Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2

Since $136=8\cdot 17$ and the factors are coprime,Hence

The required integer is the exponent of the unit group. Since $\lambda (8)=2$ and $\lambda (17)=16$,

The residue $3$ is a primitive root modulo $17$. ChooseWriting $g=1+8k$ gives $k\equiv 13(\mathrm{mod}17)$, so one may takeIts Chinese-remainder components have orders $1$ and $16$, hence its order modulo $136$ is $16$.

Chebyshev's equal-ripple criterion says that a polynomial $q$ of degree at most $m$ is a best uniform approximation to a continuous function $F$ if and only if the error $F-q$ attains alternating extrema of maximum magnitude at $m+2$ ordered points.

Since $T_n$ has leading coefficient $2^{n-1}$, the polynomial $2^{1-n}{T}_n$ is monic and alternates between $\pm 2^{1-n}$ at $t_j=\cos (j\pi /n)$. Therefore

At each $t_j$, both $T_n-f$ and $T_n+f$ have sign $(-1{)}^j$, because $|f(t_j)|<1$. Each has a root in every interval between consecutive extrema, hence all its at most $n$ roots lie in $(-1,1)$. For $t>1$ both polynomials remain positive, giving $|f(t)|<T_n(t)$. For $t<-1$ both have sign $(-1{)}^n$, giving $|f(t)|<|T_n(t)|$. On $[-1,1]$ the hypothesis gives $|f|<1$. Thus

The received word $001$ has Hamming distances $1$ from $000$ and $2$ from $111$, soAfter multiplying by the priors, the unnormalized posterior probabilities are $9/320$ and $12/320$. Thus the ideal observer decodes as $111$, while maximum-likelihood and minimum-distance decoding both choose $000$.

The ideal observer minimizes average error but needs priors and channel statistics. Maximum likelihood needs the channel law but not priors, and can be suboptimal for unequal codeword probabilities. Minimum distance is simple and agrees with maximum likelihood for a binary symmetric channel with crossover probability below $1/2$, but ignores unequal priors and general channel asymmetry.

A language $L$ satisfies the context-free pumping lemma if there is an integer $p\ge 1$ such that every $z\in L$ with $|z|\ge p$ has a decompositionwithandEvery context-free language has this property, although the property by itself is not sufficient for context-freeness.

This language is not context-free. If $p$ were a pumping length, apply the lemma toBecause $|vwx|\le p$, the pumped region meets at most two adjacent blocks and cannot meet both zero blocks. Pumping with $i=0$ or $i=2$ changes at least one symbol count. If it changes either zero block, that block no longer agrees with the untouched zero block; if it changes only the middle block, the number of ones no longer agrees with the two zero counts. In every case the pumped word leaves the language, contradicting the lemma.

The language is context-free. A grammar with start symbol $S$ isEvery derivation adds matching symbols to the two ends and starts with a matching central pair, so it produces exactly the nonempty even palindromes $w{w}^R$.

This language is not context-free. Intersect it with the regular languageAn even palindrome in $R$ has the form $0^r{1}^{2s}{0}^r$, and its first half is $w=0^r{1}^s$. The required equality of the numbers of zeros and ones in $w$ forces $r=s$. Hence the intersection isThis is not context-free: applying the pumping lemma to $0^p{1}^{2p}{0}^p$, a pumped substring of length at most $p$ cannot meet both zero blocks. Pumping therefore either destroys equality of the two zero counts or changes the middle count away from twice their common value.

Context-free languages are closed under intersection with regular languages, so a context-free $L$ would make $L\cap R$ context-free. This contradiction proves the claim.

This language is context-free. LetThe nonterminal $T$ produces exactly $1^m{0}^m$ with $m>0$. The base production for $S$ supplies one leading and one trailing zero, and each recursive production supplies one more of each. Thus $S$ produces exactly

Write $Y_{gj}$ for the yield in replicate $j$ of genotype $g\in \{aa,Aa,AA\}$. Model 1 is the one-way cell-means modelThus its three coefficients estimate the three genotype means directly.

Put $D_i=1\{{\text{count}}_i\ge 1\}$. Model 2 isIt imposes ${\mu }_{Aa}={\mu }_{AA}={\beta }_0+{\beta }_1$ and ${\mu }_{aa}={\beta }_0$. The reported finite-sample p-values assume independent normal errors, a common unknown variance in every genotype, fixed full-rank design matrices, and correctness of the relevant mean model. Random assignment of genotypes supports interpreting the fitted differences as treatment effects.

The displayed analysis of variance compares Model 2 with the unrestricted three-mean Model 1. It testswhich is precisely full dominance: once an $A$ allele is present, a second copy has no further effect. Under $H_0$,The p-value $0.01249$ rejects full dominance at the $5\%$ level, providing evidence that the heterozygote and $AA$ means differ.

No dominance says that mean yield is affine in the number of $A$ alleles:Equivalently, $2{\mu }_{Aa}={\mu }_{aa}+{\mu }_{AA}$. Fit that reduced model and compare it with Model 1 using

r
model3 <- lm(yield ~ count, data = potato)
anova(model3, model1)

The resulting one-degree-of-freedom F-test tests the no-dominance constraint against unrestricted genotype means.

The term $-IS$ removes healthy foxes through mass-action contact with infectives. The Laplacian ${\nabla }^{2}I$ models random migration of infected foxes. The matching term $+IS$ creates newly infected animals, while $-\mu I$ removes infectives at per-capita rate $\mu$. The equation $R_t=\mu I$ records those removed from infectious circulation, for example foxes that have died from rabies; they neither migrate nor return to susceptibility in this model.

In a homogeneous population,An infected fox remains infectious for mean time $1/\mu$ and creates new infections at initial rate $S_0$, so the basic reproductive ratio isInitially $I$ grows exactly when $S_0-\mu >0$, equivalently ${\mathcal{R}}_0>1$. Below that threshold each infective produces fewer than one replacement on average and the infection decays.

With $\xi =x-ct$, time derivatives become $-c$ times $\xi$ derivatives and $I_{xx}=I^{\prime \prime }$. Henceor equivalently

At the leading edge $S\simeq 1$ and $I$ is small, soSeek a decaying mode $I\sim e^{-\lambda \xi }$ with $\lambda >0$. Its characteristic equation isA nonoscillatory positive leading edge requires real positive roots, henceFor an invasive infection, $mu<1$, and the selected minimum speed is thereforeIf $\mu \ge 1$, the state ahead of the wave is below the epidemic threshold and no growing infection front is selected.

The partial-fraction decomposition isThe residues at $i$ and $-i$ are respectively $1/(2i)$ and $-1/(2i)$. If two paths from $0$ to $z$ have winding-number differences $n_{+}$ and $n_{-}$ about those poles, their integrals differ byLoops can wind around either pole independently, so the value changes by arbitrary integer multiples of $\pi$ and $F$ is multivalued.

Along the straight real segment,Consequently the complete set of values is

Let $\Gamma$ be any closed curve in $\mathbb{C}\setminus B$. Since the arc $B$ joins $-i$ to $i$ without meeting $\Gamma$, those two points lie in the same component of $\mathbb{C}\setminus \Gamma$. Their winding numbers about $\Gamma$ are therefore equal. The residue theorem givesThus every closed-path period vanishes. Integrals from $0$ to $z$ along any two paths avoiding $B$ agree, so $F_B$ is single-valued. Notice that this argument uses cancellation of the two residues; simple connectedness of the slit domain is not required.

On the positive real axis, whose path does not cross the left-displaced cut,Since $F_B^{\prime }(z)=1/(1+z^2)=O(|z{|}^{-2})$ near infinity, integrating this derivative along large arcs extends the same expansion uniformly there:Thereforeand hence

For a one-dimensional closed energy orbit, defineAfter locally inverting this relation to write $E=H(I)$, use the Hamilton--Jacobi generating functionand define the angle $\varphi =\partial W/\partial I$ modulo $2\pi$. The transformation is canonical, the Hamiltonian depends only on $I$, and

The adiabatic-invariance principle says that if an external parameter changes on a time scale much longer than the orbital period, and no separatrix is crossed, the action $I$ remains approximately constant. Its small within-cycle changes do not accumulate at leading adiabatic order.

Put $ϵ=|E|>0$. On the energy surface,soEquality occurs at a turning point, so this is the smallest possible bound.

Using the natural collision continuation through the singular point, a full orbit runs between both turning points. HenceSet $q=x^2/ϵ$. ThenThereforeIf instead one identifies the singularity as a reflecting endpoint and regards one half-line as the orbit, the action is half this value; the following scaling is identical.

Adiabatic invariance keeps $m/|E|$ constant. Thus doubling $m$ doubles $|E|$:The energy becomes twice as negative.

Chemical equilibrium in $n+{\nu }_e\leftrightarrow p+e^{-}$ givesThe negligible lepton chemical potentials therefore imply ${\mu }_n\simeq {\mu }_p$. Nonrelativistic Maxwell--Boltzmann number densities satisfyNeutrons and protons have equal spin degeneracy and nearly equal masses, so their prefactor ratio is approximately one. Hence

The ratiofalls rapidly as the universe cools. Below $k_{B}T\simeq 0.7\mathrm{MeV}$, weak reactions occur less than once per expansion time and can no longer track the falling equilibrium ratio. The neutron-to-proton ratio therefore freezes at approximately $1/6$.

Before nuclei can form efficiently, free neutrons continue to beta-decay,During the deuterium bottleneck this lowers the ratio to about $1/7$ by $k_{B}T\simeq 0.07\mathrm{MeV}$.

Once deuterium survives photodissociation, nuclear reactions rapidly assemble nearly every available neutron into tightly bound helium-4. Protons are more abundant, so neutrons are the limiting constituent. If $r=n_n/n_p=1/7$, thenNeglecting the neutron--proton mass difference,Thus

Increasing $\kappa$ makes ${\Gamma }_W/H$ smaller at every fixed temperature, so freeze-out occurs earlier, at a higher $T$. The equilibrium value $e^{-Q/(k_{B}T)}$ is then closer to one. Although some neutrons subsequently decay, the final neutron fraction is larger, andincreases, approaching unity if the surviving neutron and proton abundances become comparable.

Such a universe begins with less hydrogen and more helium. It has less fuel for long-lived hydrogen-burning stars; stellar evolution would tend to proceed through hotter, faster helium-burning stages, reducing typical stellar lifetimes. The shorter stable-energy window, altered heavy-element production, and lower primordial hydrogen abundance would make familiar water-rich, slowly evolving habitats less likely, although a detailed conclusion would require a stellar-evolution model.

By the Born rule,andEach is the expectation of the projector onto the span of the two displayed matching-label product states.

In the $A$ basis, the transformations whose columns are the new basis vectors areThus $U_C$ is a rotation through $-\pi /3$, while $U_B$ is a real orthogonal reflection, equivalently a rotation with one basis-vector phase reversed.

For any real orthogonal matrix $U$ with columns $|u_j⟩$,Applying this to $U_B$ and $U_C$ and normalizing proves

For $|\psi ⟩=|00⟩$, independence of the product state givesFor the remaining pair,Therefore