Past exam of the mathematics course of the University of Cambridge / 2026 / ib

Bilinearity and symmetry follow from the integral. Also $⟨p,p⟩={\int }_0^{1}p(t{)}^{2}dt\ge 0$, and equality forces the continuous polynomial $p$ to vanish identically.

Gram–Schmidt applied to $1,t$ gives the orthonormal basis $1$ and $\sqrt{12}(t-1/2)$.

A neighbourhood of $x$ contains an open set containing $x$; $x_n\to x$ when every neighbourhood eventually contains every $x_n$. If two metrics have the same topology they plainly have the same convergent sequences. Conversely, if a $d$-open set $U$ were not $d\prime$-open, some $x\in U$ would have points $x_n\notin U$ with $d\prime (x_n,x)<1/n$; the common-sequence assumption would imply $d(x_n,x)\to 0$, contradicting $d$-openness. Symmetry finishes the proof. This fails generally: on an uncountable set the discrete and cocountable topologies differ, but in both every convergent sequence is eventually constant.

The residue is the coefficient $a_{-1}$ in the Laurent expansion. At a double pole, $h(z)=(z-z_0{)}^{2}f(z)$ is holomorphic and $h(z)=h(z_0)+h\prime (z_0)(z-z_0)+\cdots$, so $\mathrm{Res}(f,z_0)=h\prime (z_0)$.

Partial fractions giveFor $1<|z|<2$ this becomes

At zero: $\sin z$ is holomorphic with a zero; $1/\sin z$ has a simple pole of residue $1$; $z/\sin z$ has a removable singularity with value $1$; $\sin (1/z)$ is essential with residue $1$; $z\sin (1/z)$ is essential with residue $0$. Finally $1/\sin (1/z)$ has poles accumulating at zero, so zero is a non-isolated singularity and no residue there is defined.

Fermat’s principle makes $\int ds/c$ stationary. Writing the ray as $x=x(y)$ gives a Lagrangian independent of $y$, and its first integral is $\sin \theta /c=K$. At the origin $K=(1/\sqrt{2})/(1/\sqrt{3})=\sqrt{3/2}$. At $x=1$, $c=1/\sqrt{2}$, so $\sin \theta =\sqrt{3}/2$ and the forward angle is $\theta =\pi /3$.

Successive substitution shows every stage is a polynomial in $z=h\lambda$ times $y_n$, of degree at most its stage number. Hence $y_{n+1}=r(z)y_n$ with $r$ a degree-$\nu$ polynomial.

Order $\nu$ requires $r(z)-e^z=O(z^{\nu +1})$ on the test equation, fixing all $\nu +1$ coefficients: $r(z)={\sum }_{l=0}^{\nu }{z}^l/l!$.

For $\nu =2$, $|r(iy){|}^2=1+y^4/4$, so only $y=0$ is stable. For $\nu =3$, $|r(iy){|}^2=1-y^4/12+y^6/36$, so the stable imaginary segment is $|y|\le \sqrt{3}$.

Factorisation of the likelihood $(1-\theta {)}^{-n}{1}_{\{\theta \le X_{(1)}\}}$ shows that $T=X_{(1)}$ is sufficient: conditional on it, the sample law is independent of $\theta$.

The Rao–Blackwell theorem says that for convex loss, conditioning any estimator on a sufficient statistic cannot increase risk; for squared error it preserves the mean and weakly reduces variance.

Since $E{X}_{(1)}=\theta +(1-\theta )/(n+1)$, take $\hat{\theta }=((n+1)X_{(1)}-1)/n$. It is unbiased, and the supplied variance formula gives $\mathrm{var}\hat{\theta }=(1-\theta {)}^2/[n(n+2)]$, which is its MSE.

Replace the quadratic constraint by the two linear inequalities $-2\le x_1-x_2+x_3\le 2$. Simplex pivots lead to the active constraints $x_1=0$, $-x_2+x_3=2$, and $3x_2+x_3=4$. Thus $x^{\ast }=(0,1/2,5/2)$ and the value is $3$. With right-hand side $4+\epsilon$, the same basis remains optimal for small nonzero $\epsilon$, giving $x^{\ast }=(0,1/2+\epsilon /4,5/2+\epsilon /4)$.

A Jordan form is block diagonal with blocks $J_k(\lambda )$ having $\lambda$ on the diagonal and ones on the superdiagonal. Over $\mathbb{C}$ every square matrix is similar to one, uniquely up to permutation of blocks.

$D$ lowers total degree, so it is nilpotent and has only eigenvalue zero. Solving $({\partial }_z+{\partial }_w)f=0$ gives $f=h(z-w)$; the degree restrictions give the eigenspace $\mathrm{span}\{1,z-w,(z-w{)}^2\}$.

Direct degree counting gives $\dim \ker D^k=3,5,7,8,9$ for $k=1,\dots ,5$. Hence the Jordan block sizes are $5,3,1$.

On each Jordan block of $A$, $g(A)$ is triangular with diagonal entry $g(\lambda )$; hence the eigenvalues, with multiplicity, are $g({\lambda }_i)$.

The shift matrix obeys $A^4=I$ and $B=aI+bA+c{A}^2+d{A}^3$. Evaluating $g(t)=a+bt+c{t}^2+d{t}^3$ at $1,i,-1,-i$ gives

If $a^m=b^n=0$, then every term in $(a+b{)}^{m+n-1}$ vanishes; also $(ra{)}^m=0$. Thus the nilpotents form an ideal $N$. The quotient $R/N$ has no nonzero nilpotents.

One direction follows because finitely many nilpotent coefficients generate a nilpotent ideal. Conversely, reducing a nilpotent polynomial modulo $N$ gives a nilpotent polynomial over the reduced ring $R/N$; comparing its highest nonzero coefficient shows it must be zero, so every coefficient lay in $N$.

If $b^k=0$, then $(u+b{)}^{-1}=u^{-1}{\sum }_{j=0}^{k-1}(-b{u}^{-1}{)}^j$. For the polynomial, its positive-degree part is nilpotent by (ii), so adding it to the unit $a_0$ gives a unit.

Choose nonzero $g$ of least degree with $fg=0$. If $b$ is its leading coefficient, then the top coefficient equation gives $a_{n}b=0$; minimality applied after removing the top term forces $bf=0$. In $(\mathbb{Z}/m\mathbb{Z})[X]$, a nonzero annihilator $r$ therefore satisfies $r{a}_i=0$ for every $i$; a prime divisor of $m/\gcd (r,m)$ divides all $a_i$ and $m$, so the displayed gcd exceeds one.

Uniform convergence means: for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|f_n(x)-f(x)|<\epsilon$ for every $x\in E$.

If $|f_n(x)|\le M_n$ and $\sum M_n<\infty$, then every tail is bounded uniformly by the corresponding numerical tail. The uniform Cauchy criterion therefore proves uniform convergence.

For each $a>1$, $n^{-x}\le n^{-a}$ on $[a,\infty )$ and the p-series converges. The M-test gives locally uniform convergence there, so the sum $\zeta$ is continuous on $(1,\infty )$.

The series for $S$ converges uniformly by the M-test. Therefore termwise integration is valid and every term integrates to $\sin (n^3\pi )/n^7=0$, so the integral is zero.

It is not uniform. For a small fixed $c>0$, set $x_N=c/N^3$. On a block $N\le n\le (1+\delta )N$, all $n^3{x}_N$ lie in a compact subinterval of $(0,\pi )$, so the corresponding tail has a positive lower bound independent of $N$; this violates the uniform Cauchy criterion.

Apply the maximum modulus principle to $f(z)/z$, with its removable value $f\prime (0)$, to obtain $|f(z)|\le |z|$. Equality at a nonzero interior point makes $f(z)/z$ constant, so $f(z)=az$ with $|a|=1$.

Let $a=e^{i\pi /3}$ and $b=e^{-i\pi /3}$. The Möbius map $T(z)=(z-a)/(z-b)$ sends the two circular arcs through $a,b$ to rays bounding a wedge of angle $2\pi /3$. After choosing the branch and a unimodular constant so the interior has positive argument, $f(z)=e^{i\varphi }T(z{)}^{3/2}$ maps that wedge bijectively onto the upper half-plane.

Off the slit choose the stated analytic square-root branch. Differentiation gives $f\prime (z)=c^{-1}(1+z/\sqrt{z^2-c^2})$, which does not vanish there, so the map is conformal.

Multiplying $(z+\sqrt{z^2-c^2})(z-\sqrt{z^2-c^2})=c^2$ gives the reciprocal identity and hence $z=c(\zeta +{\zeta }^{-1})/2$.

For $\zeta =R{e}^{i\theta }$,
$x=\frac{c}{2}(R+R^{-1})\cos \theta$ and $y=\frac{c}{2}(R-R^{-1})\sin \theta$. Thus $a=c(R+R^{-1})/2$, $b=c(R-R^{-1})/2$. Nested level curves and the behaviour at infinity show that exteriors correspond.

The Möbius map sends $|\zeta |=R$ to the imaginary axis and $\infty$ to $1$; testing one exterior point shows that $|\zeta |>R$ maps to $\Re w>0$.

The required harmonic function isOn the upper and lower boundary its argument is respectively $\pi /2$ and $-\pi /2$, while at infinity the logarithm tends to $0$.

For the odd two-periodic extensions, $\bar{f}=\sum b_n\sin (n\pi x)$ with $b_n=2[1+(-1{)}^n]/(n\pi )$, while $\bar{F}$ has $b_n=4[1-(-1{)}^n]/(n^3{\pi }^3)$. Thus the first has $O(n^{-1})$ convergence because its periodic extension jumps, while the smoother second extension has $O(n^{-3})$ coefficients.

Writing $y=e^{-x}v$ gives $-v\prime \prime +v=\lambda v$, hence ${\lambda }_n=n^2{\pi }^2+1$ and $y_n=e^{-x}\sin (n\pi x)$. The Sturm–Liouville form is $-(e^{2x}y\prime )\prime =\lambda e^{2x}y$, so ${\int }_0^1{e}^{2x}{y}_n{y}_m=0$ for $n\ne m$. Expanding the source gives coefficients $c_n=2{\int }_0^{1}f\sin (n\pi x)dx$, hence

The nonrelativistic Schrödinger Hamiltonian uses $p^2/(2m)$ and therefore has no massless limit; massless particles require a relativistic wave equation or quantum field theory.

The walls give $\omega =n\pi /(2a)$; normalisation gives $A=1/\sqrt{a}$, and $E_n={\hslash }^2{\pi }^2{n}^2/(8m{a}^2)$. Product-to-sum integration over $(-a,a)$ gives $⟨{\psi }_n,{\psi }_m⟩={\delta }_{nm}$.

Using the quadratic form of the Hamiltonian,
$E_{av}=({\hslash }^2/2m)\int |\psi \prime {|}^{2}dx=2{\pi }^2{\hslash }^2/(m{a}^2)$.

Only odd stationary states contribute, and for $j=2n+1$ their squared coefficients are $128/[{\pi }^2(j^2-16{)}^2]$. Since $E_{av}/E_1=16$, substitution givesso $B=8$.