Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3

If a finite group is not simple, choose a maximal proper normal subgroup and repeat inside it. Orders strictly decrease, so the process terminates; maximality makes every quotient simple.

For $S_4$ one composition series isThe successive quotients have orders $2,3,2,2$, hence are simple. As subgroups of $S_4$, the groups $S_4,A_4,V_4$, and $1$ are normal. The selected order-two subgroup is not normal, since conjugation permutes the three double transpositions.

For the Möbius action on the upper half-plane, a noncentral element is elliptic if it has one fixed point in the interior and none on the boundary, parabolic if it has one boundary fixed point, and hyperbolic if it has two boundary fixed points. Examples areFixed points satisfy a real quadratic. Its discriminant is $(\mathrm{tr}g{)}^2-4$, so the three mutually exclusive cases $|\mathrm{tr}g|<2$, $=2$, and $>2$ give precisely the three types.

If $g^n=I$ and $g\ne \pm I$, its eigenvalues are nonreal reciprocal roots of unity. Thus $|\mathrm{tr}g|<2$ and $g$ is elliptic; no such element is parabolic or hyperbolic.

Uniqueness of Laurent coefficients applied to $f(z)=f(-z)$ givesHence $a_n=(-1{)}^n{a}_n$, so every coefficient with odd $n$ vanishes.

The poles inside $C_n$ are zero and $\pm k\pi$ for $1\le k\le n$. At a nonzero pole,and the residue at $-k\pi$ is its negative. At zero, the Laurent series contains only even powers because the function is even, so its residue is zero. All residues cancel and therefore

The Euler–Lagrange equations areFor $\omega \ne 0$ their general solution iswhile for $\omega =0$, $x=c_{1}t+c_2$.

The action has continuous translation symmetry in $x$; it is also invariant up to endpoint terms under adding homogeneous Jacobi solutions $at+b$ to $x$ and $a\cos t+b\sin t$ to $y$. For $\omega =0$ it additionally has continuous time-translation symmetry; for nonzero $\omega$ only the corresponding discrete period remains.

A complete set of four independent first integrals isIn particular $({\dot{y}}^2+y^2)/2=(C_3^2+C_4^2)/2$ is conserved; when $\omega =0$, the usual total energy is conserved as well.

Insert the transform definitions and use Fubini:For the indicator of $(-1,1)$,Putting $g=f$ and $x=0$ in the convolution theorem givesand hence

Position and momentum areDifferentiating the expectation, substituting Schrödinger's equation, and integrating by parts gives Ehrenfest's relationThus the mean momentum changes according to the classical mean force $-kx$.

Take $x$ vertically downward and $y$ outward from the wall, with the film occupying $0<y<h$. The ambient pressure isAt $y=h$, normal stress continuity gives $p=p_a$ and zero ambient shear gives $\mu U^{\prime }(h)=0$; at the wall, no slip gives $U(0)=0$.

For steady parallel flow $u=(U(y),0)$, incompressibility is automatic and Navier–Stokes reduces toSince $p_x={\rho }_{a}g$, integration yields

A state is recurrent when, starting there, the chain returns to it with probability one. An irreducible chain is recurrent when every state is recurrent.

A return to zero is possible only at even times, andStirling's estimate makes this asymptotic to a positive constant timesA state is recurrent exactly when the sum of its return probabilities diverges. If $p=1/2$, the resulting $k^{-1/2}$ series diverges. If $p\ne 1/2$, then $4p(1-p)<1$ and the series converges geometrically. Thus the walk is recurrent exactly when $p=1/2$.

The minimal polynomial $m_{\varphi }$ is the unique monic polynomial of least positive degree with $m_{\varphi }(\varphi )=0$. Uniqueness follows because division of one monic annihilator by another would produce a lower-degree annihilator. Cayley–Hamilton says that the characteristic polynomial annihilates $\varphi$; division then shows

The map $\theta :\mathbb{C}[x]\to V$, $p\mapsto p(\varphi )a_1$, is surjective because $1,x,\dots ,x^{n-1}$ map to the given basis. Its kernel is generated by the minimal polynomial and the quotient has dimension $n$, so $\mathrm{deg}{m}_{\varphi }=n$. The relation for $\varphi (a_n)$ gives

If $V=V_1\oplus V_2$ were a nontrivial invariant decomposition, each restricted minimal polynomial would be a power of $x-\alpha$ of degree at most $\dim V_i<n$. Their least common multiple is the minimal polynomial of $\varphi$, contradicting its degree $n$.

Since $m_{\varphi }$ has degree $n$, $\varphi$ has a cyclic vector $b_1$. Put $b_k={\varphi }^{k-1}{b}_1$. Then $\varphi (b_k)=b_{k+1}$ for $k<n$. In this basis the matrix is the companion matrix with ones immediately below the diagonal and last columnobtained by expanding $(x-\alpha {)}^n$.

The content $c(f)$ is a greatest common divisor of the coefficients, defined up to a unit. A polynomial is primitive when its content is a unit.

If a prime divided every coefficient of $fg$, reducing modulo that prime would make the product of the two nonzero reduced primitive polynomials zero in a domain, impossible. Thus a product of primitive polynomials is primitive. Factoring the contents from arbitrary $f,g$ then givesfor a unit $u$.

Write the rational root in lowest terms as $a/b$. The factor $bx-a$ is primitive and divides $f$ over $\mathbb{Q}[x]$; Gauss's lemma makes it divide in $\mathbb{Z}[x]$. Comparison of leading coefficients makes $b$ divide the leading coefficient one, so $b=\pm 1$ and the root is integral.

Regard the polynomial as an element of $\mathbb{C}[x][y]$:The prime $x-1$ divides every nonleading coefficient, its square does not divide the constant coefficient, and it does not divide the leading coefficient $x^3$. Eisenstein's criterion proves irreducibility over $\mathbb{C}(x)[y]$; the polynomial is primitive in $\mathbb{C}[x][y]$, so Gauss's lemma proves irreducibility there.

A space is connected when it is not a union of two disjoint nonempty open subsets. If $f(X)$ had such a separation, its inverse images would separate connected $X$, so continuous images of connected spaces are connected. Equivalently, a nonconstant continuous map to discrete $\{0,1\}$ records a separation, and every separation defines such a map.

If $\mathbb{R}$ were separated, points in opposite pieces and the intermediate value theorem applied to the associated $\{0,1\}$-valued map would give a contradiction. Thus $\mathbb{R}$ is connected. Its quotient by $x\sim y$ is connected because the quotient map is continuous and surjective.

Finally let $A\subset B\subset \mathrm{Cl}(A)$ and suppose $B=U\cup V$ were a separation. Connectedness puts $A$ wholly in one side, say $U$. Every point of $V$ nevertheless lies in the closure of $A$, while $V$ is relatively open and disjoint from $A$, a contradiction. Hence $B$ is connected.

For a closed surface, $\chi =V-E+F$ for any finite cell decomposition. Gauss–Bonnet for a geodesic polygon saysand for a closed smooth surface it says ${\int }_{S}KdA=2\pi \chi (S)$.

For a large radius, the graph surface bounded over that circle is flat near its boundary. The boundary geodesic-curvature integral is $2\pi$, so Gauss–Bonnet on the disc gives $\int K=0$. Since $K\ge 0$, continuity forces $K\equiv 0$.

For the torus,Thus $T_{+}$ has $\cos u>0$ and $T_{-}$ has $\cos u<0$. Cap the two boundary circles of the outer half $T_{+}$ by flat discs. The result is a sphere, and the caps contribute no Gaussian curvature, so Gauss–Bonnet gives ${\int }_{T_{+}}K=4\pi$. The full torus has Euler characteristic zero, hence total curvature zero and

Cauchy's formula gives $|f^{\prime }(z_0)|\le M_R/R$ for an entire function bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to \infty$ gives $f^{\prime }=0$, proving Liouville's theorem.

Set $h(w)=f(1/w)$. Its limit at zero makes the singularity removable, soConsequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2{f}^{\prime }(z)\to -c_1=:b$.

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to 1$. Thus $m=n$. The functionhas removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence

Sincethe divergence theorem proves Green's second identity.

If $x_0=(x_0,y_0,z_0)$ and $x_0^{\ast }=(x_0,y_0,-z_0)$, the image solution isIt vanishes on the plane and has the required unit delta source.

Green's identity yields the Poisson-kernel integralOn the axis this becomes

With signature $(+---)$ one convenient convention iswhich gives the stated dual after index lowering. The invariants areFor the boost,with zero $x$ components.

In the final configuration, parallelism givesThe physical root isThus $v\sim c\theta /2$ for small angle. As $\theta \to \pi /2$, $v\to c$, so exactly crossed equal-magnitude null fields cannot be aligned by any finite inertial boost.

The initially irrotational inviscid flow remains irrotational, so $u=\nabla \varphi$; incompressibility gives ${\nabla }^2\varphi =0$. At the plates, normal velocity matches their motion:With $\varphi =r^{2}f(\theta )$, Laplace's equation gives $f^{\prime \prime }+4f=0$. The symmetric solution isHenceA streamfunction isso streamlines are $r^2\sin 2\theta =\text{constant}$ and fluid is expelled radially as the plates close. The outward flux through $r=R$ is

If the monic orthogonal polynomial $p_n$ had fewer than $n$ distinct sign-changing zeros in $(-1,1)$, multiply the factors corresponding to those zeros to form a polynomial $q$ of degree below $n$. Then $p_{n}q$ has one sign and is not identically zero, so its positive-weight integral cannot vanish, contradicting orthogonality. Thus all $n$ zeros are distinct and lie in the open interval.

Expansion of $\det (xI-A_n)$ along the last row giveswith $D_0=1$, $D_1=x-{\alpha }_1$. This is exactly the recurrence defining $P_n$, so induction gives $P_n=\det (xI-A_n)$.

The three-term recurrence theorem with the displayed coefficients identifies $P_n$ with the monic orthogonal polynomial $p_n$. Part (b) says its zeros are precisely the eigenvalues of the real symmetric Jacobi matrix $A_n$, and part (a) says those zeros are distinct and all belong to $(-1,1)$.

A statistic $T$ is sufficient when the conditional law of the sample given $T$ is independent of the parameter. The factorisation criterion says this holds exactly whenIf factorisation holds, cancellation in the conditional mass proves parameter independence. Conversely, multiply the parameter-free conditional mass given $T=t$ by the mass of $T=t$ to obtain the factorisation.

Rao–Blackwell says that if $U$ estimates a parameter and $T$ is sufficient, then $U^{\ast }=E(U\mid T)$ has the same mean and no larger variance, with strict improvement unless $U$ is already a function of $T$ almost surely. The mean statement is the tower property, whileproves the variance claim.

The joint mass factorizes through $T={\sum }_i{X}_i$, so $T$ is sufficient for $q$. If $\lambda =\sqrt{q}$, thenConditionally on $T$, $X_1\sim \mathrm{Bin}(T,1/n)$. Rao–Blackwell therefore gives the unbiased estimatorFor $n\ge 2$ the original estimator is not a function of $T$, so the variance reduction is strict.

For probability vectors $p,q$, they are optimal with value $v$ exactly whenand $p^{T}Aq=v$. These are the mutual best-response and security-level conditions.

For every probability vector $x$, antisymmetry gives $x^{T}Ax=0$. Applying the minimax inequalities with the same strategy on both sides shows the lower value is at most zero and the upper value at least zero; transposition also changes the value to its negative. Hence the value is zero.

Starting from any optimal strategy, transfer all probability on row $i_0$ to row $i_1$. Every payoff against a pure column weakly increases, so the security level remains optimal. The resulting strategy has $p_{i_0}=0$.

The matrix is antisymmetric, so the value is zero. For a common candidate $p=(a,b,c,d{)}^T$, the column-player inequalities are $Ap\le 0$:The first three imply $a+d\le b\le c\le a-2d$, hence $d=0$ and $a=b=c$. Normalization givesIndeed $Ap=(0,0,0,-1{)}^T/3\le 0$, and antisymmetry gives the corresponding row inequalities.