Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2

The möbius function is $mu(1)=1$, $mu(n)=0$ if a prime square divides $n$, and $mu(n)=(-1{)}^r$ when $n$ is a product of $r$ distinct primes. The Riemann zeta function is $zeta(s)={\sum }_{n\ge 1}{n}^{-s}$ for $\Re s>1$.

Both sides of the proposed identity are multiplicative. At $n=p^a$, the right side is $1$ for $a=0,1$ and $1+\mu (p)=0$ for $a\ge 2$, exactly $mu(p^a{)}^2$. Absolute convergence permits rearrangement, so

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Runge theorem says that if $K\subset \mathbb{C}$ is compact and $\mathbb{C}\setminus K$ is connected, every function holomorphic near $K$ is uniformly approximable on $K$ by polynomials.

Choose a compact exhaustion $K_j$ of the slit disc $\Delta \setminus L^{+}$, rounding the two sides of the slit so that $0\notin K_j$ and $\mathbb{C}\setminus K_j$ is connected. Runge applied to $1/z$ gives a polynomial $p_j$ with ${\sup }_{K_j}|p_j-1/z|<1/j$. Every compact subset of the slit disc lies in all sufficiently large $K_j$, proving locally uniform convergence.

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The source is Bernoulli when the $X_n$ are independent and identically distributed with a common law $p$ on the finite alphabet.

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It is reliably encodable at rate $r$ if there are block encoders into at most $2^{nr}$ codewords and decoders whose block error probability tends to zero as $n\to \infty$.

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The information rate is ${\lim }_{n\to \infty }{n}^{-1}H(X_1,\dots ,X_n)$ when the limit exists.

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The asymptotic equipartition property is $-n^{-1}{\log }_{2}p(X_1,\dots ,X_n)\to H$ in probability. For an i.i.d. source this is the weak law applied to $-{\log }_{2}p(X_i)$, and independence gives $H(X_1,\dots ,X_n)=nH(X)$. Thus its information rate and AEP constant are both $H(X)$.

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A state $q$ is inaccessible when no word $w\in {\Sigma }^{\ast }$ has $\hat{\delta }(q_0,w)=q$.

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Set $p\sim q$ when $\hat{\delta }(p,w)\in F⟺\hat{\delta }(q,w)\in F$ for every word $w$. Put ${\delta }^{\prime }([q],a)=[\delta (q,a)]$ and $F^{\prime }=\{[q]:q\in F\}$. If $p\sim q$, then testing continuations $aw$ shows $\delta (p,a)\sim \delta (q,a)$, so ${\delta }^{\prime }$ is well-defined.

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The Myhill-Nerode theorem says this quotient is, up to isomorphism, the unique accessible DFA with the least possible number of states accepting the language.

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Take states $0,\dots ,2n-1$, transition $j\mapsto j+1(\mathrm{mod}2n)$, initial state $0$, and make every even state accepting. Every state is reached by some $a^j$, and two states are equivalent exactly when they have the same parity. The quotient therefore has two states and accepts the even-length words.

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Logistic regression assumes independent $Y_i\mid X_i\sim \mathrm{Bernoulli}({\pi }_i)$ withFor the binomial exponential family, $\ell =y\theta -m\log (1+e^{\theta })+\text{constant}$, so the mean is $m{e}^{\theta }/(1+e^{\theta })$ and the canonical parameter is its logit. It may be fitted by iteratively reweighted least squares.

Under LDA, Bayes' rule giveswhich is a logistic model with an intercept.

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Conservation requires zero flux: $n_x(0,t)=n_x(L,t)=0$. Indeed $d{\int }_0^{L}n,dx/dt=D[n_x{]}_0^L=0$.

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Add exponential growth: $n_t=D{n}_{xx}+\alpha n$.

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With hostile Dirichlet boundaries, expand in $\sin (k\pi x/L)$. The $k$th amplitude grows at rate $\alpha -D(k\pi /L{)}^2$. Persistence requires the principal rate to be nonnegative, hence $\alpha \ge D{\pi }^2/L^2$; strict inequality gives growth.

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The laplace transform is $\hat{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t),dt$, and inversion iswhere the vertical bromwich contour lies to the right of all singularities. For $t>0$ it is closed leftwards, and for $t<0$ rightwards, where the exponential decays.

Transforming the PDE and imposing boundedness as $x\to \infty$ giveswith the square root cut conventionally along $(-\infty ,0]$. ThereforeThere are poles at $p=0,-1$ (with cancellation assessed in the complete expression) and a branch point at $0$; close left for $t>0$ and wrap the cut while including the relevant residues.

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With $x=\ell \sin \theta$ and $y=-a{t}^2/2-\ell \cos \theta$, discard a total derivative from the kinetic energy to obtainThus $\ddot{\theta }+(g-a)\sin \theta /\ell =0$. If $a=g$, then $\ddot{\theta }=0$: the pendulum is weightless in the lift and rotates uniformly (or remains fixed).

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The equilibria are $\theta =0,\pi$ modulo $2\pi$ unless $a=g$, when every angle is an equilibrium. Linearisation shows $heta=0$ stable for $a<g$ and $\theta =\pi$ stable for $a>g$; the other is unstable.

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Since $p\propto L^{-1}=V^{-1/3}$, $dp/dV=-p/(3V)$. At fixed occupations, $dE=\int {\mathcal{E}}^{\prime }(p)(dp/dV)\bar{n}(p),dp,dV$. Equating this to $-P,dV$ yields $P=(3V{)}^{-1}\int p{\mathcal{E}}^{\prime }(p)\bar{n}(p),dp$.

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For $\mathcal{E}=pc$, $p{\mathcal{E}}^{\prime }=\mathcal{E}$, so $P=E/(3V)=\rho /3$. Adiabatic energy conservation gives $\rho \propto a^{-4}$, equivalently conserved particle number gives $n\propto a^{-3}$; with $\rho \propto T^4$ or $n\propto T^3$, both imply $T\propto a^{-1}$.

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Here $\mathcal{E}=m{c}^2+p^2/(2m)+\cdots$, so $p{\mathcal{E}}^{\prime }=p^2/m$. Consequently $P=(3V{)}^{-1}\int p^2/m,\bar{n},dp=2U/(3V)$. For an adiabatic monatomic gas, $u\propto a^{-5}$ and $n\propto a^{-3}$, hence $T\propto u/n\propto a^{-2}$.

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The reflection is $I_{|\psi ⟩}=I-2|\psi ⟩⟨\psi |$. Hence $I_{|x_0⟩}|x⟩=-|x_0⟩$ for $x=x_0$ and equals $|x⟩$ otherwise.

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The marked-subspace reflection is $I_G=I-2{\sum }_{j=1}^M|g_j⟩⟨g_j|$.

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Write $|G⟩=M^{-1/2}{\sum }_g|g⟩$ and $|B⟩=(N-M{)}^{-1/2}{\sum }_{x\notin G}|x⟩$. Then $|{\psi }_0⟩=\sin \alpha |G⟩+\cos \alpha |B⟩$. Both reflections preserve $\mathrm{span}\{|G⟩,|B⟩\}$, hence so does their product.

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In the orthonormal basis $(|G⟩,|B⟩)$, direct multiplication gives(up to the equivalent opposite orientation convention).

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The displayed orthogonal matrix is a rotation through $2\alpha$, where ${\sin }^2\alpha =M/N$.

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After $r$ iterations the marked amplitude is $\sin ((2r+1)\alpha )$. Choose $r$ nearest to $\pi /(4\alpha )-1/2$; for $M\ll N$, this is about $(\pi /4)\sqrt{N/M}$ oracle iterations.

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For a closed path avoiding zero, choose a continuous lift $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$ and set $w(\gamma ;0)=(\theta (1)-\theta (0))/(2\pi )$. Conjugation replaces $\theta$ by $-\theta$, as does reversing the path, so both winding numbers are $-w(\gamma ;0)$.

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Homotopy invariance of winding number states that a homotopy $H(s,t)$ through closed paths in ${\mathbb{C}}^{\ast }$ preserves winding number. Uniform continuity divides the $s$ interval so successive paths differ pointwise by less than the minimum modulus of the earlier path; the supplied perturbation lemma makes their winding numbers equal. Chaining the subdivisions proves the result.

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If no $z\in S^1$ obeyed $z^n=-g(z)$, the loops $z^n$ and $-g(z)$ could be joined without crossing zero by the straight-line homotopy, so they would have equal winding number. The first has winding $n$. Since $g{|}_{S^1}$ extends over the disc, it is null-homotopic and has winding $0$; multiplication by $-1$ does not change it. This contradicts $n\ne 0$.

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Shannon noiseless coding theorem says the minimum expected binary prefix-code length $L$ satisfies $H(X)\le L<H(X)+1$. Kraft and Gibbs give $H\le L$ for every prefix code. Taking lengths ${\ell }_i=\lceil -{\log }_2{p}_i\rceil$ satisfies Kraft because $\sum 2^{-{\ell }_i}\le 1$, and gives $L<\sum p_i(-{\log }_2{p}_i+1)=H+1$.

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Equiprobability gives $H={\log }_{2}m$.

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For $m=2^b$, assign every outcome one of the $b$-bit words. This complete fixed-length prefix code has expected length $b$ and meets the information entropy lower bound.

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If two lengths differed by at least two, replacing two deepest sibling leaves by their parent and splitting a shallower leaf lowers total length, contradicting optimality. Thus lengths are $b=\lfloor {\log }_{2}m\rfloor$ and $b+1$. Kraft equality gives $n_b=2^{b+1}-m$ and $n_{b+1}=2m-2^{b+1}$, hence

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A branch point is a point around which analytic continuation changes a value; a branch cut removes curves so continuation becomes single-valued, and a branch is one such single-valued analytic choice. Defineon $\mathbb{C}\setminus ((-\infty ,-1]\cup [1,\infty ))$, choosing the square root equal to $1$ at zero. Monodromy around $\pm 1$ explains multivaluedness.

On the upper lip $0+$ of the cut convention, the branch remains positive on $(0,1)$, so ${\int }_{0+}^{1}dt/\sqrt{1-t^2}=\pi /2$. Define $\mathrm{Arccos}z=\pi /2-\mathrm{Arcsin}z$ on the same slit domain.

AlsoChanging logarithm branches changes the value by multiples of $\pi$ (already visible at $z=1$). Principal logarithms give a single branch on the plane cut from $\pm i$ outward. Differentiating both sides and checking the value at zero proveswhere the compatible principal branches are defined.

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Let $r$ be the cylindrical radius of the point of contact and $\varphi$ its azimuth. Rolling without slip or twist gives $|\omega |=v/a$ and thereforeup to an irrelevant additive constant. The factor $7/5$ combines translational and rotational inertia.

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Time and azimuthal invariance give energyand angular momentum $J=(7m/5)r^2\dot{\varphi }$. They represent mechanical energy and vertical angular momentum.

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The radial effective potential is $V_{\mathrm{eff}}=mgr\cot \alpha +5J^2/(14m{r}^2)$. Initially $r=8a$, $J=(7m/5)(8a)v_0$. Comparing the initial energy with $V_{\mathrm{eff}}(2a)$ giveswhich is exactly $v_0^2\ge (4/7)ga\cot \alpha$. Larger speed raises the centrifugal barrier at the neck faster than it raises the initial kinetic energy.

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A holonomic constraint is $F(q,t)=0$. In the multiplier formulation its generalized constraint force is $Q_i=\lambda {\partial }_{i}F$. Along an allowed motion, its power is ${\sum }_i{Q}_i{\dot{q}}_i=\lambda (\dot{F}-{\partial }_{t}F)$, which vanishes when the constraint has no explicit time dependence.

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At the circular minimum, $v^2=(5/7)gr\cot \alpha$ and $r=8a$. Projecting the centre-of-mass equation on the cone normal givesThis is the magnitude of the normal constraint force.

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After transmission Alice and Bob announce bases, retain matching-basis bits, reveal a random sample, and abort if its error rate is too large; privacy amplification and error correction produce the key. A Breidbart intercept-resend causes error $1/4$ conditional on matching Alice and Bob bases. Since matching occurs with probability $1/2$, one transmitted qubit detects Eve with probability $1/8$.

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Conditional on Alice and Bob using the same basis, Eve chooses the wrong basis with probability $2/3$, then Bob disagrees with probability $1/2$, so the conditional error is $1/3$. Alice and Bob match with probability $1/3$, hence the per-transmission detection probability is $1/9$.

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The channel is $\rho \mapsto (1-p)\rho +pI/2$, soWithout Eve the sifted error is $p/2$. If Eve alone produces sifted error $e$ ($1/4$ for BB84, $1/3$ for six-state), the combined error is $(1-p)e+p/2$, strictly above the noise baseline exactly when $p<1$. Thus with known stationary $p$ and enough test bits, Eve is statistically detectable for $0\le p<1$ in either protocol; at $p=1$ the channel is completely mixed and erases the distinction.

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Using the definition twice gives $P(a,b)=1$ when $a\le b$; when $a>b$, $a\to b=b$, then $b\to a=1$, and $1\to a=a$.

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If $b\le c$, then $b\to c=1$ and $T=1\to a=a$. If $b>c$, then $b\to c=c$, so $T=c\to a$, which is $1$ exactly when $c\le a$ and is $a$ otherwise.

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The implicational Hilbert axioms are $A\to (B\to A)$, $(A\to (B\to C))\to ((A\to B)\to (A\to C))$, and Peirce's law $((A\to B)\to A)\to A$.

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A direct case split in the linearly ordered chain verifies types 1 and 2. Peirce's value is $P(a,b)$ from part (a); for $n\ge 3$, choose $0<a>b$, obtaining $P(a,b)=a<1$. Thus type 3 need not be $n$-valid.

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One example isApplying the two case formulas from (a) and (b) checks the three possible values in $H_3$ and always gives $1$. In $H_4$, the valuation $p=1/2,q=1/3$ gives $v(\phi )=1/2$, so it is not 4-valid.

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Ramsey theorem follows from $R(s,t)\le R(s-1,t)+R(s,t-1)$: at a vertex, either at least $R(s-1,t)$ neighbours or at least $R(s,t-1)$ non-neighbours produce the desired clique or independent set. Induction with $R(2,t)=R(s,2)=t$ gives $R(s,t)\le \left(\genfrac{}{}{0ex}{}{s+t-2}{s-1}\right)$, hence $R(t)\le \left(\genfrac{}{}{0ex}{}{2t-2}{t-1}\right)<2^{2t}$.

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Apply Ramsey's theorem to a sufficiently large complete graph on integer vertices, colouring edge $\{i,j\}$ by the colour of $|i-j|$. Choose a monochromatic clique $y_0<y_1<\cdots <y_{n-1}$. Set $x_i=y_i-y_{i-1}$ for $1\le i\le n-1$ and $x_n=y_{n-1}-y_0$. Every $x_i$ is the difference along an edge of the clique, so all have one colour, and the differences telescope to $x_1+\cdots +x_{n-1}=x_n$.

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Alternating suitable half-open vertical strips of width $\sqrt{3}/2$ gives a two-colouring in which the horizontal projections forced by a unit equilateral triangle cannot all occupy strips of one parity; boundary strips are assigned consistently. For three colours, embed the Moser spindle, a finite unit-distance graph of chromatic number four, in the plane. Any three-colouring therefore has a monochromatic unit edge.

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Define ${\Phi }_n(X)={\prod }_{1\le k\le n,(k,n)=1}(X-e^{2\pi ik/n})$. From $X^n-1={\prod }_{d\mid n}{\Phi }_d(X)$, induction and monic division show ${\Phi }_n\in \mathbb{Z}[X]$. Every ${\Phi }_n$ is irreducible over $\mathbb{Q}$. For prime $p$, ${\Phi }_p(X+1)=((X+1{)}^p-1)/X$ is Eisenstein at $p$, proving the assertion.

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