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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4.bigb
= Paper 4
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperib_4_2024.pdf
= 1G
{parent=Paper 4}
{scope}
{title2=Linear Algebra}
= Solution
{parent=1G}
The <Jordan normal form> theorem says that every complex square <matrix> is similar to a direct sum of Jordan blocks, uniquely up to their order. For an <eigenvalue> $\lambda$, if its blocks have sizes $n_1,\ldots,n_r$, then its <algebraic multiplicity> and <geometric multiplicity> are
$$
a_\lambda=\sum_{j=1}^r n_j,
\qquad
g_\lambda=r,
$$
and its contribution to the <minimal polynomial> is $(t-\lambda)^{\max_j n_j}$. Thus
$$
m_\alpha(t)=\prod_\lambda(t-\lambda)^{s_\lambda},
$$
where $s_\lambda$ is the largest $\lambda$-block size.
For the given <matrix>,
$$
\det(tI-A)=(t-2)(t+1)^2.
$$
Moreover,
$$
\dim\ker(A-2I)=1,
\qquad
\dim\ker(A+I)=1.
$$
Hence
$$
\boxed{a_2=g_2=1,
\qquad a_{-1}=2,
\qquad g_{-1}=1}.
$$
The <eigenvalue> $-1$ therefore has one Jordan block of size two, so
$$
\boxed{m_\alpha(t)=(t-2)(t+1)^2}.
$$
Solved by gpt-5.6-sol high.
= 2F
{parent=Paper 4}
{scope}
{title2=Analysis and Topology}
= Solution
{parent=2F}
A subset $A\subseteq X$ is connected if it cannot be written as $A=U\cup V$, where $U,V$ are disjoint, nonempty sets open in the subspace topology on $A$.
If $f(X)$ were disconnected as $U\cup V$, continuity would make $f^{-1}(U)$ and $f^{-1}(V)$ a disconnection of $X$. Thus the <continuous image of a connected space> is connected.
For $Y=\{0,1\}$ with the discrete topology, a nonconstant continuous $h:X\to Y$ gives the disconnection
$$
X=h^{-1}(\{0\})\cup h^{-1}(\{1\}).
$$
Conversely, a disconnection $X=U\cup V$ defines a continuous nonconstant <function> by assigning $0$ on $U$ and $1$ on $V$. Hence $X$ is connected exactly when every such $h$ is constant.
Finally suppose $C$ is connected but $\operatorname{Cl}(C)=U\cup V$ is a disconnection. The intersections $C\cap U$ and $C\cap V$ cannot both be nonempty, so, after interchanging $U,V$, we have $C\subseteq U$. For any $v\in V$, the relatively open neighbourhood $V$ of $v$ in $\operatorname{Cl}(C)$ is disjoint from $C$, contradicting $v\in\operatorname{Cl}(C)$. Therefore the <closure of a connected set> is connected.
Solved by gpt-5.6-sol high.
= 3F
{parent=Paper 4}
{scope}
{title2=Complex Analysis}
= Solution
{parent=3F}
A <function> $f:U\to\mathbb C$ is holomorphic when it is complex <differentiable> at every point of the open connected set $U$. <Morera's theorem> states that a <continuous function> on a domain is holomorphic if its <integral> around every triangle whose interior lies in the domain is zero.
The integrand $e^{tz}/(1+t^2)$ is continuous jointly in $(t,z)$ on $[0,1]\times\mathbb C$, so the displayed <integral> defines a <continuous function>. For any triangle $T$, Fubini's theorem and the Cauchy <integral> theorem give
$$
\int_{\partial T}f(z)\,dz
=\int_0^1\frac1{1+t^2}
\left(\int_{\partial T}e^{tz}\,dz\right)dt
=0.
$$
Morera's theorem therefore proves that $f$ is entire.
The <function> $1/z$ is holomorphic on $\mathbb C\setminus\{0\}$ but has no antiderivative there, because an antiderivative would integrate to zero around every closed curve whereas
$$
\int_{|z|=1}\frac{dz}{z}=2\pi i.
$$
This is the standard <period obstruction to a holomorphic antiderivative>.
Solved by gpt-5.6-sol high.
= 4A
{parent=Paper 4}
{scope}
{title2=Quantum Mechanics}
= Solution
{parent=4A}
The time-independent <Schrodinger equation> is
$$
-\frac{\hbar^2}{2m}\psi''(x)+V(x)\psi(x)=E\psi(x).
$$
Integrating it over $(a-\varepsilon,a+\varepsilon)$ and letting $\varepsilon\to0$ shows that $\psi'$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in $\psi'$ and hence a <derivative> of a delta in $\psi''$, so $\psi$ is also continuous.
For a <bound state> $-V_0<E<0$, define
$$
\eta^2=-\frac{2mE}{\hbar^2},
\qquad
k^2=\frac{2m(E+V_0)}{\hbar^2}.
$$
Then the proposed even <wavefunction> solves the equation away from the interfaces, and
$$
\boxed{k^2+\eta^2=\frac{2mV_0}{\hbar^2}}.
$$
Continuity at $x=a$ gives $B\cos(ka)=Ae^{-\eta a}$, while continuity of the <derivative> gives $-Bk\sin(ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relation
$$
\boxed{\eta=k\tan(ka)}.
$$
Thus the lowest even state of the <finite square well> is the first-quadrant intersection, with $0<ka<\pi/2$, of the circle $k^2+\eta^2=2mV_0/\hbar^2$ and the curve $\eta=k\tan(ka)$.
Solved by gpt-5.6-sol high.
= 5C
{parent=Paper 4}
{scope}
{title2=Electromagnetism}
= Solution
{parent=5C}
<Faraday's law> is
$$
\mathcal E=\oint_C\mathbf E\cdot d\mathbf l
=-\frac{d\Phi_B}{dt},
\qquad
\Phi_B=\int_S\mathbf B\cdot d\mathbf S.
$$
Choose the positive <circulation> to be anticlockwise as viewed from $+z$. Since $\Phi_B=\pi Br^2$,
$$
\boxed{I=\frac{\mathcal E}{R}
=-\frac{2\pi Br\dot r}{R}}.
$$
If $\dot r>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf F=I\,d\mathbf l\times\mathbf B$, the force due to the imposed field is radially inward. If $\dot r<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with <Lenz's law>.
Solved by gpt-5.6-sol high.
= 6D
{parent=Paper 4}
{scope}
{title2=Numerical Analysis}
= Solution
{parent=6D}
On each interval of width $h=1/N$, Taylor expansion about its midpoint shows that the local midpoint-rule error is $O(h^3)$. Summing over $N$ intervals gives the <composite midpoint rule error>
$$
\boxed{I(f)-I_N(f)=O(N^{-2})}.
$$
To remove the endpoint singularity, set $x=s^2$. Then
$$
I(f)=\int_0^1 2s f(s^2)\,ds
=\int_0^1 2g(s^2)\,ds,
$$
whose transformed integrand is analytic. Applying the composite midpoint rule in $s$ with $s_n=(n+\tfrac12)/N$ gives
$$
I(f)\approx\frac2N\sum_{n=0}^{N-1}s_nf(s_n^2).
$$
This has the required form when
$$
\boxed{y_n=s_n^2
=\left(\frac{n+\tfrac12}{N}\right)^2}.
$$
The <quadratic substitution for a square-root endpoint singularity> therefore restores second-order convergence.
Solved by gpt-5.6-sol high.
= 7H
{parent=Paper 4}
{scope}
{title2=Markov Chains}
= a
{parent=7h}
{scope}
= Solution
{parent=a}
With states ordered as $(A,B,C)$ and rows representing the current state, the <Markov chain> has transition <matrix>
$$
\boxed{
P=\begin{pmatrix}
0&1/2&1/2\\
3/4&0&1/4\\
3/4&1/4&0
\end{pmatrix}}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=7h}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Starting from the airport, after one step the distribution is $(0,1/2,1/2)$. Multiplying once more by $P$ gives
$$
\boxed{\mathbb P(X_2=A)=\frac34,
\qquad
\mathbb P(X_2=B)=\mathbb P(X_2=C)=\frac18}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Let $p_n=\mathbb P(X_n=A)$. Whenever the taxi is away from the airport it returns with probability $3/4$, whereas it must leave whenever it is at the airport. Hence
$$
p_{n+1}=\frac34(1-p_n),
\qquad p_0=1.
$$
Subtracting the fixed point $3/7$ gives
$$
p_{n+1}-\frac37=-\frac34
\left(p_n-\frac37\right).
$$
Therefore
$$
\boxed{
p_n=\frac37+\frac47\left(-\frac34\right)^n
}
$$
for every $n\geq0$, and in particular for the requested $n\geq1$.
Solved by gpt-5.6-sol high.
= 8G
{parent=Paper 4}
{scope}
{title2=Linear Algebra}
= a
{parent=8g}
{scope}
= Solution
{parent=a}
Two <matrices> are equivalent when $A'=PAQ$ for invertible <matrices> $P$ and $Q$ of the appropriate sizes. A change of <basis> in the codomain multiplies a <matrix> on the left, while a change of <basis> in the domain multiplies it on the right. Hence two <matrices> of the same <linear map> in two pairs of <bases> are equivalent.
Conversely, start with the map $\alpha:F^n\to F^m$ having <matrix> $A$ in the standard <bases>. Given $A'=PAQ$, choose domain and codomain <bases> whose change-of-coordinate <matrices> produce $Q$ and $P$; then $A'$ represents the same map in those <bases>. This proves the equivalence.
The column rank of $A$ is the dimension of the span of its columns, and its row rank is the dimension of the span of its rows. If $A$ represents $\alpha:F^n\to F^m$, its column rank is $\operatorname{rank}\alpha$. The transpose represents the dual map
$$
\alpha^*:(F^m)^*\to(F^n)^*,
\qquad \phi\mapsto\phi\circ\alpha,
$$
so the row rank is $\operatorname{rank}\alpha^*$.
If $\operatorname{rank}\alpha=r$, then
$$
\ker\alpha^*=(\operatorname{im}\alpha)^0,
$$
whose dimension is $m-r$: extend a <basis> of $\operatorname{im}\alpha$ to one of $F^m$ and use the dual <basis>. Rank-nullity now gives $\operatorname{rank}\alpha^*=r$. This proves the <equality of row rank and column rank>.
Solved by gpt-5.6-sol high.
= b
{parent=8g}
{scope}
= Solution
{parent=b}
For $I\subseteq[m]$, the independent <vectors> $\{v_i:i\in I\}$ lie in the coordinate subspace supported on
$$
S_I=\bigcup_{i\in I}\operatorname{supp}(v_i),
$$
which has dimension $|S_I|$. Hence $|I|\leq|S_I|$, and the stated form of <Hall marriage theorem> gives an injection $f$ with $f(i)\in\operatorname{supp}(v_i)$.
Let $V$ be the $n\times m$ <matrix> whose $i$th column is $v_i$. Its column rank is $m$. By equality of row and column rank, it has $m$ linearly independent rows. Let $B$ be their indices. The $m\times m$ submatrix $V_B$ is invertible, so its columns are independent. Those columns are exactly the nonzero coordinates of $Bv_1,\ldots,Bv_m$, and therefore these truncated <vectors> are linearly independent.
Expanding $\det V_B$ gives a permutation $f:[m]\to B$ for which
$$
\prod_{i=1}^m (v_i)_{f(i)}\ne0.
$$
Thus $f(i)\in\operatorname{supp}(v_i)$. Finally, order the coordinates with $B$ first. The <matrix> whose columns are the $v_i$ together with the $e_j$ for $j\notin B$ is block triangular, with diagonal blocks $V_B$ and an identity <matrix>. Its <determinant> is nonzero. Consequently
$$
\boxed{
\bigl(\{e_j:j\in[n]\}\setminus\{e_{f(i)}:i\in[m]\}\bigr)
\cup\{v_i:i\in[m]\}
}
$$
is a <basis> of $\mathbb C^n$. This is the <simultaneous basis exchange from a nonzero minor>.
Solved by gpt-5.6-sol high.
= 9E
{parent=Paper 4}
{scope}
{title2=Groups, Rings and Modules}
= a
{parent=9e}
{scope}
= Solution
{parent=a}
An $R$-module is free if it has a <basis>: a subset $B$ such that every element has a unique expression as a finite $R$-linear combination of elements of $B$.
Choose a maximal <ideal> $\mathfrak m$ of the nonzero <ring> $R$. If $R^n\cong R^m$, quotienting by $\mathfrak m$ gives
$$
(R/\mathfrak m)^n\cong(R/\mathfrak m)^m.
$$
These are <vector spaces> over the field $R/\mathfrak m$, so equality of dimension gives $n=m$. This proves <invariant basis number for a commutative ring>.
A direct summand of a free module need not be free. Take $R=\mathbb Z/6\mathbb Z$. Its <ideals> $P=(2)$ and $Q=(3)$ satisfy
$$
P\cap Q=0,
\qquad P+Q=R,
$$
so $P\oplus Q\cong R$. But $P$ has three elements, whereas a finite-rank free $R$-module has $6^r$ elements; hence $P$ is not free.
Solved by gpt-5.6-sol high.
= b
{parent=9e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Let $P$ be free with <basis> $(e_i)_{i\in I}$. Given a surjection $f:M\to N$ and a map $g:P\to N$, choose $m_i\in M$ satisfying $f(m_i)=g(e_i)$. There is a unique <linear map> $h:P\to M$ with $h(e_i)=m_i$, and on every <basis> <vector>
$$
(f\circ h)(e_i)=g(e_i).
$$
Thus $f\circ h=g$, proving that <free modules are projective>.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
We prove by induction on $n$ that every submodule $N\subseteq R^n$ is free. The case $n=0$ is immediate. Let $\pi:R^n\to R$ be projection onto the first coordinate. Since $R$ is a principal <ideal> domain,
$$
\pi(N)=dR
$$
for some $d$. If $d=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise choose $x\in N$ with $\pi(x)=d$. For every $y\in N$, write $\pi(y)=rd$; then $y-rx\in\ker(\pi|_N)$. Also $Rx\cap\ker(\pi|_N)=0$ because $R$ is a domain and $d\ne0$. Therefore
$$
N=Rx\oplus\ker(\pi|_N).
$$
The kernel is a submodule of $R^{n-1}$ and is free by induction, while $Rx\cong R$. Hence $N$ is free. This is the <submodule theorem for free modules over a principal ideal domain>.
If $P$ is finitely generated and projective, choose a surjection $f:R^n\to P$. Projectivity supplies $h:P\to R^n$ with $f\circ h=\operatorname{id}_P$, so
$$
R^n=\ker f\oplus h(P).
$$
Thus $P\cong h(P)$ is a submodule of the finitely generated free module $R^n$, and the result just proved shows that $P$ is free.
Solved by gpt-5.6-sol high.
= 10F
{parent=Paper 4}
{scope}
{title2=Analysis and Topology}
= Solution
{parent=10F}
Differentiability at $x\in\mathbb R^2$ means that there is a <linear map> $Df|_x:\mathbb R^2\to\mathbb R$ such that
$$
f(x+h)=f(x)+Df|_x(h)+o(\|h\|).
$$
The <partial derivatives> are $D_if(x)=Df|_x(e_i)$ when the <derivative> exists; independently, they may be defined by the corresponding one-variable difference quotients.
Suppose the <partial derivatives> exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points $\xi_h,\eta_h\to x$ such that
$$
f(x+h)-f(x)
=D_1f(\xi_h)h_1+D_2f(\eta_h)h_2.
$$
Continuity of the partials makes the remainder after subtracting
$$
D_1f(x)h_1+D_2f(x)h_2
$$
equal to $o(\|h\|)$. Hence $f$ is <differentiable> and this is its <derivative>.
For the given <function>, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we have
$$
\boxed{
D_1f=2x\sin(1/r)-\frac{x}{r}\cos(1/r),
\qquad
D_2f=2y\sin(1/r)-\frac{y}{r}\cos(1/r)}.
$$
At the origin both <partial derivatives> are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a <limit>. Nevertheless
$$
|f(x,y)|\leq r^2=o(r),
$$
so $f$ is <differentiable> at the origin with <derivative> zero. This illustrates that <existence of partial derivatives does not imply their continuity>.
The final assertion is false. Define
$$
g(x,y)=
\begin{cases}
(x^2+y^2)\sin\!\left(\dfrac1{x^2+y^2}\right),&(x,y)\ne(0,0),\\
0,&(x,y)=(0,0).
\end{cases}
$$
Again $|g|\leq r^2$, so $g$ is <differentiable> at the origin and it is smooth elsewhere. But along the $x$-axis,
$$
D_1g(x,0)=2x\sin(1/x^2)-\frac2x\cos(1/x^2),
$$
is unbounded near zero, and similarly $D_2g(0,y)$ is unbounded. Thus both <partial derivatives> can be unbounded in every neighbourhood of a point even when the <function> is <differentiable> everywhere.
Solved by gpt-5.6-sol high.
= 11G
{parent=Paper 4}
{scope}
{title2=Geometry}
= a
{parent=11g}
{scope}
= Solution
{parent=a}
The surface is obtained by revolving the graph $z=f(x)$ about the $x$-axis. It has a central cylindrical section of radius $b$, two smooth transition collars, and unit spherical caps centred at $(\pm3,0,0)$; its projection onto the $(x,z)$-plane is the region $|z|\leq f(x)$.
For a surface of revolution, the <Gaussian curvature of a surface of revolution> is
$$
K=-\frac{f''}{f(1+f'^2)^2}.
$$
Thus the cylindrical region has $K=0$. On each transition collar, $K<0$ where $f''>0$, vanishes at the unique inflection circle, and is positive where $f''<0$. The spherical caps have $K>0$.
Solved by gpt-5.6-sol high.
= b
{parent=11g}
{scope}
= Solution
{parent=b}
The area element is
$$
dA=f\sqrt{1+f'^2}\,dx\,d\theta.
$$
Consequently
$$
K\,dA
=-\frac{f''}{(1+f'^2)^{3/2}}\,dx\,d\theta
=-d\!\left(\frac{f'}{\sqrt{1+f'^2}}\right)d\theta.
$$
At $x=2-a$, smooth matching to the cylinder gives $f'=0$. At $x=2+a$, the spherical formula gives
$$
f'=\frac{1-a}{\sqrt{2a-a^2}},
\qquad
\frac{f'}{\sqrt{1+f'^2}}=1-a.
$$
The <total Gaussian curvature of a surface-of-revolution strip> is therefore
$$
\boxed{\int_RK\,dA=-2\pi(1-a)}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=11g}
{scope}
= Solution
{parent=c}
The two curves produced by $y=0$ are meridians, and meridians of a surface of revolution are geodesics. A boundary circle $x=x_0$ is a geodesic precisely when $f'(x_0)=0$, as follows either from the geodesic equations or from its geodesic curvature
$$
k_g=\frac{|f'(x_0)|}{f(x_0)\sqrt{1+f'(x_0)^2}}.
$$
The circle at $x=2-a$ is geodesic because it joins the cylinder smoothly. At the other boundary,
$$
f'(2+a)=\frac{1-a}{\sqrt{2a-a^2}},
$$
which vanishes exactly when $a=1$. Hence the cut pieces are geodesic polygons only for
$$
\boxed{a=1}.
$$
Solved by gpt-5.6-sol high.
= 12B
{parent=Paper 4}
{scope}
{title2=Complex Methods}
= i
{parent=12b}
{scope}
= Solution
{parent=i}
For $\operatorname{Re}s>0$,
$$
\boxed{
\mathcal L\{H(t-t_0)\}(s)
=\int_{t_0}^{\infty}e^{-st}\,dt
=\frac{e^{-st_0}}s}.
$$
This is the <Laplace transform time-shift rule> in its simplest form.
Solved by gpt-5.6-sol high.
= ii
{parent=12b}
{scope}
= Solution
{parent=ii}
Fourier transformation gives
$$
(k^2+m^2)\widehat G(k)=1,
\qquad
\widehat G(k)=\frac1{k^2+m^2}.
$$
For $x>0$, close the inverse-transform contour in the upper half-plane and take the residue at $k=im$; for $x<0$, close it in the lower half-plane. This gives the <one-dimensional modified Helmholtz Green function>
$$
\boxed{G(x)=\frac{e^{-m|x|}}{2m}}.
$$
The same formula works for complex $m$ with $\operatorname{Re}m>0$: it decays at both ends, is continuous at zero, and its <derivative> has the jump $G'(0+)-G'(0-)=-1$ required by the delta source.
Convolution therefore yields
$$
\boxed{
u(x)=\frac1{2m}\int_{-\infty}^{\infty}
e^{-m|x-y|}f(y)\,dy}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=12b}
{scope}
= Solution
{parent=iii}
Let $U(s,x)=\mathcal L_t\{u(t,x)\}$. The initial data turn the wave equation into
$$
-\frac{\partial^2U}{\partial x^2}+s^2U=f(x).
$$
Using the Green <function> from part (ii) with $m=s$ gives
$$
U(s,x)=\frac1{2s}\int_{-\infty}^{\infty}
e^{-s|x-y|}f(y)\,dy.
$$
Since
$$
\mathcal L^{-1}\!\left\{\frac{e^{-as}}s\right\}=H(t-a),
$$
we obtain the <D'Alembert formula with initial velocity>
$$
\boxed{
u(t,x)=\frac12\int_{-\infty}^{\infty}
H(t-|x-y|)f(y)\,dy
=\frac12\int_{x-t}^{x+t}f(y)\,dy}.
$$
Solved by gpt-5.6-sol high.
= 13C
{parent=Paper 4}
{scope}
{title2=Variational Principles}
= Solution
{parent=13C}
Write
$$
\mathbf u=\nabla\phi+\beta\nabla\alpha,
\qquad
\mathcal L=-\beta\alpha_t-\frac12\mathbf u\cdot\mathbf u.
$$
Variation of $\phi$ gives $\nabla\cdot\mathbf u=0$. Variation of $\alpha$ gives
$$
\beta_t+\nabla\cdot(\beta\mathbf u)=0,
$$
and variation of $\beta$ gives
$$
\alpha_t+\mathbf u\cdot\nabla\alpha=0.
$$
Using incompressibility, the middle equation becomes
$$
\beta_t+\mathbf u\cdot\nabla\beta=0.
$$
These are the three required Euler-Lagrange equations.
Let $D_t=\partial_t+\mathbf u\cdot\nabla$. Since $D_t\alpha=D_t\beta=0$, differentiating $u_i=\partial_i\phi+\beta\partial_i\alpha$ gives
$$
D_tu_i
=\partial_i(D_t\phi)-u_j\partial_i u_j
=\partial_i\left(D_t\phi-\frac12\mathbf u^2\right).
$$
Now
$$
D_t\phi
=\phi_t+\mathbf u\cdot\nabla\phi
=\phi_t+\mathbf u^2+\beta\alpha_t,
$$
because $\mathbf u\cdot\nabla\alpha=-\alpha_t$. Hence
$$
D_tu_i
=\partial_i\left(\phi_t+\beta\alpha_t+\frac12\mathbf u^2\right)
=-\partial_i p,
$$
where
$$
\boxed{p=-\frac12\mathbf u^2-\phi_t-\beta\alpha_t},
\qquad
\boxed{f(\phi_t,\alpha_t,\beta)=-\phi_t-\beta\alpha_t}.
$$
This is the <Clebsch-potential variational derivation of incompressible Euler flow>.
Solved by gpt-5.6-sol high.
= 14B
{parent=Paper 4}
{scope}
{title2=Methods}
= i
{parent=14b}
{scope}
= Solution
{parent=i}
For $a=0$ and $\kappa>0$, the <heat kernel> gives
$$
\boxed{u(t,x)=\int_{-\infty}^{\infty}K_t(x-y)u_0(y)\,dy}.
$$
As $t\downarrow0$, $K_t$ is a Gaussian of total mass one whose width is of order $\sqrt{\kappa t}$; it becomes a narrow spike at zero and converges to the delta distribution. Thus the convolution tends to $u_0(x)$.
Solved by gpt-5.6-sol high.
= ii
{parent=14b}
{scope}
= Solution
{parent=ii}
When $\kappa=0$, the characteristics satisfy $x-at=\text{constant}$ and $u$ is constant along them. Therefore
$$
\boxed{u(t,x)=u_0(x-at)}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=14b}
{scope}
= Solution
{parent=iii}
Set $v(t,z)=u(t,z+at)$. Then $v_t=\kappa v_{zz}$ and $v(0,z)=u_0(z)$. Applying the heat kernel and returning to $x$ gives the <advection-diffusion heat-kernel solution>
$$
\boxed{
u(t,x)=\int_{-\infty}^{\infty}
K_t(x-at-y)u_0(y)\,dy}.
$$
Solved by gpt-5.6-sol high.
= 15A
{parent=Paper 4}
{scope}
{title2=Quantum Mechanics}
= i
{parent=15a}
{scope}
= Solution
{parent=i}
For $\psi=Ae^{-Bx^2}$,
$$
\psi_t=\left(\frac{A'}A-B'x^2\right)\psi,
\qquad
\psi_{xx}=(-2B+4B^2x^2)\psi.
$$
Substitution into the <Schrodinger equation> and comparison of the constant and $x^2$ coefficients gives
$$
\boxed{A'=-i\hbar AB,
\qquad
B'=-\frac{i}{2\hbar}-2i\hbar B^2}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=15a}
{scope}
= Solution
{parent=ii}
Put $B=\xi\tan(\phi+\alpha t)$. Matching the constant and $\tan^2$ coefficients in the <Riccati equation> requires
$$
\xi\alpha=-\frac{i}{2\hbar},
\qquad
\xi\alpha=-2i\hbar\xi^2.
$$
One convenient choice is therefore
$$
\boxed{\xi=\frac1{2\hbar},
\qquad \alpha=-i},
$$
so
$$
\boxed{B(t)=\frac1{2\hbar}\tan(\phi-it)}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=15a}
{scope}
= Solution
{parent=iii}
Since $A'/A=-i\hbar B=-\tfrac i2\tan(\phi-it)$ and
$$
\frac d{dt}\log\cos(\phi-it)=i\tan(\phi-it),
$$
integration gives
$$
\boxed{A(t)=A_0[\cos(\phi-it)]^{-1/2}},
$$
with the branch chosen continuously from the initial value. The constant $A_0$ fixes normalization.
Solved by gpt-5.6-sol high.
= iv
{parent=15a}
{scope}
= Solution
{parent=iv}
Normalizability requires $\operatorname{Re}B>0$. Since
$$
|\psi|^2=|A|^2e^{-(B+B^*)x^2},
$$
the given Gaussian <integral> gives
$$
\boxed{\langle \hat x^2\rangle
=\frac1{2(B+B^*)}
=\frac1{4\operatorname{Re}B}}.
$$
Also $\psi_x=-2Bx\psi$, so integration by parts yields
$$
\langle\hat p^2\rangle
=\hbar^2\int|\psi_x|^2dx
=4\hbar^2|B|^2\langle x^2\rangle.
$$
Therefore
$$
\boxed{\langle\hat p^2\rangle
=\frac{\hbar^2|B|^2}{\operatorname{Re}B}}.
$$
These are the <second moments of a complex Gaussian wave packet>.
Solved by gpt-5.6-sol high.
= 16D
{parent=Paper 4}
{scope}
{title2=Fluid Dynamics}
= Solution
{parent=16D}
Taking the vertical component of the curl of the <momentum> equation gives
$$
\omega_t+f\nabla\cdot\mathbf u=0.
$$
The continuity equation gives $\eta_t+h_0\nabla\cdot\mathbf u=0$, and hence
$$
\boxed{\frac\partial{\partial t}
\left(\omega-\frac f{h_0}\eta\right)=0}.
$$
Since initially $\mathbf u=0$ and $\eta=\eta_0$,
$$
\omega=\frac f{h_0}(\eta-\eta_0).
$$
Taking the divergence of the <momentum> equation and using
$$
\nabla\cdot(\mathbf f\times\mathbf u)=-f\omega
$$
gives
$$
(\nabla\cdot\mathbf u)_t-f\omega=-g\nabla^2\eta.
$$
Differentiate continuity in time and substitute the last two identities to obtain
$$
\boxed{\eta_{tt}-gh_0\nabla^2\eta+f^2\eta=f^2\eta_0}.
$$
Let the <Rossby deformation radius> be
$$
L_R=\frac{\sqrt{gh_0}}{|f|}.
$$
The even, decaying steady solution of
$$
-L_R^2\eta_\infty''+\eta_\infty=\eta_0
$$
with continuous value and <derivative> at $x=\pm a$ is
$$
\boxed{
\eta_\infty(x)=
\begin{cases}
\epsilon\left[1-e^{-a/L_R}\cosh(x/L_R)\right],&|x|<a,\\
\epsilon\sinh(a/L_R)e^{-|x|/L_R},&|x|>a.
\end{cases}}
$$
The steady <momentum> balance is a <geostrophic balance>:
$$
\boxed{u=0,
\qquad v=\frac g f\frac{d\eta_\infty}{dx}}.
$$
Thus the flow is parallel to the two edges of the raised strip, in opposite $y$-directions on the two sides; for $f>0$, it points toward $+y$ on the left and $-y$ on the right. Its magnitude is concentrated within a few deformation radii of the edges.
Solved by gpt-5.6-sol high.
= 17H
{parent=Paper 4}
{scope}
{title2=Statistics}
= a
{parent=17h}
{scope}
= Solution
{parent=a}
Let
$$
S_{xx}=\sum_{i=1}^n(x_i-\bar x)^2,
\qquad
S_{xy}=\sum_{i=1}^n(x_i-\bar x)(Y_i-\bar Y).
$$
Differentiating the Gaussian log-likelihood, equivalently minimizing the residual sum of squares, gives the <ordinary least squares estimators>
$$
\boxed{\hat\beta=\frac{S_{xy}}{S_{xx}},
\qquad
\hat\alpha=\bar Y-\hat\beta\bar x},
$$
assuming $S_{xx}>0$.
Solved by gpt-5.6-sol high.
= b
{parent=17h}
{scope}
= Solution
{parent=b}
The centered model has residual sum of squares
$$
\sum_i\{Y_i-\alpha'-\beta'(x_i-\bar x)\}^2.
$$
Its slope normal equation gives
$$
\hat\beta'
=\frac{\sum_i(x_i-\bar x)(Y_i-\bar Y)}
{\sum_i(x_i-\bar x)^2}
=\boxed{\hat\beta}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=17h}
{scope}
= Solution
{parent=c}
The intercept normal equation is
$$
\sum_i\{Y_i-\hat\alpha'-\hat\beta'(x_i-\bar x)\}=0.
$$
Because the centered predictors sum to zero,
$$
\boxed{\hat\alpha'=\bar Y}.
$$
Meanwhile $\hat\alpha=\bar Y-\hat\beta\bar x$, so the two intercept estimates are unequal in general. The parameters themselves satisfy $\alpha'=\alpha+\beta\bar x$.
Solved by gpt-5.6-sol high.
= d
{parent=17h}
{scope}
= Solution
{parent=d}
Averaging the centered model gives
$$
\hat\alpha'=\bar Y
=\alpha'+\bar\epsilon.
$$
Therefore
$$
\boxed{\hat\alpha'\sim N\left(\alpha',\frac{\sigma^2}{n}\right)}.
$$
Writing $z_{0.975}=\Phi^{-1}(0.975)$, an exact 95% confidence interval is
$$
\boxed{\bar Y\pm z_{0.975}\frac\sigma{\sqrt n}}.
$$
Solved by gpt-5.6-sol high.
= e
{parent=17h}
{scope}
= Solution
{parent=e}
Use the residual variance estimator
$$
s^2=\frac1{n-2}\sum_{i=1}^n
\{Y_i-\hat\alpha'-\hat\beta'(x_i-\bar x)\}^2.
$$
Under the Gaussian linear model, $\bar Y$ is independent of $s^2$ and
$$
\frac{(n-2)s^2}{\sigma^2}\sim\chi^2_{n-2}.
$$
Consequently the <Student t confidence interval for a centered regression intercept> follows from
$$
\frac{\bar Y-\alpha'}{s/\sqrt n}\sim t_{n-2}.
$$
It is
$$
\boxed{\bar Y\pm t_{n-2,0.975}\frac{s}{\sqrt n}}.
$$
The displayed pivotal quantity lies between its 2.5% and 97.5% quantiles with probability $0.95$, which proves the stated coverage.
Solved by gpt-5.6-sol high.
= 18H
{parent=Paper 4}
{scope}
{title2=Optimisation}
= a
{parent=18h}
{scope}
= Solution
{parent=a}
Newton's method for minimization uses
$$
\boxed{x_{k+1}=x_k-[\nabla^2f(x_k)]^{-1}\nabla f(x_k)}.
$$
For a quantitative local bound, suppose on a convex neighbourhood containing the iterates that
$$
mI\preceq\nabla^2f(x)\preceq LI,
\qquad
\|\nabla^2f(x)-\nabla^2f(y)\|\leq M\|x-y\|,
$$
with $m>0$, and let $x^*$ be the minimizer. The <integral> form of the <gradient> and the Hessian Lipschitz bound give the <quadratic convergence bound for Newton's method>
$$
\|x_{k+1}-x^*\|
\leq\frac M{2m}\|x_k-x^*\|^2.
$$
If $q=M\|x_0-x^*\|/(2m)<1$, induction yields
$$
\|x_k-x^*\|\leq\frac{2m}{M}q^{2^k}.
$$
Since $f(x)-f(x^*)\leq L\|x-x^*\|^2/2$,
$$
\boxed{f(x_k)-f(x^*)
\leq\frac{2Lm^2}{M^2}q^{2^{k+1}}}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=18h}
{scope}
= Solution
{parent=b}
If $x_k\geq1$, then positivity and the <arithmetic-geometric mean inequality> give
$$
x_{k+1}=\frac12\left(x_k+\frac a{x_k}\right)
\geq\sqrt a\geq1.
$$
Thus all iterates remain in $[1,\infty)$.
Consider the strictly convex <function>
$$
f(x)=\frac{x^3}{3}-ax
$$
on $[1,\infty)$. Since $f'(x)=x^2-a$ and $f''(x)=2x$, its Newton minimization step is
$$
x-\frac{f'(x)}{f''(x)}
=\frac12\left(x+\frac ax\right).
$$
Its unique minimizer is $x^*=\sqrt a$.
Solved by gpt-5.6-sol high.
= c
{parent=18h}
{scope}
= Solution
{parent=c}
Every $x_0\in[1,\infty)$ works. Indeed, $x_1\geq\sqrt a$, and whenever $x_k\geq\sqrt a$,
$$
0\leq x_{k+1}-\sqrt a
=\frac{(x_k-\sqrt a)^2}{2x_k}
\leq x_k-\sqrt a.
$$
The iterates from $k=1$ onward therefore decrease to a <limit>, and the recurrence forces that <limit> to be $\sqrt a$.
There is also an explicit error bound. Put
$$
q=\left|\frac{x_0-\sqrt a}{x_0+\sqrt a}\right|<1.
$$
A direct calculation gives
$$
\frac{x_{k+1}-\sqrt a}{x_{k+1}+\sqrt a}
=\left(\frac{x_k-\sqrt a}{x_k+\sqrt a}\right)^2.
$$
Hence, for $k\geq1$,
$$
\boxed{
|x_k-\sqrt a|
=\frac{2\sqrt a\,q^{2^k}}{1-q^{2^k}}
\leq\frac{2\sqrt a}{1-q}q^{2^k}}.
$$
This double-exponential decay is consistent with the Newton bound in part (a). For the chosen objective, the supplied factorization also gives
$$
f(x)-f(\sqrt a)
=\frac13(x-\sqrt a)^2(x+2\sqrt a),
$$
so convergence of the objective and convergence of the iterates are equivalent on $[1,\infty)$.
Solved by gpt-5.6-sol high.