= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/Paperib_1_2026.pdf

= 1G
{parent=Paper 1}
{scope}
{title2=Linear Algebra}

= a
{parent=1g}
{scope}

= Solution
{parent=a}

Bilinearity and symmetry follow from the <integral>. Also $\langle p,p\rangle=\int_0^1p(t)^2dt\ge0$, and equality forces the continuous <polynomial> $p$ to vanish identically.

Solved by gpt-5.6-sol high.

= b
{parent=1g}
{scope}

= Solution
{parent=b}

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

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 1}
{scope}
{title2=Topological Spaces}

= Solution
{parent=2f}

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 $d\prime(x_n,x)\lt 1/n$; the common-sequence assumption would imply $d(x_n,x)\to0$, 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.

Solved by gpt-5.6-sol high.

= 3
{parent=Paper 1}
{scope}
{title2=Complex Analysis OR Complex Methods}

= 3.1G
{parent=3}
{scope}

= a
{parent=3.1g}
{scope}

= Solution
{parent=a}

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

Solved by gpt-5.6-sol high.

= b
{parent=3.1g}
{scope}

= Solution
{parent=b}

Partial fractions give
$$f=\frac1{2z^2}-\frac3{4z}+\frac1{z+1}-\frac1{4(z+2)}.$$
For $1\lt |z|\lt 2$ this becomes
$$\frac1{2z^2}-\frac3{4z}+\sum_{n\ge0}(-1)^nz^{-n-1}-\frac18\sum_{n\ge0}(-z/2)^n.$$

Solved by gpt-5.6-sol high.

= 3.2C
{parent=3}
{scope}

= Solution
{parent=3.2c}

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.

Solved by gpt-5.6-sol high.

= 4B
{parent=Paper 1}
{scope}
{title2=Variational Principles}

= Solution
{parent=4b}

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 $K=(1/\sqrt2)/(1/\sqrt3)=\sqrt{3/2}$. At $x=1$, $c=1/\sqrt2$, so $\sin\theta=\sqrt3/2$ and the forward angle is $\theta=\pi/3$.

Solved by gpt-5.6-sol high.

= 5C
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= i
{parent=5c}
{scope}

= Solution
{parent=i}

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>.

Solved by gpt-5.6-sol high.

= ii
{parent=5c}
{scope}

= Solution
{parent=ii}

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

Solved by gpt-5.6-sol high.

= iii
{parent=5c}
{scope}

= Solution
{parent=iii}

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 $|y|\le\sqrt3$.

Solved by gpt-5.6-sol high.

= 6H
{parent=Paper 1}
{scope}
{title2=Statistics}

= a
{parent=6h}
{scope}

= Solution
{parent=a}

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 $\theta$.

Solved by gpt-5.6-sol high.

= b
{parent=6h}
{scope}

= Solution
{parent=b}

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.

Solved by gpt-5.6-sol high.

= c
{parent=6h}
{scope}

= Solution
{parent=c}

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 $\operatorname{var}\hat\theta=(1-\theta)^2/[n(n+2)]$, which is its MSE.

Solved by gpt-5.6-sol high.

= 7H
{parent=Paper 1}
{scope}
{title2=Optimisation}

= Solution
{parent=7h}

Replace the quadratic constraint by the two linear inequalities $-2\le x_1-x_2+x_3\le2$. Simplex pivots lead to the active constraints $x_1=0$, $-x_2+x_3=2$, and $3x_2+x_3=4$. Thus $x^*=(0,1/2,5/2)$ and the value is $3$. With right-hand side $4+\varepsilon$, the same <basis> remains optimal for small nonzero $\varepsilon$, giving $x^*=(0,1/2+\varepsilon/4,5/2+\varepsilon/4)$.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 1}
{scope}
{title2=Linear Algebra}

= a
{parent=8e}
{scope}

= Solution
{parent=a}

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.

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

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

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

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

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

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

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The shift <matrix> obeys $A^4=I$ and $B=aI+bA+cA^2+dA^3$. Evaluating $g(t)=a+bt+ct^2+dt^3$ at $1,i,-1,-i$ gives
$$a+b+c+d,quad a-c+i(b-d),quad a-b+c-d,quad a-c-i(b-d).$$

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 1}
{scope}
{title2=Groups, Rings and Modules}

= i
{parent=9e}
{scope}

= Solution
{parent=i}

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.

Solved by gpt-5.6-sol high.

= ii
{parent=9e}
{scope}

= Solution
{parent=ii}

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 $N$.

Solved by gpt-5.6-sol high.

= iii
{parent=9e}
{scope}

= Solution
{parent=iii}

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

Solved by gpt-5.6-sol high.

= iv
{parent=9e}
{scope}

= Solution
{parent=iv}

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

Solved by gpt-5.6-sol high.

= 10G
{parent=Paper 1}
{scope}
{title2=Analysis II}

= a
{parent=10g}
{scope}

= Solution
{parent=a}

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

Solved by gpt-5.6-sol high.

= b
{parent=10g}
{scope}

= Solution
{parent=b}

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

Solved by gpt-5.6-sol high.

= c
{parent=10g}
{scope}

= Solution
{parent=c}

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

Solved by gpt-5.6-sol high.

= d
{parent=10g}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

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.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

It is not uniform. For a small fixed $c\gt 0$, set $x_N=c/N^3$. On a block $N\le n\le(1+\delta)N$, all $n^3x_N$ 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.

Solved by gpt-5.6-sol high.

= 11
{parent=Paper 1}
{scope}
{title2=Complex Analysis OR Complex Methods}

= 11.1G
{parent=11}
{scope}

= a
{parent=11.1g}
{scope}

= Solution
{parent=a}

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 $|a|=1$.

Solved by gpt-5.6-sol high.

= b
{parent=11.1g}
{scope}

= Solution
{parent=b}

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, $f(z)=e^{i\phi}T(z)^{3/2}$ maps that wedge bijectively onto the upper half-plane.

Solved by gpt-5.6-sol high.

= 11.2C
{parent=11}
{scope}

= i
{parent=11.2c}
{scope}

= Solution
{parent=i}

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.

Solved by gpt-5.6-sol high.

= ii
{parent=11.2c}
{scope}

= Solution
{parent=ii}

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

Solved by gpt-5.6-sol high.

= iii
{parent=11.2c}
{scope}

= Solution
{parent=iii}

For $\zeta=Re^{i\theta}$,
$x=\tfrac c2(R+R^{-1})\cos\theta$ and $y=\tfrac c2(R-R^{-1})\sin\theta$. 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.

Solved by gpt-5.6-sol high.

= iv
{parent=11.2c}
{scope}

= Solution
{parent=iv}

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

Solved by gpt-5.6-sol high.

= v
{parent=11.2c}
{scope}

= Solution
{parent=v}

The required harmonic <function> is
$$u(z)=\frac2\pi\operatorname{Im}\log\!\left(\frac{f(z)-R}{f(z)+R}\right).$$
On the upper and lower boundary its argument is respectively $\pi/2$ and $-\pi/2$, while at infinity the logarithm tends to $0$.

Solved by gpt-5.6-sol high.

= 12D
{parent=Paper 1}
{scope}
{title2=Methods}

= i
{parent=12d}
{scope}

= Solution
{parent=i}

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.

Solved by gpt-5.6-sol high.

= ii
{parent=12d}
{scope}

= Solution
{parent=ii}

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 $\int_0^1e^{2x}y_ny_m=0$ for $n\ne m$. Expanding the source gives coefficients $c_n=2\int_0^1f\sin(n\pi x)dx$, hence
$$a_n=\frac{2[1+(-1)^n]}{n\pi(n^2\pi^2+1)}.$$

Solved by gpt-5.6-sol high.

= 13B
{parent=Paper 1}
{scope}
{title2=Quantum Mechanics}

= i
{parent=13b}
{scope}

= Solution
{parent=i}

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.

Solved by gpt-5.6-sol high.

= ii
{parent=13b}
{scope}

= Solution
{parent=ii}

The walls give $\omega=n\pi/(2a)$; normalisation gives $A=1/\sqrt a$, and $E_n=\hbar^2\pi^2n^2/(8ma^2)$. Product-to-sum integration over $(-a,a)$ gives $\langle\psi_n,\psi_m\rangle=\delta_{nm}$.

Solved by gpt-5.6-sol high.

= iii
{parent=13b}
{scope}

= Solution
{parent=iii}

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

Solved by gpt-5.6-sol high.

= iv
{parent=13b}
{scope}

= Solution
{parent=iv}

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 gives
$$\pi^2=8\sum_{n\ge0}\frac{(2n+1)^2}{[(2n+1)^2-16]^2},$$
so $B=8$.

Solved by gpt-5.6-sol high.

= 14D
{parent=Paper 1}
{scope}
{title2=Electromagnetism}

= a
{parent=14d}
{scope}

= Solution
{parent=a}

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.

Solved by gpt-5.6-sol high.

= b
{parent=14d}
{scope}

= Solution
{parent=b}

The negative-$z$ segment integrates to the stated azimuthal field. The negative-$y$ segment contributes
$$\frac{\mu_0I}{4\pi}\frac{x\hat z-z\hat x}{x^2+z^2}\left(1-\frac y{\sqrt{x^2+y^2+z^2}}\right).$$
Adding 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 $-\mu_0I\hat x/(2\pi)$ as $z\to-\infty$.

Solved by gpt-5.6-sol high.

= 15A
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics}

= a
{parent=15a}
{scope}

= Solution
{parent=a}

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

Solved by gpt-5.6-sol high.

= b
{parent=15a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

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

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

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

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

$\omega(t)=\omega_0\exp(\int_0^t\alpha(s)ds)$. Positive $\alpha$ stretches material lines vertically; conservation of vortex flux through the shrinking transverse area amplifies <vorticity>.

Solved by gpt-5.6-sol high.

= 16C
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= i
{parent=16c}
{scope}

= Solution
{parent=i}

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.

Solved by gpt-5.6-sol high.

= ii
{parent=16c}
{scope}

= Solution
{parent=ii}

The recurrence $T_{k+1}=2xT_k-T_{k-1}$ shows that $T_{n+1}$ has leading coefficient $2^n$. Since its roots are the interpolation nodes, $T_{n+1}=2^n\Pi$, and $|T_{n+1}|\le1$ on $[-1,1]$.

Solved by gpt-5.6-sol high.

= iii
{parent=16c}
{scope}

= Solution
{parent=iii}

Since $|f^{(n+1)}|\le k^{n+1}$, the error is at most $k^{n+1}/[2^n(n+1)!]\to0$ uniformly.

Solved by gpt-5.6-sol high.

= iv
{parent=16c}
{scope}

= Solution
{parent=iv}

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

Solved by gpt-5.6-sol high.

= 17H
{parent=Paper 1}
{scope}
{title2=Statistics}

= i
{parent=17h}
{scope}

= Solution
{parent=i}

With $P_0=X_0(X_0^TX_0)^{-1}X_0^T$, $\hat\beta_0=(X_0^TX_0)^{-1}X_0^TY$ and $\hat\sigma^2=\|(I-P_0)Y\|^2/n$.

Solved by gpt-5.6-sol high.

= ii
{parent=17h}
{scope}

= Solution
{parent=ii}

The statistic is $F=[(RSS_0-RSS_1)/p_1]/[RSS_1/(n-p)]$, with an $F_{p_1,n-p}$ null distribution.

Solved by gpt-5.6-sol high.

= iii
{parent=17h}
{scope}

= Solution
{parent=iii}

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

Solved by gpt-5.6-sol high.

= iv
{parent=17h}
{scope}

= Solution
{parent=iv}

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

Solved by gpt-5.6-sol high.

= v
{parent=17h}
{scope}

= Solution
{parent=v}

By (iv) and the union bound, $P(T\ge t)\le p_1e^{-t^2/2}$. Setting $t=\sqrt{2\log(p_1/\alpha)}$ bounds the size by $\alpha$.

Solved by gpt-5.6-sol high.

= vi
{parent=17h}
{scope}

= Solution
{parent=vi}

Scale cancels in
$$T=\max_j\frac{|X_{1j}^TR|}{\|R\|}.$$
Because $R=\sigma(I-P_0)Z$ under the null, it has exactly the requested distribution; its quantiles can be simulated without knowing $\sigma$.

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 1}
{scope}
{title2=Markov Chains}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

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

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

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

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

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 $h_B=39/4$.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

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 $10m$.

Solved by gpt-5.6-sol high.