past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-4.bigb

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperib_4_2022.pdf

= 1F
{parent=Paper 4}
{scope}
{title2=Linear Algebra}

= Solution
{parent=1F}

A <Hermitian form> is conjugate-linear in its first argument, linear in its second, and satisfies $\phi(w,v)=\overline{\phi(v,w)}$. For $H=\phi-\psi$, the hypothesis says $H(v,v)=0$. The complex <polarization identity>
$$
4H(v,w)=\sum_{k=0}^3i^kH(v+i^kw,v+i^kw)
$$
therefore gives $H(v,w)=0$ for all $v,w$.

The displayed matrix is Hermitian. By <Sylvester's criterion>, its leading principal minors are
$$
4>0,
\qquad \det A=12-(2i)(-2i)=8>0.
$$
Hence the form is
$$
\boxed{\text{positive definite}.}
$$

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 4}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=2G}

The <closure> $\overline Z$ is the intersection of all closed subsets containing $Z$; equivalently, every neighbourhood of a point of $\overline Z$ meets $Z$. The subspace is <dense subset>[dense] when $\overline Z=X$. A space is <Hausdorff space>[Hausdorff] when distinct points have disjoint neighbourhoods.

For Hausdorff $Y$, the diagonal $\Delta_Y$ is closed in $Y\times Y$. Thus
$$
E=\{x:f(x)=g(x)\}=(f,g)^{-1}(\Delta_Y)
$$
is closed. Since it contains the dense set $Z$, it equals $X$.

The conclusion fails without Hausdorffness. Let $X=Y=\{0,1\}$ have the <Sierpinski space> topology $\{\varnothing,\{1\},X\}$ and let $Z=\{1\}$, which is dense. The identity map and the constant map with value $1$ are continuous and agree on $Z$, but differ at $0$.

Solved by gpt-5.6-sol high.

= 3G
{parent=Paper 4}
{scope}
{title2=Complex Analysis}

= Solution
{parent=3G}

If such an $f$ existed, boundedness would make its isolated singularity at zero removable. Let $F$ be the holomorphic extension to the disc. Continuity gives $1\leq|F(0)|\leq2$. Equality at either endpoint contradicts the minimum-modulus principle applied to $F$ or the <maximum modulus principle>, since $F$ is nonconstant. Hence $F(0)\in A$. Surjectivity of $f$ supplies some nonzero $z$ with $f(z)=F(0)$, contradicting injectivity of the extension. Therefore
$$
\boxed{\text{no such biholomorphic map exists}.}
$$

Solved by gpt-5.6-sol high.

= 4B
{parent=Paper 4}
{scope}
{title2=Quantum Mechanics}

= a
{parent=4b}
{scope}

= Solution
{parent=a}

The first term is radial kinetic energy from the Laplacian, the second is the attractive <Coulomb's law>[Coulomb potential], and the third is the centrifugal energy from the angular-momentum eigenvalue $L^2=l(l+1)\hbar^2$. Single-valued spherical harmonics require
$$
\boxed{l=0,1,2,\ldots}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=4b}
{scope}

= Solution
{parent=b}

Substitute $g=r^ae^{-br}$ into the radial <Schrödinger equation> and equate the coefficients of $r^{-2}$, $r^{-1}$, and the constant term. Regularity selects
$$
a=l,
\qquad
b=\frac{me^2}{4\pi\varepsilon_0\hbar^2(l+1)}.
$$
The corresponding bound-state energy is
$$
\boxed{
E_l=-\frac{me^4}{2(4\pi\varepsilon_0)^2\hbar^2(l+1)^2}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=4b}
{scope}

= Solution
{parent=c}

The higher state has principal quantum number $l+2$ and the lower one $l+1$. Energy conservation for the emitted <photon> gives
$$
\boxed{
\nu=\frac{me^4}{2h(4\pi\varepsilon_0)^2\hbar^2}
\left(\frac1{(l+1)^2}-\frac1{(l+2)^2}\right)}.
$$

Solved by gpt-5.6-sol high.

= 5D
{parent=Paper 4}
{scope}
{title2=Electromagnetism}

= a
{parent=5d}
{scope}

= Solution
{parent=a}

In vacuum, curl <Faraday's law> and use the <Ampère-Maxwell equation>:
$$
\nabla\times(\nabla\times\mathbf B)
=\mu_0\varepsilon_0\frac{\partial}{\partial t}
(\nabla\times\mathbf E)
=-\mu_0\varepsilon_0\mathbf B_{tt}.
$$
Since $\nabla\cdot\mathbf B=0$, the left side is $-\nabla^2\mathbf B$. Hence
$$
\boxed{\mathbf B_{tt}=c^2\nabla^2\mathbf B,
\qquad c=(\mu_0\varepsilon_0)^{-1/2}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=5d}
{scope}

= Solution
{parent=b}

The divergence equation requires $B_3=0$, and the wave equation requires
$$
\boxed{\omega^2=c^2k^2}.
$$
For propagation in the positive $z$ direction, Faraday's law gives
$$
\boxed{
\mathbf E=\frac\omega k(B_2,-B_1,0)\cos(kz-\omega t)}.
$$
Thus $\mathbf E$, $\mathbf B$, and the propagation direction are mutually perpendicular.

Solved by gpt-5.6-sol high.

= c
{parent=5d}
{scope}

= Solution
{parent=c}

The <Poynting vector> is
$$
\boxed{
\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B
=\frac{\omega}{\mu_0k}(B_1^2+B_2^2)
\cos^2(kz-\omega t)\,\mathbf e_z}.
$$
It is the electromagnetic energy flux in the propagation direction; its time average is $c(B_1^2+B_2^2)/(2\mu_0)$.

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 4}
{scope}
{title2=Numerical Analysis}

= a
{parent=6c}
{scope}

= Solution
{parent=a}

Let $\ell_i$ be the degree-$n-1$ cardinal polynomial with $\ell_i(c_j)=\delta_{ij}$. Exactness applied to the nonnegative polynomial $\ell_i^2$ gives
$$
b_i=\sum_jb_j\ell_i(c_j)^2
=\int_a^bw(x)\ell_i(x)^2\,dx>0,
$$
because $w>0$ and $\ell_i$ is nonzero. Thus every <Gaussian quadrature> weight is positive.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

Symmetry gives nodes $\pm c$ with equal weights $b$. Exactness for $1$ and $x^2$ requires
$$
2b=\int_{-1}^1x^2dx=\frac23,
\qquad
2bc^2=\int_{-1}^1x^4dx=\frac25.
$$
Therefore
$$
\boxed{b_1=b_2=\frac13,
\qquad c_1=-\sqrt{\frac35},
\quad c_2=\sqrt{\frac35}}.
$$
Odd moments vanish automatically, so the rule is exact through degree three.

Solved by gpt-5.6-sol high.

= 7H
{parent=Paper 4}
{scope}
{title2=Markov Chains}

= a
{parent=7h}
{scope}

= Solution
{parent=a}

Let $q=\mathbb P_i(T_i<\infty)$. By the <strong Markov property>, after every return to $i$ the chance of another return is again $q$. Hence
$$
\boxed{\mathbb P_i(V_i=k)=(1-q)q^{k-1},
\qquad k=1,2,\ldots}.
$$
Thus $V_i$ is geometric with success parameter $1-q$; when $q=1$, it is infinite almost surely.

Solved by gpt-5.6-sol high.

= b
{parent=7h}
{scope}

= Solution
{parent=b}

In a transient irreducible chain, every state $j$ is transient, so the expected number of visits satisfies
$$
\sum_{n=0}^\infty P^n(i,j)=\mathbb E_iV_j<\infty.
$$
The terms are nonnegative, and every convergent series has terms tending to zero. Therefore
$$
\boxed{P^n(i,j)\to0}.
$$

Solved by gpt-5.6-sol high.

= 8F
{parent=Paper 4}
{scope}
{title2=Linear Algebra}

= Solution
{parent=8F}

The rank is $r(\alpha)=\dim\operatorname{im}\alpha$ and the nullity is $n(\alpha)=\dim\ker\alpha$. Extending a basis of the kernel to a basis of $V$, the images of the added vectors form a basis of the image, proving the <rank-nullity theorem>
$$
\dim V=r(\alpha)+n(\alpha).
$$

Since $\operatorname{im}(\alpha+\beta)\subseteq\operatorname{im}\alpha+\operatorname{im}\beta$, the upper sum bound follows. Applying it to $\alpha=(\alpha+\beta)-\beta$ and interchanging $\alpha,\beta$ gives the lower bound. For products, the image lies in $\operatorname{im}\alpha$ and is the image under $\alpha$ of $\operatorname{im}\beta$, giving the upper bound; rank-nullity on that restriction gives
$$
r(\alpha\beta)\geq r(\beta)-n(\alpha)=r(\alpha)+r(\beta)-n.
$$

All four bounds are sharp. In a fixed basis, nested diagonal projections attain the product upper bound; placing $\operatorname{im}\beta$ with the smallest possible intersection with $\ker\alpha$ attains its lower bound. Taking $\beta=-\alpha$ on a subspace of dimension $\min(r(\alpha),r(\beta))$ attains the sum lower bound, while choosing the two images in general position and avoiding cancellation attains the sum upper bound.

Both upper bounds need not be attainable simultaneously over every field. Over $\mathbb F_2$ with $n=1$ and both ranks one, necessarily $\alpha=\beta=I$. Then $r(\alpha\beta)=1$ but $r(\alpha+\beta)=0$, below its upper bound one.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 4}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

A polynomial over a <unique factorization domain> is primitive when its coefficients have no common nonunit factor. If a prime divided every coefficient of a product of two primitive polynomials, reduction modulo that prime would turn the product of two nonzero polynomials over an integral domain into zero, impossible. This proves <Gauss lemma for polynomials>.

If positive-degree $f,g\in R[X]$ became non-coprime in $F[X]$, they would share a positive-degree primitive factor after clearing content. Gauss's lemma then makes that factor divide both in $R[X]$, contrary to coprimality. Hence they remain coprime over the fraction field.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

Regard $f,g$ as polynomials in $Y$ over $\mathbb C(X)$. Their coprimality and the preceding argument let Euclid's algorithm produce
$$
A(X,Y)f+B(X,Y)g=1
$$
with coefficients in $\mathbb C(X)$. Clearing denominators gives a nonzero $h(X)\in(f,g)$, so $I\cap\mathbb C[X]\ne0$. Interchanging $X,Y$ gives a nonzero $k(Y)\in I$.

In the quotient, $h(X)=k(Y)=0$. Reducing powers by these two univariate relations shows that the finitely many monomials
$$
X^iY^j,
\qquad 0\leq i<\deg h,
\quad0\leq j<\deg k,
$$
span. Therefore $\mathbb C[X,Y]/I$ is finite-dimensional.

Solved by gpt-5.6-sol high.

= 10G
{parent=Paper 4}
{scope}
{title2=Analysis and Topology}

= a
{parent=10g}
{scope}

= Solution
{parent=a}

A space is connected when it has no separation into two nonempty disjoint open sets; the connected subspaces of $\mathbb R$ are exactly the intervals. A space is path connected if every two points are joined by a continuous path. A separation would pull back along a path to a separation of $[0,1]$, so path connectedness implies connectedness.

Each of the $n$ interval components of $I_n$ must map into one of the $m$ components of $I_m$. This assignment is constant on every connected component of the uniform function space. Conversely, for a fixed assignment, straight-line interpolation within each target interval gives paths between maps. Hence
$$
\boxed{C(I_n,I_m)\text{ has }m^n\text{ connected components}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=10g}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Take the <connected space>[topologist's sine curve]
$$
S=\{(x,\sin(1/x)):0<x\leq1\}
\cup(\{0\}\times[-1,1]).
$$
It is closed and bounded, hence compact. The oscillating graph is connected and accumulates on every point of the vertical segment, so its closure $S$ is connected. No path joins the graph to the vertical segment, as continuity of the first coordinate and the unbounded oscillation of $\sin(1/x)$ would fail at the first time the path reaches $x=0$. Thus $S$ is not path connected.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Constant maps identify a copy of $S$ inside $C([0,1],S)$. If the function space were path connected, evaluating a path of functions at $0$ would give a path in $S$ between any two constant values. Since $S$ is not path connected,
$$
\boxed{C([0,1],S)\text{ is not path connected}.}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The constant maps form a connected subspace homeomorphic to $S$. Every $f\in C([0,1],S)$ is joined to the constant map $f(0)$ by
$$
H(s)(t)=f((1-s)t).
$$
Uniform continuity of $f$ makes $s\mapsto H(s)$ continuous in the uniform metric. Thus the function space is a union of connected sets, each meeting the connected set of constant maps, and so
$$
\boxed{C([0,1],S)\text{ is connected}.}
$$

Solved by gpt-5.6-sol high.

= 11E
{parent=Paper 4}
{scope}
{title2=Geometry}

= a
{parent=11e}
{scope}

= Solution
{parent=a}

The <Poincare disc model> has metric
$$
ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.
$$
Rotations about the origin are hyperbolic isometries, so a Euclidean circle centred there is a hyperbolic circle. Its radius is the radial length
$$
\boxed{R=\int_0^r\frac{2,dt}{1-t^2}
=\log\frac{1+r}{1-r}=2\operatorname{artanh}r}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=11e}
{scope}

= Solution
{parent=b}

The <hyperbolic law of sines> is
$$
\frac{\sinh a}{\sin\alpha}
=\frac{\sinh b}{\sin\beta}
=\frac{\sinh c}{\sin\gamma}.
$$
Set $\gamma=\pi/2$ in the cosine laws for sides and angles and eliminate $\cos\alpha$ and $\cos\beta$. Straight algebra gives the right-triangle identities
$$
\boxed{
\tan\alpha=\frac{\sinh a}{\cosh a\sinh b},
\qquad
\tan\alpha\tan\beta\cosh c=1}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=11e}
{scope}

= Solution
{parent=c}

<Gauss-Bonnet theorem> gives the area of a curvature-$-1$ triangle as $\pi-(\alpha+\beta+\gamma)$. Split the regular polygon into $2n$ congruent right triangles. If $\theta$ is a polygon half-angle, the right-triangle identity and part (a) give
$$
\tan\theta
=\frac1{\tan(\pi/n)\cosh R}
=\frac{1-r^2}{1+r^2}\cot\frac\pi n.
$$
The polygon area is $(n-2)\pi-2n\theta$. Rewriting $\theta=\pi/2-\cot^{-1}(\tan\theta)$ yields
$$
\boxed{
A_n(r)=2n\left[
\cot^{-1}\!\left(\frac{1-r^2}{1+r^2}\cot\frac\pi n\right)
-\frac\pi n\right]}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=11e}
{scope}

= Solution
{parent=d}

A regular hyperbolic hexagon has angle tending to $2\pi/3$ as its radius tends to zero and to zero as its vertices approach the ideal boundary. Continuity therefore gives a radius at which every angle is $\pi/2$.

Two congruent right-angled hexagons glued along three alternating sides form a hyperbolic pair of pants with geodesic boundary. Glue $2g-2$ such pairs of pants along boundary curves in the usual pants decomposition. The right angles match smoothly and produce a closed hyperbolic surface of genus $g$ for every $g\geq2$.

Solved by gpt-5.6-sol high.

= 12A
{parent=Paper 4}
{scope}
{title2=Complex Methods}

= a
{parent=12a}
{scope}

= Solution
{parent=a}

Repeated integration by parts gives
$$
\mathcal L\{t^n\}(s)=\frac n s\mathcal L\{t^{n-1}\}(s),
\qquad \mathcal L\{1\}=\frac1s,
$$
so
$$
\boxed{\mathcal L\{t^n\}=\frac{n!}{s^{n+1}}}.
$$
Multiplication by $e^{at}$ shifts the transform:
$$
\boxed{\mathcal L\{e^{at}t^n\}(s)=\frac{n!}{(s-a)^{n+1}}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=12a}
{scope}

= Solution
{parent=b}

The <Bromwich contour>[Bromwich inversion integral] closes to the left for $t>0$. Equivalently, partial fractions at the two double poles give
$$
\frac1{s^2(s+2)^2}
=-\frac1{4s}+\frac1{4s^2}
+\frac1{4(s+2)}+\frac1{4(s+2)^2}.
$$
Taking residues therefore yields
$$
\boxed{f(t)=\frac14\bigl(t-1+(t+1)e^{-2t}\bigr)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=12a}
{scope}

= Solution
{parent=c}

The transform factors as $s^{-2}(s+2)^{-2}$, whose inverse factors are $t$ and $te^{-2t}$. The <convolution theorem>[Laplace convolution theorem] gives
$$
f(t)=\int_0^t\tau(t-\tau)e^{-2(t-\tau)},d\tau,
$$
and direct integration produces
$$
f(t)=\frac14\bigl(t-1+(t+1)e^{-2t}\bigr),
$$
confirming part (b).

Solved by gpt-5.6-sol high.

= d
{parent=12a}
{scope}

= Solution
{parent=d}

The first three initial values vanish and the third derivative at zero equals one. Transforming the differential equation gives
$$
(s^4+4s^3+4s^2)F(s)-1=0,
$$
so $F(s)=1/[s^2(s+2)^2]$. Hence
$$
\boxed{f(t)=\frac14\bigl(t-1+(t+1)e^{-2t}\bigr)}.
$$

Solved by gpt-5.6-sol high.

= 13D
{parent=Paper 4}
{scope}
{title2=Variational Principles}

= a
{parent=13d}
{scope}

= Solution
{parent=a}

For variations $y+\varepsilon\eta$ with $\eta=\eta'=0$ at both endpoints, integrating the $\eta'$ term once and the $\eta''$ term twice gives
$$
\delta I=\int_a^b
\left(f_y-\frac d{dx}f_{y'}+\frac{d^2}{dx^2}f_{y''}\right)\eta,dx.
$$
The fundamental lemma yields the higher-order <Euler-Lagrange equation>
$$
\boxed{f_y-\frac d{dx}f_{y'}+\frac{d^2}{dx^2}f_{y''}=0}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=13d}
{scope}

= Solution
{parent=b}

Differentiate
$$
Q=f-\left(f_{y'}-\frac d{dx}f_{y''}\right)y'-f_{y''}y''.
$$
The product rule cancels all terms except
$$
Q'=y'\left(f_y-\frac d{dx}f_{y'}+\frac{d^2}{dx^2}f_{y''}\right)+f_x.
$$
On an extremal with no explicit $x$ dependence both terms vanish, proving the stated conserved quantity.

Solved by gpt-5.6-sol high.

= c
{parent=13d}
{scope}

= Solution
{parent=c}

The Euler-Lagrange equation is
$$
y^{(4)}-2y''-y=0.
$$
Put $A=\sqrt{1+\sqrt2}$, $B=\sqrt{\sqrt2-1}$, and $L=\pi/2$. The conditions at zero reduce the general solution to
$$
y=C(\cosh Ax-\cos Bx)
+D\left(\sinh Ax-\frac AB\sin Bx\right).
$$
Define
$$
U=\cosh(AL)-\cos(BL),
\quad V=\sinh(AL)-\frac AB\sin(BL),
$$
$$
U'=A\sinh(AL)+B\sin(BL),
\quad V'=A(\cosh(AL)-\cos(BL)).
$$
The remaining boundary conditions give
$$
\boxed{C=\frac{LV'-V}{UV'-U'V},
\qquad D=\frac{U-LU'}{UV'-U'V}},
$$
which determines the unique extremal explicitly.

Solved by gpt-5.6-sol high.

= 14B
{parent=Paper 4}
{scope}
{title2=Methods}

= a
{parent=14b}
{scope}

= Solution
{parent=a}

Integration by parts, using decay of $m$, gives the <Fourier transform of a derivative>
$$
\boxed{
\widetilde h(k)=\int m'(x)e^{-ikx}dx
=ik\widetilde m(k)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=14b}
{scope}

= Solution
{parent=b}

The standard contour transform
$$
\int_{-\infty}^{\infty}\frac{x}{x^2+\alpha^2}e^{-ikx}dx
=-i\pi\operatorname{sgn}(k)e^{-\alpha|k|}
$$
holds for $\operatorname{Re}\alpha>0$, with the appropriate upper or lower pole chosen according to the sign of $k$. Therefore
$$
\boxed{m(x)=\frac{x}{x^2+\alpha^2}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=14b}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Fourier transformation in $x$ changes Laplace's equation into
$$
\widetilde u_{yy}-k^2\widetilde u=0.
$$
Imposing the two boundary values gives
$$
\boxed{
\widetilde u(k,y)=
\frac{\widetilde f(k)\sinh(|k|(a-y))
+\widetilde g(k)\sinh(|k|y)}{\sinh(|k|a)}}.
$$
The value at $k=0$ is understood by continuity.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Part (b) gives
$$
\widetilde g(k)=-i\pi\operatorname{sgn}(k)
(e^{-a|k|}-e^{-3a|k|}).
$$
Since $(e^{-a q}-e^{-3aq})/\sinh(aq)=2e^{-2aq}$,
$$
\widetilde u=-i\pi\operatorname{sgn}(k)
\left(e^{-(2a-y)|k|}-e^{-(2a+y)|k|}\right).
$$
Inverting with part (b) yields
$$
\boxed{
u(x,y)=
\frac{x}{x^2+(2a-y)^2}
-\frac{x}{x^2+(2a+y)^2}}.
$$

Solved by gpt-5.6-sol high.

= 15B
{parent=Paper 4}
{scope}
{title2=Quantum Mechanics}

= a
{parent=15b}
{scope}

= Solution
{parent=a}

The <time-dependent Schrödinger equation> is $i\hbar\partial_t\psi=\widehat H\psi$. Differentiate $\langle\widehat A\rangle=\langle\psi|\widehat A|\psi\rangle$, use the equation and its adjoint, and include explicit time dependence of the operator. This gives <Ehrenfest theorem>
$$
\boxed{i\hbar\frac d{dt}\langle\widehat A\rangle
=\langle[\widehat A,\widehat H]\rangle
+i\hbar\left\langle\frac{\partial\widehat A}{\partial t}\right\rangle}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=15b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Substitution of the stated phase into
$$
i\hbar\psi_t=-\frac{\hbar^2}{2m}\psi_{xx}+mgx\psi
$$
cancels the potential and scalar phase terms, leaving
$$
\boxed{i\hbar\Phi_t
=-\frac{\hbar^2}{2m}\Phi_{xx}+i\hbar gt\Phi_x}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

With $X=x+gt^2/2$ and $T=t$, the chain rule gives $\Phi_t=\Theta_T+gt\Theta_X$ and $\Phi_x=\Theta_X$. The drift terms cancel, so
$$
i\hbar\Theta_T=-\frac{\hbar^2}{2m}\Theta_{XX}.
$$
Applying Ehrenfest's theorem with $[X,P]=i\hbar$ and $[P,P^2]=0$ gives
$$
\boxed{\frac d{dT}\langle X\rangle=\frac1m\langle P\rangle,
\qquad \frac d{dT}\langle P\rangle=0}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The phase has unit modulus and $X=x+gt^2/2$, hence
$$
\langle X\rangle_\Theta=\langle x\rangle_\psi+\frac12gt^2.
$$
The free-particle equations imply $\langle X\rangle=a+vt$. Therefore
$$
\boxed{\langle x\rangle_\psi=a+vt-\frac12gt^2}.
$$
The wavepacket centre obeys the classical constant-acceleration trajectory, an instance of the <correspondence principle>.

Solved by gpt-5.6-sol high.

= 16C
{parent=Paper 4}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=16C}

Irrotationality gives $\mathbf u_i=\nabla\phi_i$ and $\nabla^2\phi_i=0$. At $z=\zeta$, the exact kinematic conditions are
$$
\zeta_t+\phi_{ix}\zeta_x=\phi_{iz},
$$
and pressure continuity gives $p_1=p_2$. <Bernoulli equation> in each fluid is
$$
p_i=-\rho_i\left(\phi_{it}+\tfrac12|\nabla\phi_i|^2+gz\right)+C_i(t).
$$

Linearize at $z=0$ and write $\zeta=Ze^{i(kx-\omega t)}$. Decay at infinity gives
$$
\phi_1=A_1e^{-kz}e^{i(kx-\omega t)},
\qquad
\phi_2=A_2e^{kz}e^{i(kx-\omega t)}.
$$
The kinematic conditions give $-i\omega Z=-kA_1=kA_2$, while pressure continuity gives
$$
\rho_1(-i\omega A_1+gZ)=
ho_2(-i\omega A_2+gZ).
$$
Elimination yields
$$
\boxed{
\omega^2=gk\frac{\rho_2-\rho_1}{\rho_1+\rho_2},
\qquad F(r)=\frac{1-r}{1+r}}.
$$
Stable waves require the lower fluid to be denser, $\rho_2>\rho_1$; otherwise this is the <Rayleigh-Taylor instability>.

Solved by gpt-5.6-sol high.

= 17H
{parent=Paper 4}
{scope}
{title2=Statistics}

= a
{parent=17h}
{scope}

= Solution
{parent=a}

A statistic is sufficient when the conditional law of the sample given it is parameter-free; it is minimal sufficient when it is a function of every sufficient statistic. The likelihood factors as
$$
L=\bigl[\pi+(1-\pi)e^{-\lambda}\bigr]^{n_0}
(1-\pi)^{n-n_0}e^{-\lambda(n-n_0)}\lambda^S
\prod_i\frac1{X_i!},
$$
so the factorization theorem makes $(n_0,S)$ sufficient. The likelihood-ratio criterion shows that the ratio for two samples is parameter-independent exactly when both $n_0$ and $S$ agree. Thus
$$
\boxed{(n_0,S)\text{ is minimal sufficient}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=17h}
{scope}

= Solution
{parent=b}

For known $\lambda_0$, the likelihood ratio of $\pi=1/2$ to $\pi=0$ is
$$
\Lambda=2^{-n}(1+e^{\lambda_0})^{n_0},
$$
which increases with $n_0$. Reject for $n_0>c$, with randomization at $n_0=c$ chosen so that the null rejection probability is exactly $\alpha$. Since the zero indicators form a Bernoulli family with success probability increasing in $\pi$ and have a monotone likelihood ratio in $n_0$, the <Karlin-Rubin theorem> makes this test uniformly most powerful for $\pi>0$.

Solved by gpt-5.6-sol high.

= c
{parent=17h}
{scope}

= Solution
{parent=c}

The null value $\pi=1/2$ is interior and the model is regular. <Wilks' theorem> therefore gives
$$
\boxed{W\xrightarrow{d}\chi_1^2}
$$
under the null, because the alternative adds one free parameter. An asymptotic size-$\alpha$ test rejects when
$$
\boxed{W>\chi^2_{1,1-\alpha}}.
$$

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 4}
{scope}
{title2=Optimisation}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

A two-player zero-sum game has a payoff gained by one player and lost by the other. A pure strategy chooses one action deterministically; a mixed strategy is a probability distribution over pure actions.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

A pure optimum is a saddle point: an entry minimal in its row and maximal in its column. In $A_1$, $c$ at $(2,1)$ is a saddle; in $A_2$, $c$ at $(2,2)$ is a saddle. Matrix $A_3$ has row maximin $b$ and column minimax $c$, so no saddle exists. Thus
$$
\boxed{A_1,A_2\text{ admit pure optimal strategies; }A_3\text{ does not}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Every payoff is positive, so every pair of mixed strategies has positive expected payoff and the game value satisfies $v>0$. Player 1 maximizes $v$ subject to $A^Tp\geq ve$, $e^Tp=1$, $p\geq0$. Setting $x=p/v$ turns these into
$$
A^Tx\geq e,
\qquad x\geq0,
\qquad e^Tx=1/v.
$$
Maximizing $v$ is therefore equivalent to the stated minimization of $e^Tx$.

Solved by gpt-5.6-sol high.

= d
{parent=18h}
{scope}

= Solution
{parent=d}

The dual is
$$
\boxed{\text{maximize }e^Ty
\quad\text{subject to }Ay\leq e, y\geq0},
$$
which is in standard maximization form after adding nonnegative slack variables. The simplex pivots first introduce the variable with coefficient seven and then the remaining payoff column; the optimum occurs where both constraints bind:
$$
y_1+5y_2=1,
\qquad7y_1+3y_2=1.
$$
Thus $y=(1/16,3/16)$ and the dual value is $1/4$. Complementary slackness gives $x=(1/8,1/8)$. Hence $v=4$ and the normalized optimal strategies are
$$
\boxed{p=(1/2,1/2),
\qquad q=(1/4,3/4),
\qquad v=4}.
$$

Solved by gpt-5.6-sol high.