$, since its probabilities must sum to one. On this constraint its full log-likelihood satisfieswhere both likelihoods include the same $-\sum \log (y_{ij}!)$ terms. Thus they differ by a constant on the admissible multinomial parameter set.

If the Poisson maximum-likelihood estimator has $S(\hat{\beta })=n$, it is feasible for the multinomial problem, and no feasible parameter can have a larger likelihood than it. Consequently it is also a multinomial maximum. With the usual uniqueness and identifiability assumption,Without uniqueness, the precise conclusion is that $\hat{\beta }$ is a multinomial maximizer; arbitrary independently chosen maximizers need not be identical. This qualification is necessary for a general unspecified parametrization.

In a log-linear generalized linear model with an unrestricted intercept, differentiation with respect to that intercept gives $\sum (y_{ij}-{\hat{\mu }}_{ij})=0$. Thus the fitted total automatically equals $n$. One can fit independent Poisson cell counts using the usual GLM machinery and recover the multinomial fitted probabilities as ${\hat{\mu }}_{ij}/n$. The result is constrained Poisson and multinomial likelihood equivalence, not an assertion that the sampling experiments themselves are identical.

The pair-birth rate here is the constant immigration rate $\lambda$, while deaths occur at total rate $\beta n$. This interpretation is fixed by the requested mean formula. With $P_n(t)=\mathrm{Pr}(N_t=n)$ and $P_j=0$ for $j<0$, the master equation isThe Markov jump-process generator acts as $Lf(n)=\lambda [f(n+2)-f(n)]+\beta n[f(n-1)-f(n)]$. Taking $f(n)=n$ gives $m^{\prime }=2\lambda -\beta m$, henceFor the second moment $s=\mathbb{E}{N}_t^2$, the two jumps give $s^{\prime }=4\lambda m+4\lambda -2\beta s+\beta m$. Differentiating $V=s-m^2$ therefore yields $V^{\prime }=4\lambda +\beta m-2\beta V$. Substitution of the mean and solution of this scalar linear equation giveThus $V(t)\to 3\lambda /\beta$, assuming $\beta >0$ and finite initial second moment. This is pair immigration with linear deaths; births at rate $\lambda n$ would be a different model and would not give the printed mean.

A form of the Poincaré-Bendixson theorem states that a nonempty omega-limit set of a forward orbit of a $C^1$ planar autonomous vector field, contained in a compact region and containing no equilibrium, is a periodic orbit.

Set $\rho =x^2+y^2$. Direct calculation givesbecause $0\le x^2{y}^2\le {\rho }^2/4$. At the inner boundary $\rho =2/(2+k^2)$ the flow points into or along the annulus, and at the outer boundary $\rho =1$ it does likewise. Smoothness and these inequalities make the compact annulus forward invariant.

For nonzero $(x,y)$, write $x=\sqrt{\rho }\cos \theta$, $y=\sqrt{\rho }\sin \theta$. Thenthroughout the annulus. Hence it contains no equilibrium. Any orbit starting there remains compactly trapped, has a nonempty omega-limit set, and the Poincaré-Bendixson theorem supplies the required periodic orbit in the stated annulus. For $k=0$ the annulus reduces to $\rho =1$ itself, and there $\dot{\theta }=1$, giving the periodic orbit directly.

Let $u_1,u_2$ be a local fundamental pair at zero, including a logarithmic Frobenius solution if the exponents are resonant. Any other independent pair on a common branch domain has $w_1=a{u}_1+b{u}_2$ and $w_2=c{u}_1+d{u}_2$ with constant coefficients and $ad-bc\ne 0$. Bilinearity and antisymmetry of the Wronskian giveThus changing the independent branches multiplies the principal-pair Wronskian by a nonzero constant, once the branches are continued onto the same domain.

For the second-order Fuchsian equation represented by the Papperitz symbol, the coefficient of $w^{\prime }$ isThis follows from the sums of local exponents at its two finite regular singular points. Abel's Wronskian identity, obtained by differentiating $w_1{w}_2^{\prime }-w_1^{\prime }{w}_2$ and using the differential equation, gives $W^{\prime }=-pW$. Thus, on every branch domain avoiding zero and one,The normalized expression is locally constant. Its analytic continuation around either singularity is unchanged: the power factors cancel the determinant of the local monodromy, equivalently the integral of $-p$ around that singularity. It therefore defines a single-valued holomorphic function on the punctured plane, locally constant near both punctures. Those punctures are removable, so the normalized expression extends to an entire function. This argument also covers resonant exponents and avoids assuming two pure power solutions in that case.

Let $H=z^{1-\alpha -{\alpha }^{\prime }}(1-z{)}^{1-\gamma -{\gamma }^{\prime }}W$. The exponent-sum relation gives $2-\alpha -{\alpha }^{\prime }-\gamma -{\gamma }^{\prime }=1+\beta +{\beta }^{\prime }$. Therefore the assumed boundedness of $z^{1+\beta +{\beta }^{\prime }}W$ at infinity makes $H$ bounded there, up to a fixed nonzero branch factor. By part (ii), $H$ is entire and hence bounded also on compact sets. Liouville theorem makes it a constant $A$, givingIndependence forces $A\ne 0$: if the Wronskian vanished, uniqueness for the second-order linear equation at an ordinary point would make the two solutions dependent. In fact the differential identity in part (ii) already proves constancy; the boundedness hypothesis supplies the requested Liouville route as well.

Put $I_j=m_j{r}_j^2$ and $K={\omega }^2{r}_1{r}_2$. The Euler-Lagrange equations are $I_1{\ddot{\varphi }}_1=K\sin \psi$ and $I_2{\ddot{\varphi }}_2=-K\sin \psi$, where $\psi ={\varphi }_2-{\varphi }_1$. Adding shows that $I_1{\dot{\varphi }}_1+I_2{\dot{\varphi }}_2$ is conserved. Subtracting gives the equation of a simple pendulumThe relative equilibria are $\psi =0$ and $\psi =\pi$ modulo $2\pi$. The relative energy ${\dot{\psi }}^2/2+k^2(1-\cos \psi )$ has a strict minimum at zero, proving stability there. Linearization gives $\ddot{\eta }=-k^2\eta$ near zero and $\ddot{\eta }=k^2\eta$ near $\pi$, so the latter is unstable. The small-oscillation period is $2\pi /k$. These are equilibria of the relative angle; a nonzero conserved total angular momentum permits common rotation of both particles.

Chemical equilibrium gives ${\mu }_{\mathrm{He}}=2{\mu }_p+2{\mu }_n$, since the photon chemical potential is zero. Set $A=(2\pi m_{p}kT/h^2{)}^{3/2}$. The degeneracies and approximate masses give prefactors $2A,2A,8A$ for protons, neutrons and helium. ThusHere the binding energy uses $2m_p+2m_n-m_{\mathrm{He}}$; the PDF prints $2n_n$ in that mass relation, a dimensional typo.

Baryon conservation gives $n_B\propto a^{-3}$. In adiabatic expansion without photon-producing entropy release, $T\propto a^{-1}$ and $n_{\gamma }\propto T^3\propto a^{-3}$, so $\eta =n_B/n_{\gamma }$ stays constant. This statement is for that epoch; entropy transfer such as electron-positron annihilation modifies the photon normalization.

Since the density ratio equals $n_B^3{X}_p^2{X}_n^2/X_{\mathrm{He}}$, substitution of the given photon density yieldsAt $m_p{c}^2\simeq 938$ MeV, $B_{\mathrm{He}}=30$ MeV and $kT=0.3$ MeV, the right side is about $1.22\times 10^{-29}{\eta }^{-3}$. Unity therefore corresponds to $\eta \simeq 2.3\times 10^{-10}$, illustrating the stated very rough temperature estimate for a cosmological baryon-to-photon ratio of order $10^{-10}$ to $10^{-9}$. A numerical value of $\eta$ is necessary; it is not specified in the question. The exponential makes the temperature estimate far less sensitive than the ratio itself to order-one prefactors.

Actual helium synthesis must proceed through light intermediates. The deuterium bottleneck persists because deuterium's much smaller binding energy allows the abundant high-energy photon tail to destroy it. Only at a lower temperature do enough deuterons survive to initiate efficient reactions leading to helium. Direct four-nucleon assembly at the equilibrium temperature is ineffective.

The Chinese remainder theorem says that for pairwise coprime positive moduli $m_i$, the map $\mathbb{Z}/(\prod m_i)\to {\prod }_i\mathbb{Z}/m_i$ is an isomorphism; in particular prescribed residues have a unique solution modulo the product.

Write $n=p^{e}m$, with $e\ge 2$ and $(p,m)=1$. Choose $z\equiv 1+p^{e-1}(\mathrm{mod}{p}^e)$ and $z\equiv 1(\mathrm{mod}m)$. The Chinese remainder theorem supplies $z$. The binomial theorem gives $z^p\equiv 1(\mathrm{mod}{p}^e)$, since every nonconstant term is divisible by $p^e$; also $z^p\equiv 1(\mathrm{mod}m)$. Thus $z^p\equiv 1(\mathrm{mod}n)$, whereas $z\not\equiv 1(\mathrm{mod}n)$. Its multiplicative order is exactly $p$.

For odd $n=\prod p_i^{e_i}$ define the Jacobi symbol $(a/n)={\prod }_i(a/p_i{)}^{e_i}$, using the Legendre symbol, with $(a/1)=1$. It is zero if $a$ shares a prime factor with $n$. For example $(2/9)=(2/3{)}^2=1$, but the square residues modulo nine are $0,1,4,7$, so two is not a square.

For positive odd coprime $u,v$, Jacobi reciprocity isFactor $u=\prod p_i^{a_i}$ and $v=\prod q_j^{b_j}$ and multiply quadratic reciprocity for each prime pair. Its sign exponent is ${\sum }_{i,j}{a}_i{b}_j(p_i-1)(q_j-1)/4$. For odd numbers, $(rs-1)/2\equiv (r-1)/2+(s-1)/2(\mathrm{mod}2)$, so this exponent has the same parity as $(u-1)(v-1)/4$. Multiplicativity of the Legendre symbol gives the two Jacobi products, proving the formula. If $u,v$ are not coprime, both sides of the displayed reciprocity equality are zero.

For the previously constructed $z$, every prime divisor of $n$ sees residue one, so $(z/n)=1$. But $p\nmid (n-1)/2$, because $p\mid n$ and $p$ is odd. The order-$p$ unit therefore has $z^{(n-1)/2}\not\equiv 1(\mathrm{mod}n)$. This supplies a unit failing the proposed test.

Finally the passing units form the kernel of the homomorphismMultiplicativity makes this a homomorphism, and the failing unit shows its kernel is proper. A proper subgroup has index at least two. Hence the test fails on at least half the units modulo $n$.

The Liouville approximation theorem states that if a real irrational algebraic number $\alpha$ has degree $d$, there is $c>0$ such that $|\alpha -p/q|\ge c/q^d$ for every integer $p$ and positive integer $q$.

Choose its irreducible polynomial $F\in \mathbb{Z}[X]$ of degree $d$. Since $\alpha$ is irrational, $F(p/q)\ne 0$; after multiplication by $q^d$ it is a nonzero integer. Hence $|F(p/q)|\ge q^{-d}$. On $[\alpha -1,\alpha +1]$ the polynomial derivative has a finite bound $M>0$. For $|p/q-\alpha |<1$, the mean value theorem gives $|F(p/q)|\le M|p/q-\alpha |$, proving the required bound with $c=1/M$. For more distant rationals, the bound holds with $c\le 1$. Taking $c=\min \{1,1/M\}$ covers every rational and proves the theorem.

Let ${\xi }_{n+1}=a_{n+1}+1/(a_{n+2}+\cdots )$ be the complete quotient. Iterating the fractional-linear formula for a continued fraction givesThe supplied determinant identity and ${\xi }_{n+1}>a_{n+1}$ therefore implyThe positive partial quotients ensure increasing denominators tending to infinity; the same determinant identity makes the even and odd convergents nested, giving their common limit $x$. The displayed formula follows by passing to the limit in the finite-tail identity.

An explicit choice is $a_n=10^{n!}$ for every $n\ge 1$. The recurrence gives $q_n\le {\prod }_{j=1}^n(a_j+1)\le 2^{n}10^{{\sum }_{j=1}^{n}j!}$. For $n\ge 2$, ${\sum }_{j=1}^{n}j!\le 2n!$, so ${\log }_{10}{q}_n\le 3n!$. Consequently $a_{n+1}=10^{(n+1)!}\ge q_n^{(n+1)/3}$ andFirst $x$ is irrational: if $x=P/Q$, the nonzero difference from a convergent is at least $1/(Q{q}_n)$, contradicting the error bound once $q_{n+1}>Q$. If $x$ were algebraic of any fixed degree $d$, the last displayed upper bound would eventually be smaller than $c/q_n^d$ for the positive constant in part (i), since $q_n\to \infty$. This contradicts the Liouville approximation theorem. Thus this explicit continued fraction is transcendental.

The term $\gamma n(n_0-n)$ is logistic population growth, with carrying density $n_0$ and low-density growth rate $\gamma n_0$. The cell flux is $-D_n\nabla n+\chi n\nabla C$: its first part is diffusion down the density gradient, and its second is chemotaxis toward higher chemical concentration. Taking minus its divergence gives the two transport terms. Chemical production is proportional to cell density at rate $\alpha n$, degradation is first order at rate $\beta C$, and the chemical diffuses with coefficient $D_c$. All the given positive signs correspond to growth, attraction, production, degradation and ordinary diffusion respectively.

The positive homogeneous equilibrium is $n=n_0$, $C=\alpha n_0/\beta$. For spatially constant perturbations $u,v$, linearization gives $\dot{u}=-\gamma n_{0}u$ and $\dot{v}=\alpha u-\beta v$. The triangular matrix has eigenvalues $-\gamma n_0$ and $-\beta$, both negative, proving asymptotic stability to homogeneous perturbations. The other homogeneous equilibrium, zero population and zero chemical, has positive population eigenvalue $\gamma n_0$ and is not the desired stable state.

For a Laplacian mode with ${\nabla }^2=-k^2$, the linearized chemotaxis system isIts trace is negative for all $k$. Writing $s=k^2$, its determinant isInstability occurs when this determinant is negative, because then one real eigenvalue is positive. A quadratic with positive leading and constant coefficients is negative for some $s>0$ precisely when its middle coefficient is negative and its discriminant positive. These conditions combine toThis proves the printed threshold. In a bounded spatial domain, only its permitted Laplacian eigenmodes are available; the continuous-$k$ threshold describes the onset when wavenumbers can vary freely.

At the threshold, the determinant quadratic has a double zero. Its vertex lies at $s=(\chi \alpha n_0-\beta D_n-\gamma n_0{D}_c)/(2D_n{D}_c)$. Substituting the equality case of part (iii) givesThe mode is neutral exactly at onset and grows when the chemotactic coefficient increases beyond that threshold.

Choose $A=B=\sigma /r$ in the proposed Lyapunov function. The cubic terms cancel and direct differentiation givesFor $0<r<1$ this is negative definite, while $V$ is positive definite and radially unbounded. Finite-dimensional equivalence of these positive quadratic forms gives $\dot{V}\le -cV$ for some $c>0$. Thus $V(t)\le V(0)e^{-ct}$, every solution remains bounded and exists for all forward times, and every solution tends to zero. Therefore the origin is globally asymptotically stable, indeed exponentially stable.

The centre manifold theorem says that near an equilibrium of a sufficiently smooth finite-dimensional vector field, there is a locally invariant centre manifold tangent to the generalized eigenspace of eigenvalues with zero real part. It can be represented as a graph over that space, with graph value and derivative zero at the equilibrium; the dynamics on it determine the local centre-direction behaviour. For parameter-dependent systems the parameters may be adjoined as variables with zero derivative, producing an extended centre manifold.

At $r=1$, the linearization has the $x,y$ block $\left(\begin{array}{cc}-\sigma & \sigma \\ 1& -1\end{array}\right)$ and the independent eigenvalue $-b$. Its other eigenvalues are zero and $-(\sigma +1)$. Hence the equilibrium is nonhyperbolic, the stable subspace has dimension two, and the non-extended centre subspace has dimension one. A centre eigenvector is $(1,1,0)$.

With $\sigma =1$, substitute $x=(u+v)/2$, $y=(u-v)/2$, $r=1+\mu$. The equations becomeThe extended centre variables are $u,\mu$. The centre manifold theorem gives smooth graph functions $V,Z$ with $V(0,0)=Z(0,0)=0$ and all first derivatives zero. Their invariance equations are $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$, after substitution of the graph.

Count $\mu$ as order $u^2$. Then $\dot{u}=O(u^3)$ on the graph, so the left side of the $Z$ equation contributes only at higher order. Its quadratic terms give $0=u^2/4-bZ$, hence $Z=u^2/(4b)+O(u^3)$. The cubic terms of the $V$ equation similarly give $0=-2V-\mu u/2+uZ/2$. ThusTerms purely in $\mu$ vanish because the origin remains an equilibrium for every parameter; these expansions describe the weighted scaling $\mu =O(u^2)$, not an unrestricted one-variable Taylor series. Substitution gives the reduced equationIts cubic coefficient is negative. For $\mu <0$ the origin is attracting in the centre direction; for $\mu >0$ it loses stability and two attracting branches emerge with $u\sim \pm 2\sqrt{b\mu }$. Therefore the bifurcation is a supercritical pitchfork. These branches agree with the exact Lorenz equilibria $x=y=\pm \sqrt{b\mu }$, $z=\mu$.

By the gravitational shell theorem, the inward acceleration at radius $r$ is $Gm(r)/r^2$, where $m(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Force balance for a thin fluid shell gives $-\nabla P=\rho \nabla \Phi$, and henceAssembling shells from infinity gives $E_{\mathrm{grav}}=-{\int }_0^{R}Gm(r)dm(r)/r=-4\pi G{\int }_0^{R}m\rho rdr$. Using the pressure equation and integration by parts yieldsFor a star with negligible external pressure, $P(R)=0$, this is $E_{\mathrm{grav}}=-3⟨P⟩V$. Nonzero external pressure would add the displayed surface term.

For a nonrelativistic ideal gas, the virial relation is $E_{\mathrm{grav}}=-2E_{\mathrm{kin}}$ and total energy is negative. Under adiabatic compression its kinetic energy scales as $R^{-2}$, faster than the magnitude $R^{-1}$ of gravity, providing restoring support; its adiabatic index $5/3$ exceeds $4/3$. For an ultrarelativistic gas, $E_{\mathrm{grav}}=-E_{\mathrm{kin}}$ and both terms scale as $R^{-1}$, giving marginal stability at adiabatic index $4/3$ in this idealized Newtonian model. Corrections such as general relativity can then destabilize it.

Integrating the kinetic energy $p^2/(2m_p)$ over filled momentum states givesSince $p_F=h[3n/(4\pi g_s){]}^{1/3}$, the pressure isFinally $|E_{\mathrm{grav}}|\sim G{M}^2/R$, whereas $PV\sim (h^2/m_p)[M/(m_p{R}^3){]}^{5/3}{R}^3$. Equating these by the virial theorem givesThis is an order-of-magnitude estimate for nonrelativistic degeneracy support, not a uniform-density exact stellar solution.

The recursive ordinal addition definitions are $a+0=a$, $a+(b+1)=(a+b)+1$, and $a+\lambda ={\sup }_{b<\lambda }(a+b)$ for a nonzero limit ordinal. Ordinal multiplication is $a\cdot 0=0$, $a\cdot (b+1)=a\cdot b+a$, and $a\cdot \lambda ={\sup }_{b<\lambda }a\cdot b$. For $a>0$, ordinal exponentiation is $a^0=1$, $a^{b+1}=a^b\cdot a$, and $a^{\lambda }={\sup }_{b<\lambda }{a}^b$; the exceptional zero base has $0^0=1$ and $0^b=0$ for $b>0$.

Transfinite induction gives ${\omega }^a\ge a$. At zero this is immediate; at a successor, ${\omega }^{b+1}={\omega }^b\omega \ge {\omega }^b+1\ge b+1$. At a limit, continuity and the induction hypothesis give ${\sup }_{b<a}{\omega }^b\ge {\sup }_{b<a}b=a$. Also $a\mapsto {\omega }^a$ is strictly increasing: its successor values strictly increase, and continuity preserves the strict inequality between any two distinct indices.

For $\alpha >0$, ${\omega }^{\alpha +1}>{\omega }^{\alpha }\ge \alpha$, so there is a least $\delta$ with ${\omega }^{\delta }>\alpha$. It cannot be zero and cannot be a limit: at a limit, a supremum greater than $\alpha$ has a term already greater than $\alpha$, contradicting minimality. Thus $\delta ={\alpha }_0+1$, andStrict increase proves uniqueness, and ${\alpha }_0\le {\omega }^{{\alpha }_0}\le \alpha$ proves the requested index bound.

Choose the largest positive integer $a_0$ with ${\omega }^{{\alpha }_0}{a}_0\le \alpha$. It exists because $\alpha <{\omega }^{{\alpha }_0}\omega ={\sup }_{m<\omega }{\omega }^{{\alpha }_0}m$. The remaining tail of the ordinal interval has an order type $\rho$ with $\alpha ={\omega }^{{\alpha }_0}{a}_0+\rho$; maximality implies $\rho <{\omega }^{{\alpha }_0}$. If $\rho >0$, repeat the construction on it. The leading exponents strictly decrease, and there is no infinite strictly decreasing sequence of ordinals: its set of values would have a least member followed by a smaller one. Hence the process terminates, yielding the Cantor normal formThe first exponent is determined by the preceding band, the first coefficient by maximality, and the tail by the ordinal interval left after that initial segment. Recursion therefore proves uniqueness of every exponent and coefficient. The PDF's second coefficient is printed ${\alpha }_1$ rather than $a_1$; the stated natural-number coefficients and the construction identify the intended expression.