Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1

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.

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

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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.

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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 $\mathrm{Res}(f,z_0)=h\prime (z_0)$.

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

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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.

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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 $\theta =\pi /3$.

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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.

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

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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 $|y|\le \sqrt{3}$.

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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 $\theta$.

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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.

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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.

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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 $x^{\ast }=(0,1/2+\epsilon /4,5/2+\epsilon /4)$.

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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.

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$D$ 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 $\mathrm{span}\{1,z-w,(z-w{)}^2\}$.

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

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

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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

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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.

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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 $N$.

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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.

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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.

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

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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.

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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 $(1,\infty )$.

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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.

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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.

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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 $|a|=1$.

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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.

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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.

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

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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.

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

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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 $0$.

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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.

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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

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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.

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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 $⟨{\psi }_n,{\psi }_m⟩={\delta }_{nm}$.

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

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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 $B=8$.

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Using $\nabla |x-x\prime {|}^{-1}=-\nabla \prime |x-x\prime {|}^{-1}$, the divergence is a line integral of a total derivative, zero for a closed wire or suitable vanishing endpoints. Taking the curl and moving it under the integral gives the stated Biot–Savart formula.

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The negative-$z$ segment integrates to the stated azimuthal field. The negative-$y$ segment contributesAdding this to the first field is the complete answer. On $x=0,y=1$, the field tends to $0$ as $z\to +\infty$ and to $-{\mu }_{0}I\hat{x}/(2\pi )$ as $z\to -\infty$.

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Taking the curl of Euler and using incompressibility gives ${\partial }_t\omega +(u\cdot \nabla )\omega =(\omega \cdot \nabla )u$, equivalently $D\omega /Dt=(\omega \cdot \nabla )u$.

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The divergence is $-\alpha /2-\alpha /2+\alpha =0$.

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For uniform vertical vorticity, advection vanishes and $(\omega \cdot \nabla )u=\alpha \omega \hat{z}$, hence $\dot{\omega }=\alpha (t)\omega$.

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$\omega (t)={\omega }_0\exp ({\int }_0^t\alpha (s)ds)$. Positive $\alpha$ stretches material lines vertically; conservation of vortex flux through the shrinking transverse area amplifies vorticity.

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For fixed $x$ add $K\Pi (t)$ to $p(t)-f(t)$ so the resulting function also vanishes at $t=x$. Repeated Rolle gives a point $\xi$ where its $(n+1)$st derivative vanishes, yielding the formula.

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The recurrence $T_{k+1}=2x{T}_k-T_{k-1}$ shows that $T_{n+1}$ has leading coefficient $2^n$. Since its roots are the interpolation nodes, $T_{n+1}=2^n\Pi$, and $|T_{n+1}|\le 1$ on $[-1,1]$.

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Since $|f^{(n+1)}|\le k^{n+1}$, the error is at most $k^{n+1}/[2^n(n+1)!]\to 0$ uniformly.

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At $x=1$, direct substitution gives a product of odd half-integers. Taking logarithms and applying the hinted lower Riemann-sum estimate yields $\log \Pi (1)>n\log (2/e)$, hence $\Pi (1)>(2/e{)}^n$.

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With $P_0=X_0(X_0^T{X}_0{)}^{-1}{X}_0^T$, ${\hat{\beta }}_0=(X_0^T{X}_0{)}^{-1}{X}_0^{T}Y$ and ${\hat{\sigma }}^2=\Vert (I-P_0)Y{\Vert }^2/n$.

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The statistic is $F=[(RS{S}_0-RS{S}_1)/p_1]/[RS{S}_1/(n-p)]$, with an $F_{p_1,n-p}$ null distribution.

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Under the null, $R=(I-P_0)\epsilon$, so $R/\sigma \stackrel{d}{=}(I-P_0)Z$ for $Z\sim N(0,I)$.

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The displayed linear form is centred normal with variance $n^{-1}{X}_{1j}^T(I-P_0)X_{1j}\le n^{-1}\Vert X_{1j}{\Vert }^2=1$. The normal tail bound proves the claim.

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By (iv) and the union bound, $P(T\ge t)\le p_1{e}^{-t^2/2}$. Setting $t=\sqrt{2\log (p_1/\alpha )}$ bounds the size by $\alpha$.

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Scale cancels inBecause $R=\sigma (I-P_0)Z$ under the null, it has exactly the requested distribution; its quantiles can be simulated without knowing $\sigma$.

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A stopping time $T$ has $\{T\le n\}$ determined by $X_0,\dots ,X_n$. The Strong Markov property says that, conditional on $T<\infty$ and $X_T=i$, the post-$T$ chain is a fresh chain started at $i$, independent of the pre-$T$ history.

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First-step equations $h_B=1+h_W/2$ and $h_W=1+2h_B/3$ give $h_B=9/4$.

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Augment by whether the last sound was a growl. The equations $h_B=1+(h_G+h_W)/2$, $h_G=1+2h_B/3$, $h_W=1+2h_B/3+h_G/3$ give $h_B=39/4$.

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Augment the state by the current sound and $D=N_b-N_w$, and solve the translation-invariant first-step equations. Equivalently, solve the Poisson equation for the additive reward $+1,0,-1$. The stationary probabilities are $(2/5,3/10,3/10)$, so the drift is $1/10$ per sound. The correction term has the same value at the initial and terminal bark states; optional stopping at $D=m$ therefore gives the expected produced length $10m$.

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