past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperii_3_2024.pdf

= 1F
{parent=Paper 3}
{scope}
{title2=Number Theory}

= Solution
{parent=1F}

Write $N=\prod_i p_i^{e_i}$ with odd primes $p_i$. The <Jacobi symbol> is
$$
\left(\frac aN\right)=\prod_i\left(\frac a{p_i}\right)^{e_i},
$$
where each factor on the right is a Legendre symbol. The supplementary law is
$$
\left(\frac2N\right)=(-1)^{(N^2-1)/8}.
$$
For coprime positive odd $m,n$, <quadratic reciprocity> gives
$$
\left(\frac mn\right)\left(\frac nm\right)
=(-1)^{(m-1)(n-1)/4}.
$$
Applying this with $m=3$ gives
$$
\left(\frac3N\right)=
\begin{cases}
 1,&N\equiv1,11\pmod {12},\\
-1,&N\equiv5,7\pmod {12},\\
 0,&3\mid N.
\end{cases}
$$
Finally, factor $d=2^e m$ with $m$ odd. The supplementary laws for $-1$ and $2$, followed by reciprocity on every odd prime factor of $m$, express $(-d/a)$ solely in terms of the residue class of the positive odd integer $a$ modulo $d$ when $d\equiv0$ or $3\pmod4$. If $a\equiv b\pmod d$, all those factors, including the zero cases, agree; hence
$$
\left(\frac{-d}{a}\right)=\left(\frac{-d}{b}\right).
$$

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 3}
{scope}
{title2=Topics in Analysis}

= Solution
{parent=2G}

Fix $z_0\notin\Gamma$ and put $\delta=\operatorname{dist}(z_0,\Gamma)>0$. If $|z-z_0|<\delta/2$, then
$$
\left|\frac1{w-z}-\frac1{w-z_0}\right|
\leq \frac{2|z-z_0|}{\delta^2}
$$
uniformly for $w\in\Gamma$. Multiplication by the bounded <continuous function> $f$ and integration therefore prove continuity of $F$ at $z_0$.

Take $\Gamma$ to be the positively oriented unit circle and $f\equiv1$. By the <Cauchy integral formula>, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.

For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. <Integrals> over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours $\Gamma_j\subset\Omega\setminus K$, and their sum is
$$
f(z)=\sum_j\frac1{2\pi i}\oint_{\Gamma_j}\frac{f(w)}{w-z}\,dw
$$
for every $z\in K$.

Solved by gpt-5.6-sol high.

= 3K
{parent=Paper 3}
{scope}
{title2=Coding and Cryptography}

= a
{parent=3k}
{scope}

= Solution
{parent=a}

The <Reed-Muller code> is spanned by the evaluation <vectors> of the square-free monomials
$$
\prod_{i\in I}x_i,\qquad |I|\leq r,
$$
form a <basis>: every Boolean <function> has a unique algebraic normal form, so the monomials remain independent after evaluation on $\mathbb F_2^d$. Thus
$$
\dim\operatorname{RM}(d,r)=\sum_{j=0}^r\binom dj.
$$
Induction using the decomposition $(u,u+v)$, with $u\in\operatorname{RM}(d-1,r)$ and $v\in\operatorname{RM}(d-1,r-1)$, gives the lower bound $2^{d-r}$ for every nonzero weight. The monomial $x_1\cdots x_r$ attains it, so the minimum distance is
$$
d_{\min}=2^{d-r}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=3k}
{scope}

= Solution
{parent=b}

The Mariner code is $\operatorname{RM}(5,1)$ with parameters $[32,6,16]$. Its information rate is $6/32=3/16$, and it corrects
$$
\left\lfloor\frac{16-1}{2}\right\rfloor=7
$$
errors, a proportion $7/32$ of a codeword. The binary <Hamming code> of length $31$ has parameters $[31,26,3]$: its rate $26/31$ is much higher, but it corrects only one error, a proportion $1/31$.

Solved by gpt-5.6-sol high.

= c
{parent=3k}
{scope}

= Solution
{parent=c}

A word of $\operatorname{RM}(d,1)$ is the evaluation of an affine Boolean <function>
$$
f(x)=a_0+a\cdot x.
$$
If $a=0$, the two constants have weights $0$ and $2^d$. If $a\ne0$, choose $v$ with $a\cdot v=1$. Translation $x\mapsto x+v$ pairs every input where $f=0$ with one where $f=1$. Hence every one of the other codewords has weight $2^{d-1}$.

Solved by gpt-5.6-sol high.

= 4J
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}

= a
{parent=4j}
{scope}

= Solution
{parent=a}

Let $\Sigma=\{a,b\}$, $V=\{S\}$, and
$$
P=\{S\to aSa, S\to b\}.
$$
Then $\mathcal L(G)=\{a^nba^n:n\geq0\}$. Adding $S\to\varepsilon$ also produces $a^{2n}$; for example $aa$ is generated but lies in neither $\mathcal L(G)$ nor $\{\varepsilon\}$. Thus the proposed equality can fail because the new erasing rule may be used on a noninitial occurrence of $S$.

Solved by gpt-5.6-sol high.

= b
{parent=4j}
{scope}

= Solution
{parent=b}

After the first rule $S\to S^*$, every sentential form is obtained from a sentential form of $G$ by uniformly renaming $S$ as $S^*$. Each rule of $P$ has exactly the corresponding renamed rule in $P^*$, and conversely. Terminal words contain neither symbol, so this bijection of derivations proves
$$
\mathcal L(G^+)=\mathcal L(G).
$$

Solved by gpt-5.6-sol high.

= c
{parent=4j}
{scope}

= Solution
{parent=c}

In $H^+$ the unstarred $S$ occurs only as the initial symbol and has precisely the alternatives
$$
S\to\varepsilon,\qquad S\to S^*.
$$
The first produces only $\varepsilon$. After the second, the renamed-derivation argument from part (b) produces exactly $\mathcal L(G)$. Therefore
$$
\mathcal L(H^+)=\mathcal L(G)\cup\{\varepsilon\}.
$$

Solved by gpt-5.6-sol high.

= 5L
{parent=Paper 3}
{scope}
{title2=Statistical Modelling}

= Solution
{parent=5L}

In a linear model containing an intercept, the normal equations make the residual <vector> $e$ orthogonal to every design column. In particular $e\perp\mathbf1$, so
$$
\sum_{i=1}^{50}e_i=\mathbf1^Te=0
$$
up to floating-point rounding.

The second design <matrix> has $49$ random Gaussian columns plus the intercept, hence is a random $50\times50$ <matrix>. It has full rank with probability one because the <determinant> vanishes only on a measure-zero algebraic set. Its column space is then all of $\mathbb R^{50}$, so the fitted <vector> equals $Y$, the residual sum of squares is zero, and $R^2=1$.

Solved by gpt-5.6-sol high.

= 6A
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}

= Solution
{parent=6A}

The transition rates of this <birth-death process> are $q_{n,n+1}=\lambda n$ and $q_{n,n-1}=n(\gamma+\beta n)$. Thus, with $p_{-1}=0$,
$$
\dot p_n=\lambda(n-1)p_{n-1}+(n+1)(\gamma+\beta(n+1))p_{n+1}
-[\lambda n+n(\gamma+\beta n)]p_n.
$$
Multiplying by $n$, summing, and shifting indices gives
$$
\frac d{dt}\mathbb E N=(\lambda-\gamma)\mathbb EN-\beta\mathbb E(N^2).
$$
At a finite-moment steady state with positive mean and $\lambda>\gamma$,
$$
\beta\mathbb E(N^2)=(\lambda-\gamma)\mathbb EN.
$$
Since $\mathbb E(N^2)\geq(\mathbb EN)^2$, this implies
$$
0<\mathbb EN\leq\frac{\lambda-\gamma}{\beta}.
$$
If $\lambda\leq\gamma$, both terms in the mean equation are nonpositive, and stationarity forces $\mathbb EN=0$. Thus the population is extinct almost surely in any such steady state.

Solved by gpt-5.6-sol high.

= 7D
{parent=Paper 3}
{scope}
{title2=Further Complex Methods}

= a
{parent=7d}
{scope}

= Solution
{parent=a}

For any $n$ with $\operatorname{Re}z>-n$, define
$$
F(z)=\frac{F(z+n)}{z(z+1)\cdots(z+n-1)}.
$$
The definitions agree on overlaps by the recurrence, giving a meromorphic continuation to $\mathbb C$. Its only singularities are simple poles at $0,-1,-2,\ldots$. Since $F(1)=1$, repeated use of the recurrence gives
$$
\operatorname*{res}_{z=-n}F(z)=\frac{(-1)^n}{n!}.
$$
These residues are nonzero, so all the listed singularities really are simple poles.

Solved by gpt-5.6-sol high.

= b
{parent=7d}
{scope}

= Solution
{parent=b}

The meromorphic continuations of $F$ and the <gamma function> have the same recurrence and the same value at $1$. Hence their residues at every nonpositive integer are the same. Their difference $f=F-\Gamma$ therefore has removable singularities there and is analytic elsewhere, so it extends to an <entire function>. Equivalently, $f(z+1)=zf(z)$ and $f(1)=0$ show inductively that the apparent quotient singularities are removable.

Solved by gpt-5.6-sol high.

= c
{parent=7d}
{scope}

= Solution
{parent=c}

The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,
$$
S(z+1)=f(z+1)f(-z)=-S(z).
$$
Its <modulus> is therefore periodic with period one. The assumed bound for $f$ on $0\leq\operatorname{Re}z\leq1$, together with the reflected bound for $f(1-z)$, bounds $S$ on that strip and hence on all of $\mathbb C$. By Liouville's theorem $S$ is constant, while $S(z+1)=-S(z)$ forces that constant to be zero.

If $f$ were nonzero at one point, it would be nonzero on a neighborhood, so $f(1-z)=0$ on an <open set>; the identity theorem would then give $f\equiv0$, a contradiction to that choice. Therefore $f\equiv0$ and $F=\Gamma$.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 3}
{scope}
{title2=Classical Dynamics}

= a
{parent=8e}
{scope}

= Solution
{parent=a}

In <Hamiltonian mechanics>, the canonical <momenta> are
$$
p_r=m\dot r,\qquad p_\theta=mr^2\dot\theta,\qquad
p_\phi=mr^2\sin^2\theta\,\dot\phi.
$$
Hence
$$
H=\frac{p_r^2}{2m}+\frac{p_\theta^2}{2mr^2}
 +\frac{p_\phi^2}{2mr^2\sin^2\theta}+V(r).
$$

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= Solution
{parent=b}

Because $H$ is independent of $\phi$, Hamilton's equation gives $\dot p_\phi=0$. Put
$$
L^2=p_\theta^2+\frac{p_\phi^2}{\sin^2\theta}.
$$
Using
$$
\dot\theta=\frac{p_\theta}{mr^2},\qquad
\dot p_\theta=\frac{p_\phi^2\cos\theta}{mr^2\sin^3\theta},
$$
one obtains $dL^2/dt=0$. Physically, $p_\phi=L_z$ is the component of <angular momentum> along the polar axis, while $L^2$ is the squared magnitude of the <angular momentum>.

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= Solution
{parent=c}

The orbital plane has normal parallel to the conserved <angular momentum>. If its inclination to the equatorial plane is $\psi$, then $L_z=L\cos\psi$. Therefore
$$
p_\theta^2=L^2-\frac{L_z^2}{\sin^2\theta}
=\frac{L^2}{\sin^2\theta}(\sin^2\theta-\cos^2\psi),
$$
so
$$
p_\theta(\theta)=\pm\frac{L}{\sin\theta}
\sqrt{\sin^2\theta-\cos^2\psi}.
$$
It vanishes when $\sin\theta=|\cos\psi|$, equivalently at the greatest northern and southern latitudes reached by the inclined orbit. These are the turning points of the polar motion.

Solved by gpt-5.6-sol high.

= 9D
{parent=Paper 3}
{scope}
{title2=Cosmology}

= a
{parent=9d}
{scope}

= Solution
{parent=a}

The Friedmann equation immediately gives
$$
\Omega-1=\frac{kc^2}{\mathcal H^2}.
$$
Using $P=w\rho$ in the <acceleration> equation and eliminating $\rho a^2$ with Friedmann gives
$$
2\mathcal H'+(1+3w)(\mathcal H^2+kc^2)=0.
$$
Differentiate $\Omega-1=kc^2\mathcal H^{-2}$ and substitute this equation:
$$
\Omega'=-2(\Omega-1)\frac{\mathcal H'}{\mathcal H}
=(1+3w)\mathcal H\Omega(\Omega-1),
$$
where $\mathcal H^2+kc^2=\Omega\mathcal H^2$ was used in the last step.

Solved by gpt-5.6-sol high.

= b
{parent=9d}
{scope}

= Solution
{parent=b}

For an expanding universe with $0\leq w\leq1$, the coefficient $(1+3w)\mathcal H$ is positive. Thus $\Omega=1$ is unstable toward the future: an initial $0<\Omega<1$ decreases away from one toward zero, whereas an initial $\Omega>1$ increases and eventually corresponds to recollapse. In radiation domination, $w=1/3$ and
$$
\Omega'=2\mathcal H\Omega(\Omega-1),
$$
so a phase-line sketch has arrows away from the fixed point $1$ on both sides. Consequently the observed near-flatness today requires extremely fine tuning of the early value of $\Omega$.

Solved by gpt-5.6-sol high.

= c
{parent=9d}
{scope}

= Solution
{parent=c}

During inflation $w\simeq-1$, so $1+3w\simeq-2$. For expansion,
$$
\Omega'\simeq-2\mathcal H\Omega(\Omega-1),
$$
and the arrows point toward $\Omega=1$. Equivalently, accelerated expansion makes the curvature contribution $|kc^2|/\mathcal H^2$ rapidly decay. A sufficiently long inflationary epoch therefore drives a broad range of initial conditions extremely close to spatial flatness.

Solved by gpt-5.6-sol high.

= 10E
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10e}
{scope}

= Solution
{parent=a}

If the Schmidt rank is one, the decomposition is $|\psi\rangle=|\alpha\rangle|\beta\rangle$ after absorbing its sole coefficient, so the state is a product. Conversely, a product state has coefficient <matrix> $X=ab^T$, which has rank one. Therefore a bipartite pure state is entangled exactly when its Schmidt rank is at least two.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The state is already
$$
|0\rangle_A\otimes|0\rangle_B\otimes|0\rangle_C.
$$
It therefore has Schmidt rank one across each of $A|BC$, $B|AC$, and $C|AB$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Across $A|BC$ the state is the product $|0\rangle_A\otimes|\phi^+\rangle_{BC}$, so its Schmidt rank is one. Across $B|AC$ it is
$$
\frac1{\sqrt2}\bigl(|0\rangle_B|0_A0_C\rangle
+|1\rangle_B|0_A1_C\rangle\bigr),
$$
and the two states on each side are orthonormal, so the Schmidt rank is two. The same argument with $B$ and $C$ exchanged handles $C|AB$. This is precisely the entanglement of the <Bell state> shared by $B$ and $C$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Across $A|BC$, the <GHZ state> has the Schmidt decomposition
$$
\frac1{\sqrt2}\bigl(|0\rangle_A|00\rangle_{BC}
+|1\rangle_A|11\rangle_{BC}\bigr),
$$
of rank two. Symmetry gives the same rank-two decomposition across $B|AC$ and $C|AB$, so the state is entangled across every bipartition.

Solved by gpt-5.6-sol high.

= c
{parent=10e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Projecting the <GHZ state> onto Charlie's outcome $0$ leaves the unnormalised state $|00\rangle_{AB}/\sqrt2$. After normalization Alice and Bob have $|00\rangle$, a product state.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Projecting the three-qubit $W$ state onto Charlie's outcome $0$ leaves
$$
\frac1{\sqrt3}(|01\rangle+|10\rangle).
$$
After normalization this is $(|01\rangle+|10\rangle)/\sqrt2$, a <Bell state> and hence an entangled state of Alice and Bob.

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 3}
{scope}
{title2=Number Theory}

= Solution
{parent=11F}

The reduction map sends $H_k$ onto $H_{k-1}$ with kernel of order $p$, so $|H_k|=p^{k-1}$. More explicitly, the lifting-the-exponent calculation
$$
v_p((1+p)^{p^j}-1)=j+1
$$
shows that $1+p$ has order $p^{k-1}$ modulo $p^k$. Hence $H_k=\langle1+p\rangle$ is cyclic of that order.

For the <matrix> claim, suppose $A\ne I$ and choose $m$ maximal such that $A=I+p^mB$, so some entry of $B$ is nonzero modulo $p$. Replacing $A$ by a suitable power, we may assume its order is a prime $q$. If $q\ne p$, binomial expansion modulo $p^{m+1}$ gives
$$
A^q\equiv I+qp^mB\not\equiv I\pmod {p^{m+1}}.
$$
If $q=p$, expansion modulo $p^{m+2}$ gives
$$
A^p\equiv I+p^{m+1}B\not\equiv I\pmod {p^{m+2}},
$$
because $p$ is odd. Both contradict $A^q=I$, so $A=I$.

For $p=2$, $r=1$ fails because $-I=I+2(-I)$ has order two. If $r=2$, the same argument works; in the order-two case, for maximal $m\geq2$,
$$
(I+2^mB)^2=I+2^{m+1}(B+2^{m-1}B^2)
e I,
$$
since the parenthesis is nonzero modulo two. Thus the smallest value is $r=2$.

Solved by gpt-5.6-sol high.

= 12J
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}

= a
{parent=12j}
{scope}

= Solution
{parent=a}

A language $L$ satisfies the regular pumping lemma if there is an integer $p\geq1$ such that every $w\in L$ with $|w|\geq p$ can be written $w=xyz$ with
$$
|xy|\leq p,\qquad |y|\geq1,\qquad xy^iz\in L\quad(i\geq0).
$$

Solved by gpt-5.6-sol high.

= b
{parent=12j}
{scope}

= Solution
{parent=b}

Let a deterministic automaton for $L$ have $p$ states. During the first $p$ input symbols of an accepted word $w$ of length at least $p$, the run visits $p+1$ states, so two coincide. Write $w=xyz$ so that $y$ labels the nonempty loop between those visits and $|xy|\leq p$. Traversing that loop any number of times leaves the remainder of the accepting run unchanged, giving $xy^iz\in L$ for every $i\geq0$.

Solved by gpt-5.6-sol high.

= c
{parent=12j}
{scope}

= Solution
{parent=c}

If an automaton with at most $100$ states accepted a word $w\in L$ of length $100$, the preceding loop argument would produce $xy^iz\in L$ for every $i\geq0$, with $y\ne\varepsilon$. These words have unbounded lengths and are distinct, contradicting the finiteness of $L$.

Solved by gpt-5.6-sol high.

= d
{parent=12j}
{scope}

= Solution
{parent=d}

Take pumping length $p=1$. If a nonempty word $w\in\{0,1\}^*$, pump its first symbol; every pumped word remains in $\{0,1\}^*$. Otherwise write $w=u2v$ with $v\in L$. If $u\ne\varepsilon$, pump the first symbol of $u$, leaving a word of the same form. If $u=\varepsilon$, pump the initial $2$: deleting it leaves $v\in\{0,1\}^*$, while retaining one or more copies lets the last pumped $2$ serve as the separator. Thus every required pumped word belongs to $\widehat L$.

Solved by gpt-5.6-sol high.

= e
{parent=12j}
{scope}

= Solution
{parent=e}

Choose a language $L\subseteq\{0,1\}^*$ that is not recursively enumerable; such a language exists because there are uncountably many languages but only countably many algorithms. Form $\widehat L$ as in part (d), which satisfies the pumping lemma.

If a grammar generated $\widehat L$, enumerate its terminal derivations and retain precisely words of the form $2v$ with $v\in\{0,1\}^*$. Since such a word contains only one separator,
$$
2v\in\widehat L\quad\Longleftrightarrow\quad v\in L.
$$
This would enumerate $L$, a contradiction. Hence $\widehat L$ is the required language.

Solved by gpt-5.6-sol high.

= 13A
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}

= a
{parent=13a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Open ends impose $C(0,t)=C(L,t)=0$. By <separation of variables>,
$$
C(x,t)=\sum_{n=1}^\infty b_n\sin\frac{n\pi x}{L}
\exp\left[\left(\mu-\frac{Dn^2\pi^2}{L^2}\right)t\right].
$$
Expanding the initial constant gives $b_n=2C_0(1-(-1)^n)/(n\pi)$, hence
$$
C(x,t)=\frac{4C_0}{\pi}\sum_{\substack{n\geq1\\n\ {
m odd}}}
\frac1n\sin\frac{n\pi x}{L}
\exp\left[\left(\mu-\frac{Dn^2\pi^2}{L^2}\right)t\right].
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The diffusive flux is $j=-D C_x$. The total outward flux through both ends is therefore
$$
J_{\rm out}=D C_x(0,t)-D C_x(L,t)
=\frac{8DC_0}{L}\sum_{\substack{n\geq1\\n\ {
m odd}}}
\exp\left[\left(\mu-\frac{Dn^2\pi^2}{L^2}\right)t\right].
$$
It diverges as $t\downarrow0$: the initially uniform profile is incompatible with the suddenly imposed zero boundary values, creating an idealized infinite initial <gradient>.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

At late times the first sine mode dominates, with growth exponent
$$
\mu-\frac{D\pi^2}{L^2}.
$$
Thus
$$
\mu_c(D,L)=\frac{D\pi^2}{L^2},
$$
and the population grows precisely when $\mu>\mu_c$. Faster diffusion carries organisms to the lethal ends more quickly and raises the threshold; a longer channel lowers the principal diffusion <eigenvalue> and therefore lowers the threshold.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

Immediately after $t=0$, thin boundary layers enforce zero concentration at both ends while the interior remains close to $C_0$. The profile is symmetric about $L/2$. Higher sine modes then decay rapidly and the shape approaches
$$
C(x,t)\sim \frac{4C_0}{\pi}\sin\frac{\pi x}{L}
 e^{(\mu-\mu_c)t}.
$$
For $\mu<\mu_c$ successive late-time curves have the same sine shape and decreasing height; for $\mu>\mu_c$ they have that shape and increasing height. Those features determine the two requested sketches.

Solved by gpt-5.6-sol high.

= b
{parent=13a}
{scope}

= Solution
{parent=b}

The <boundary conditions> are now $C(0,t)=0$ and $C_x(L,t)=0$. Their wave numbers are
$$
k_n=\frac{(n+1/2)\pi}{L},\qquad n=0,1,\ldots.
$$
The principal growth rate is $\mu-D\pi^2/(4L^2)$, so long-term growth occurs when
$$
\mu>\frac{D\pi^2}{4L^2}.
$$
Closing one end reduces the critical growth rate by a factor of four because there is only one absorbing escape boundary.

Solved by gpt-5.6-sol high.

= c
{parent=13a}
{scope}

= Solution
{parent=c}

Each spatial mode acquires the factor
$$
\exp\left(\int_0^t\mu(s)\,ds-Dk_n^2t\right)
=\exp\left((\mu_0-Dk_n^2)t+\mu_1\sin t\right).
$$
The oscillatory factor $e^{\mu_1\sin t}$ is bounded and does not affect the long-term exponential rate. With two open ends, growth therefore occurs exactly when
$$
\mu_0>\frac{D\pi^2}{L^2}.
$$
Only the time average of this spatially uniform fluctuating growth rate controls the asymptotic threshold.

Solved by gpt-5.6-sol high.

= 14D
{parent=Paper 3}
{scope}
{title2=Cosmology}

= a
{parent=14d}
{scope}

= Solution
{parent=a}

Since $P=P(\rho)$ and $\rho=\bar\rho(1+\delta)$,
$$
\frac{c^2}{\rho}\nabla P
=\frac{c^2}{\rho}\frac{dP}{d\rho}\nabla\rho
\simeq c_s^2\nabla\delta.
$$
To first order, continuity and the divergence of Euler give
$$
\dot\delta+\theta=0,\qquad
\dot\theta+2\frac{\dot a}{a}\theta
=-\frac{4\pi G}{c^2}\bar\rho\,\delta-c_s^2\nabla^2\delta.
$$
Eliminating $\theta$ and using $\nabla\mapsto i\mathbf k/a$ yields
$$
\ddot\delta+2\frac{\dot a}{a}\dot\delta
-\left(\frac{4\pi G}{c^2}\bar\rho-\frac{c_s^2k^2}{a^2}\right)\delta=0.
$$
The second term is Hubble damping, gravity drives growth, and <pressure> produces oscillatory restoration. Their balance defines
$$
k_J^2=\frac{4\pi G\bar\rho\,a^2}{c^2c_s^2},
\qquad \lambda_J=\frac{2\pi a}{k_J}.
$$
Modes longer than the Jeans length are gravitationally unstable; shorter modes are <pressure> supported.

Solved by gpt-5.6-sol high.

= b
{parent=14d}
{scope}

= Solution
{parent=b}

Here
$$
c_s^2=c^2\frac{dP}{d\rho}=\frac43\alpha c^2\bar\rho^{1/3}.
$$
Since $\bar\rho\propto t^{-2}$ and $a\propto t^{2/3}$, $c_s^2/a^2=L_0^2/t^2$ for a constant $L_0^2$. With $\dot a/a=2/(3t)$ and $4\pi G\bar\rho/c^2=2/(3t^2)$, substitution of $\delta=t^n$ gives
$$
n^2+\frac13n+L_0^2k^2-\frac23=0,
$$
so
$$
n_\pm=-\frac16\pm\sqrt{\frac{25}{36}-L_0^2k^2}.
$$
For $k\ll\widetilde k_J=5/(6L_0)$, the modes approach $t^{2/3}$ and $t^{-1}$. For $k\gg\widetilde k_J$, the exponents are complex with real part $-1/6$, giving logarithmic oscillations with decaying envelope $t^{-1/6}$. Taking $L_0\simeq50\,\mathrm{Mpc}$ would pressure-suppress structure on enormous, cosmologically important scales and is incompatible with the observed growth of smaller-scale structure.

Solved by gpt-5.6-sol high.

= 15E
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=15e}
{scope}

= Solution
{parent=a}

Orthogonal projection gives the unique components
$$
|\phi\rangle=P_g|\phi\rangle+P_b|\phi\rangle.
$$
Choose
$$
a=\|P_g\phi\|,quad b=\|P_b\phi\|,quad
|\psi_1\rangle=\frac{P_g|\phi\rangle}{a},quad
|\psi_2\rangle=\frac{P_b|\phi\rangle}{b}.
$$
Then $a,b\geq0$, $a^2+b^2=1$, and the required decomposition follows. If one projection vanishes, its coefficient is zero and the corresponding normalized <vector> may be chosen arbitrarily in that subspace; the nonzero projected components remain unique.

Solved by gpt-5.6-sol high.

= b
{parent=15e}
{scope}

= Solution
{parent=b}

Conjugating a reflection by a unitary reflects in the transported <vector>. Since $A|0^n\rangle=|\Omega\rangle$,
$$
AR_0A^{-1}|\Omega\rangle=-|\Omega\rangle,
$$
and every <vector> orthogonal to $|\Omega\rangle$ is fixed. Hence
$$
AR_0A^{-1}=R_{|\Omega\rangle},
$$
in particular both sides act identically on $|\Omega\rangle$.

Solved by gpt-5.6-sol high.

= c
{parent=15e}
{scope}

= Solution
{parent=c}

Using $|\psi_b\rangle$ as the horizontal axis and $|\psi_g\rangle$ as the vertical axis,
$$
|\Omega\rangle=\cos\theta|\psi_b\rangle+sin\theta|\psi_g\rangle.
$$
Thus $\sin\theta=\sqrt p$ and
$$
\theta=\arcsin\sqrt p.
$$

Solved by gpt-5.6-sol high.

= d
{parent=15e}
{scope}

= Solution
{parent=d}

The operator $R_g$ reverses the $|\psi_g\rangle$ component and fixes $|\psi_b\rangle$, so it is reflection in the horizontal $|\psi_b\rangle$ axis. The operator $R_{|\Omega\rangle}$ reverses the component parallel to $|\Omega\rangle$ and fixes its perpendicular line; multiplication by $-1$ reverses those roles. Hence $-R_{|\Omega\rangle}$ is reflection in the line spanned by $|\Omega\rangle$.

Solved by gpt-5.6-sol high.

= e
{parent=15e}
{scope}

= Solution
{parent=e}

Both operators map linear combinations of $|\psi_g\rangle$ and $|\psi_b\rangle$ to linear combinations of the same two <vectors>: $R_g$ merely changes one coefficient, while
$$
R_{|\Omega\rangle}=I-2|\Omega\rangle\langle\Omega|
$$
has $|\Omega\rangle\in P$. Therefore each leaves $P$ invariant.

Solved by gpt-5.6-sol high.

= f
{parent=15e}
{scope}

= Solution
{parent=f}

The product of reflections in two lines meeting at angle $\theta$ is a rotation through $2\theta$. Here $Q=-R_{|\Omega\rangle}R_g$ rotates toward the good axis by $2\theta$. Since $|\Omega\rangle$ initially makes angle $\theta$ with the bad axis,
$$
Q^k|\Omega\rangle
=\sin((2k+1)\theta)|\psi_g\rangle
 +\cos((2k+1)\theta)|\psi_b\rangle.
$$
A single use replaces the angle $\theta$ by $3\theta$.

Solved by gpt-5.6-sol high.

= g
{parent=15e}
{scope}

= Solution
{parent=g}

The <amplitude amplification> success probability is
$$
P_k=\sin^2((2k+1)\theta).
$$
Choose $k$ to be the nearest integer to $\pi/(4\theta)-1/2$. Then $(2k+1)\theta$ differs from $\pi/2$ by at most $\theta$, so $P_k\geq\cos^2\theta=1-p$. Since $\theta=\arcsin\sqrt p\sim\sqrt p$, this uses
$$
k=O(p^{-1/2})
$$
iterations and makes the good amplitude arbitrarily close to one when $p$ is small.

Solved by gpt-5.6-sol high.

= 16I
{parent=Paper 3}
{scope}
{title2=Logic and Set Theory}

= a
{parent=16i}
{scope}

= Solution
{parent=a}

<Zorn lemma> says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. The <axiom of choice> says that every family of nonempty sets has a choice <function>. The <well-ordering theorem> says that every set admits a well-order.

Choice gives the Hausdorff maximal principle by repeatedly choosing an element extending a chain; the union at <limit> stages is again a chain. A maximal chain has an upper bound, and that upper bound is maximal, proving Zorn. Conversely, apply Zorn to partial choice <functions> ordered by extension. A maximal partial choice <function> must have the full family as domain, proving choice.

Choice also well-orders a set by recursively choosing from the unchosen remainder; <Hartogs theorem> forces this recursion to exhaust the set before it reaches the Hartogs ordinal. Conversely, from a well-order on the union of a family, choose the least element of each member. Thus all three principles are equivalent.

Solved by gpt-5.6-sol high.

= b
{parent=16i}
{scope}

= Solution
{parent=b}

The <Godel completeness theorem> says that a first-order sentence follows semantically from a theory exactly when it is formally derivable from it. The <compactness theorem> says that a theory has a model exactly when every finite subset has a model.

Fix $i$. For each $j\ne i$, the theory $T_i\cup T_j$ has no model because the model classes form a partition. By compactness, some finite $\Delta_{ij}\subseteq T_i$ is already inconsistent with $T_j$. Let
$$
\Delta_i=\bigcup_{j\ne i}\Delta_{ij},
$$
a finite subset of $T_i$. Every model of $T_i$ models $\Delta_i$. Conversely, a model of $\Delta_i$ belongs to exactly one class $T_j$; if $j\ne i$, it would model both $T_j$ and $\Delta_{ij}$, a contradiction. Hence it belongs to $T_i$, and $\Delta_i$ finitely axiomatizes $T_i$.

Solved by gpt-5.6-sol high.

= 17I
{parent=Paper 3}
{scope}
{title2=Graph Theory}

= a
{parent=17i}
{scope}

= Solution
{parent=a}

Repeatedly delete a vertex whose current degree is less than $d/2$. If every vertex were deleted, charge each edge to the endpoint deleted first. At each deletion fewer than $d/2$ remaining edges are charged, so the original graph would have fewer than $|V|d/2$ edges, contradicting average degree at least $d$. The nonempty graph left by the process has minimum degree at least $d/2$.

Solved by gpt-5.6-sol high.

= b
{parent=17i}
{scope}

= Solution
{parent=b}

Set $N=(100\delta)^g$ and take $G\sim G(N,4\delta/N)$. Its expected number of edges is asymptotic to $2\delta N$, and a Chernoff bound makes $e(G)>3\delta N/2$ with positive probability. The expected number of cycles of length below $g$ is at most
$$
\sum_{\ell=3}^{g-1}\frac{(4\delta)^\ell}{2\ell}<(4\delta)^g,
$$
which is far below $\delta N/2$; Markov's inequality shows that, simultaneously with positive probability, there are fewer than $\delta N/2$ such cycles.

Choose such a graph and delete one edge from every cycle of length below $g$. The resulting graph has girth at least $g$ and more than $\delta N$ edges, hence average degree greater than $2\delta$. Part (a) supplies a subgraph of minimum degree at least $\delta$. It uses at most $N=(100\delta)^g$ vertices and cannot acquire any shorter cycle.

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 3}
{scope}
{title2=Galois Theory}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

The composite $L_1L_2$ is the smallest subfield of $M$ containing both $L_1$ and $L_2$. Since $L_1/K$ is separable and normal, its minimal <polynomials> remain separable and split after base change to $L_2$. Hence $L_1L_2/L_2$ is Galois.

Restriction defines
$$
\operatorname{Gal}(L_1L_2/L_2)\longrightarrow\operatorname{Gal}(L_1/K),
\qquad \sigma\longmapsto\sigma|_{L_1}.
$$
It is a homomorphism because every element of the source fixes $L_2$ and therefore $K$. Its kernel fixes both generating fields $L_1$ and $L_2$, hence fixes their composite; the map is injective.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

The primitive root $\zeta_n$ is a zero of the cyclotomic <polynomial> $\Phi_n$, of degree $\varphi(n)$. Its minimal <polynomial> over $K$ divides $\Phi_n$, so
$$
[K(\zeta_n):K]\mid\varphi(n),
$$
where divisibility also follows from the embedding of the Galois <group> of the <cyclotomic field> into $(\mathbb Z/n\mathbb Z)^\times$. Over $\mathbb Q$, Gauss's irreducibility theorem for cyclotomic <polynomials> says that $\Phi_n$ is irreducible. Therefore
$$
[\mathbb Q(\zeta_n):\mathbb Q]=\deg\Phi_n=\varphi(n).
$$

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Because $\operatorname{lcm}(9,15)=45$ and suitable <integral> powers of $\zeta_9$ and $\zeta_{15}$ generate a primitive forty-fifth root,
$$
L_1L_2=\mathbb Q(\zeta_{45}).
$$
The degree formula gives
$$
[L_1\cap L_2:\mathbb Q]
=\frac{\varphi(9)\varphi(15)}{\varphi(45)}
=\frac{6\cdot8}{24}=2.
$$
Both fields contain $\mathbb Q(\zeta_3)$, which already has degree two, so
$$
L_1\cap L_2=\mathbb Q(\zeta_3).
$$

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 3}
{scope}
{title2=Representation Theory}

= a
{parent=19h}
{scope}

= Solution
{parent=a}

Write an element as $(x,y,z)$, with multiplication
$$
(x,y,z)(x',y',z')=(x+x',y+y',z+z'+xy').
$$
The centre consists of $(0,0,z)$. Each central element is its own conjugacy class. If $(x,y)\ne(0,0)$, conjugation leaves $x,y$ fixed and changes $z$ by every element of $\mathbb F_p$, so its class is
$$
\{(x,y,z):z\in\mathbb F_p\}.
$$
Thus there are $p$ central singleton classes and $p^2-1$ noncentral classes of size $p$.

Solved by gpt-5.6-sol high.

= b
{parent=19h}
{scope}

= Solution
{parent=b}

The commutator <subgroup> equals the centre, so
$$
G/[G,G]\cong\mathbb F_p^2.
$$
Fix $\omega=e^{2\pi i/p}$. Every linear character is
$$
\chi_{a,b}(x,y,z)=\omega^{ax+by},\qquad a,b\in\mathbb F_p,
$$
and these $p^2$ distinct characters exhaust the one-dimensional representations.

Solved by gpt-5.6-sol high.

= c
{parent=19h}
{scope}

= Solution
{parent=c}

Take the abelian <subgroup>
$$
A=\{(x,0,z):x,z\in\mathbb F_p\},
$$
of order $p^2$. For each $c\in\mathbb F_p^\times$, define $\theta_c(x,0,z)=\omega^{cz}$ and form the <induced representation>
$$
\rho_c=\operatorname{Ind}_A^G\theta_c.
$$
It has degree $[G:A]=p$. Conjugation by representatives $(0,y,0)$ gives $p$ distinct characters of $A$ because $c\ne0$. The stabilizer of $\theta_c$ is therefore exactly $A$, and the Mackey irreducibility criterion shows that $\rho_c$ is irreducible. The $p-1$ choices of $c$ have distinct central characters.

Solved by gpt-5.6-sol high.

= d
{parent=19h}
{scope}

= Solution
{parent=d}

With class representatives $(0,0,z)$ and $(x,y,0)$ for $(x,y)\ne(0,0)$, the complete character table is described by
$$
\chi_{a,b}(0,0,z)=1,\qquad
\chi_{a,b}(x,y,0)=\omega^{ax+by},
$$
for $a,b\in\mathbb F_p$, and
$$
\chi_c(0,0,z)=p\omega^{cz},\qquad
\chi_c(x,y,0)=0
$$
for $c\in\mathbb F_p^\times$. There are $p^2+p-1$ rows, matching the number of conjugacy classes, and
$$
p^2\cdot1^2+(p-1)p^2=p^3=|G|,
$$
so no irreducible characters are missing.

Solved by gpt-5.6-sol high.

= 20J
{parent=Paper 3}
{scope}
{title2=Algebraic Topology}

= Solution
{parent=20J}

Because $K$ is a subcomplex, $\partial C_k(K)\subseteq C_{k-1}(K)$, so
$$
\partial(c+C_k(K))=\partial c+C_{k-1}(K)
$$
is well defined and squares to zero. The short exact <sequence> of chain complexes
$$
0\to C_\bullet(K)\to C_\bullet(L)\to C_\bullet(L,K)\to0
$$
gives the long exact <sequence>
$$
\cdots\to H_k(K)\to H_k(L)\to H_k(L,K)
\to H_{k-1}(K)\to\cdots.
$$
Taking $L=\partial\Delta^{n+1}$ and using the contractibility of $\Delta^{n+1}$ gives
$$
H_k(\partial\Delta^{n+1})\cong
\begin{cases}\mathbb Z,&k=0,n,\\0,&\text{otherwise}.
\end{cases}
$$
For $0<k<n$, exactness makes $H_k(K)$ a quotient of a <subgroup> of $H_{k+1}(L,K)$, whose rank is at most
$$
\operatorname{rank}C_{k+1}(L,K)
=\binom{n+2}{k+2}-\#\{(k+1)\text{-simplices of }K\}.
$$
The same stated inequality for $k=n$ follows from $H_n(K)\hookrightarrow H_n(L)\cong\mathbb Z$. Finally, an $n$-cycle in the boundary sphere has equal signed coefficients on all $n$-faces. A proper subcomplex omits an $n$-face, so that coefficient and hence every coefficient is zero. Therefore $H_n(K)=0$.

Solved by gpt-5.6-sol high.

= 21G
{parent=Paper 3}
{scope}
{title2=Linear Analysis}

= Solution
{parent=21G}

The real <Stone-Weierstrass theorem> says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb R)$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb R)$.

Let $\overline A$ be the uniform closure. <Polynomial> approximation to $\sqrt t$ on a bounded interval shows that $|f|\in\overline A$ whenever $f\in\overline A$. Hence $\overline A$ is closed under
$$
\max(f,g)=\frac{f+g+|f-g|}{2},\qquad
\min(f,g)=\frac{f+g-|f-g|}{2}.
$$
Given $h\in C(X)$, separation and the constants provide, for each $x,y$, a <function> in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such <functions> by minima to obtain one that agrees at $x$ and lies below $h+\varepsilon$ everywhere; a second finite cover and maxima produces $f\in\overline A$ with $|f-h|<\varepsilon$. Thus $h\in\overline A$.

The space $C_b(\mathbb R)$ is Banach because a uniform Cauchy <sequence> converges uniformly to a bounded <continuous function>. Compactness is essential: the algebra of bounded <continuous functions> having finite <limits> at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb R)$.

Solved by gpt-5.6-sol high.

= 22G
{parent=Paper 3}
{scope}
{title2=Analysis of Functions}

= Solution
{parent=22G}

For $s>n/2$ and $u\in H^s(\mathbb R^n)$, Fourier inversion and Cauchy--Schwarz give
$$
|u(x)|\leq C\int|\widehat u(\xi)|\,d\xi
\leq C\left(\int(1+|\xi|^2)^s|\widehat u|^2d\xi\right)^{1/2}
\left(\int(1+|\xi|^2)^{-s}d\xi\right)^{1/2}.
$$
The last <integral> is finite exactly when $s>n/2$. Thus the <sobolev embedding theorem> gives
$$
\|u\|_\infty\leq C_{n,s}\|u\|_{H^s}.
$$

To see why the endpoint relevant to $H^1(\mathbb R^3)$ fails, choose a smooth cutoff $\chi$ supported near the origin and equal to one there, and set
$$
u(x)=\chi(x)|x|^{-1/4}.
$$
Near zero, $|u|^2$ contributes $\int_0^1r^{3/2}dr$, while $|\nabla u|^2$ contributes a constant multiple of $\int_0^1r^{-1/2}dr$. Both are finite, so $u\in H^1(\mathbb R^3)$, but $u$ is unbounded.

Solved by gpt-5.6-sol high.

= 23H
{parent=Paper 3}
{scope}
{title2=Riemann Surfaces}

= Solution
{parent=23H}

In local coordinates around $p\in R$ and $f(p)\in S$, a nonconstant analytic map has the form
$$
f(z)-f(p)=z^m h(z),\qquad h(0)\ne0.
$$
After shrinking the chart, $h$ has an analytic $m$th root, so a change of coordinate makes the map $z\mapsto z^m$. It maps small discs onto neighborhoods of the origin; hence every nonconstant analytic map of Riemann surfaces is open. If $R$ is compact and $S$ connected, $f(R)$ is compact and therefore closed, and it is also nonempty and open. Thus $f(R)=S$.

For $f:\mathbb D\to\mathbb D$ with $f(0)=0$, the removable-singularity theorem makes $g=f/z$ analytic. On $|z|=r$, $|g(z)|=|f(z)|/r\leq1/r$, and the maximum principle gives the same bound throughout $\mathbb D_r$. Applying the <Schwarz lemma> to an automorphism $h$ and to $h^{-1}$ shows that every automorphism fixing zero is
$$
h(z)=e^{i\theta}z.
$$

If $F:\mathbb C\to\mathbb C$ is an analytic isomorphism fixing zero, continuity of $F^{-1}$ implies that the preimage of every closed disc is compact. Hence $|F(z)|\to\infty$ as $|z|\to\infty$, and defining $F(\infty)=\infty$ gives an analytic isomorphism of the Riemann sphere.

Consequently an automorphism of $\mathbb D$ fixing zero does extend to $\mathbb C$. A general disc automorphism need not: for $a\ne0$,
$$
\frac{z-a}{1-\bar a z}
$$
has a finite pole and cannot extend to an entire isomorphism.

Solved by gpt-5.6-sol high.

= 24F
{parent=Paper 3}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=24F}

The <Riemann-Roch theorem> states
$$
\ell(D)-\ell(K-D)=\deg D+1-g.
$$
On a genus-one curve $K\sim0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell(D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, so
$$
D\sim P-P_0.
$$
If also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant <function> with divisor $P-Q$ would define a degree-one map to $\mathbb P^1$, forcing the genus to be zero. Hence $P=Q$.

This identifies $X$ with $\operatorname{Pic}^0(X)$ by $P\mapsto[P-P_0]$ and defines $P+Q=R$ through
$$
[P-P_0]+[Q-P_0]=[R-P_0].
$$
For the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives
$$
A+B=(1/4:-1/8:1).
$$

Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalently
$$
3P\sim3P_0.
$$
Under the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.

Solved by gpt-5.6-sol high.

= 25J
{parent=Paper 3}
{scope}
{title2=Differential Geometry}

= a
{parent=25j}
{scope}

= Solution
{parent=a}

A subset $X\subset\mathbb R^N$ is a smooth $k$-manifold if every $x\in X$ has a neighborhood parametrized by a smooth map $\phi:U\subset\mathbb R^k\to X$ that is a homeomorphism onto its image and has <derivative> of rank $k$. Define
$$
T_xX=\operatorname{im}D\phi_u,\qquad \phi(u)=x.
$$
If $\psi$ is another parametrization, the transition maps satisfy
$$
D\phi_u=D\psi_vD(\psi^{-1}\circ\phi)_u.
$$
The transition <derivative> is invertible, so the two images coincide. Thus the tangent space is independent of the parametrization.

Solved by gpt-5.6-sol high.

= b
{parent=25j}
{scope}

= Solution
{parent=b}

The <determinant> is continuous, so
$$
GL(n,\mathbb R)=\det^{-1}(\mathbb R\setminus\{0\})
$$
is open in $\mathbb R^{n^2}$. It is therefore a smooth manifold of dimension $n^2$, and
$$
T_I GL(n,\mathbb R)=M_n(\mathbb R).
$$

Solved by gpt-5.6-sol high.

= c
{parent=25j}
{scope}

= Solution
{parent=c}

The differential of <determinant> at $A$ is
$$
D(\det)_A(X)=\det(A)\operatorname{tr}(A^{-1}X).
$$
It is nonzero at every $A\in SL(n,\mathbb R)$, so the regular-level-set theorem makes $SL(n,\mathbb R)$ a manifold of dimension $n^2-1$. At the identity,
$$
T_I SL(n,\mathbb R)=\{X:\operatorname{tr}X=0\}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=25j}
{scope}

= Solution
{parent=d}

Consider $F(A)=A^TA$ with values in the symmetric <matrices>. At the identity,
$$
DF_I(X)=X^T+X,
$$
which is onto the $n(n+1)/2$-dimensional space of symmetric <matrices>. The same rank holds at every orthogonal <matrix>. Thus $O(n)=F^{-1}(I)$ is a manifold, and its determinant-one component, the <special orthogonal group>, has dimension
$$
n^2-\frac{n(n+1)}2=\frac{n(n-1)}2.
$$
The tangent equation at the identity is $X^T+X=0$, so
$$
T_I SO(n)=\{X:X^T=-X\}.
$$

Solved by gpt-5.6-sol high.

= 26G
{parent=Paper 3}
{scope}
{title2=Probability and Measure}

= a
{parent=26g}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Lévy's convergence theorem says that probability measures $\mu_n$ on $\mathbb R^d$ converge weakly to $\mu$ if and only if their characteristic <functions> converge pointwise to the characteristic <function> of $\mu$. More generally, if the characteristic <functions> converge pointwise to a <function> continuous at zero, that <function> is the characteristic <function> of a probability measure and the corresponding weak convergence holds.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

If $X_n\Rightarrow X$, the continuous mapping theorem gives
$$
\langle u,X_n\rangle\Rightarrow\langle u,X\rangle
$$
for every $u$. Conversely, the assumed one-dimensional convergence at $t=1$ gives
$$
\varphi_{X_n}(u)=\mathbb E e^{i\langle u,X_n\rangle}
\longrightarrow
\mathbb E e^{i\langle u,X\rangle}=\varphi_X(u)
$$
for every $u\in\mathbb R^d$. Lévy's convergence theorem then yields $X_n\Rightarrow X$. This is the <Cramer-Wold theorem>.

Solved by gpt-5.6-sol high.

= b
{parent=26g}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Fubini's theorem gives
$$
\frac1u\int_{-u}^u(1-\varphi(t))dt
=\mathbb E\left[\frac1u\int_{-u}^u(1-e^{itX})dt\right].
$$
The odd imaginary part integrates to zero, while
$$
\int_{-u}^ue^{itX}dt=\frac{2\sin(uX)}{X}.
$$
Therefore the expression equals
$$
2\mathbb E\left(1-\frac{\sin(uX)}{uX}\right),
$$
with the quotient interpreted continuously at $X=0$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $|y|>2$, one has $2(1-\sin y/y)\geq1$. Applying this with $y=uX$ and using part (i),
$$
\mathbb P(|X|>2/u)
\leq2\mathbb E\left(1-\frac{\sin(uX)}{uX}\right)
=\frac1u\int_{-u}^u(1-\varphi(t))dt.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Continuity of $\varphi$ at zero and $\varphi(0)=1$ allow $u>0$ so small that
$$
\frac1u\int_{-u}^u(1-\varphi(t))dt<\varepsilon/2.
$$
<Pointwise convergence> and dominated convergence make the corresponding expression for $\varphi_n$ less than $\varepsilon$ for all sufficiently large $n$. Part (ii) then bounds
$$
\mathbb P(|X_n|>2/u)\leq\varepsilon.
$$
There are only finitely many remaining $n$, so enlarge $M\geq2/u$ until each of their tails is also at most $\varepsilon$. This proves uniform tightness.

Solved by gpt-5.6-sol high.

= 27K
{parent=Paper 3}
{scope}
{title2=Applied Probability}

= a
{parent=27k}
{scope}

= Solution
{parent=a}

Let regeneration cycles have lengths $C_i$, contain $A_i$ arrivals, and accrue queue-length reward
$$
R_i=\int_{\text{cycle }i}Q(t)dt.
$$
Fubini's geometric identity writes this area as the sum, over customers in the cycle, of their time in the system, up to boundary terms whose contribution vanishes over many cycles. The renewal-reward theorem therefore gives
$$
L=\frac{\mathbb ER_1}{\mathbb EC_1},
\qquad
\lambda=\frac{\mathbb EA_1}{\mathbb EC_1},
\qquad
W=\frac{\mathbb ER_1}{\mathbb EA_1}.
$$
Hence <Little law> is
$$
L=\lambda W.
$$

Solved by gpt-5.6-sol high.

= b
{parent=27k}
{scope}

= Solution
{parent=b}

An <M-G-1 queue> has Poisson arrivals of rate $\lambda$, independent identically distributed service times with a general law of mean $m$, and one server. During the first service time $S$, a mean $\lambda\mathbb ES=\lambda m$ customers arrive, and each initiates an independent descendant busy-period contribution. Thus
$$
\mathbb EB=m+\lambda m\,\mathbb EB,
$$
so for traffic intensity $\rho=\lambda m<1$,
$$
\mathbb EB=\frac{m}{1-\rho}.
$$
For $\rho\geq1$ the expected busy period is infinite.

Solved by gpt-5.6-sol high.

= c
{parent=27k}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For state $n\geq1$, births occur at rate $\lambda$ and departures at rate
$$
d_n=\mu+(n-1)\delta.
$$
The birth--death detailed-balance weights are
$$
\pi_n=\pi_0\frac{\lambda^n}{\mu(\mu+\delta)\cdots(\mu+(n-1)\delta)}.
$$
Their ratios tend to zero, so the <series> is summable and normalization gives an invariant probability distribution.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The chain is irreducible and nonexplosive, and it has an invariant probability distribution. Therefore every state is positive recurrent. The increasingly strong abandonment rate is what stabilizes the system even when $\lambda\geq\mu$.

Solved by gpt-5.6-sol high.

= 28L
{parent=Paper 3}
{scope}
{title2=Principles of Statistics}

= a
{parent=28l}
{scope}

= Solution
{parent=a}

Since $X/n$ is unbiased and has variance $\theta(1-\theta)/n$,
$$
R(\delta_{\rm MLE},\theta)
=\frac{\mathbb E_\theta[(X/n-\theta)^2]}{\theta(1-\theta)}
=\frac1n
$$
for $0<\theta<1$.

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= Solution
{parent=b}

The $\pi$-Bayes risk is
$$
r_\pi(\delta)=\int R(\delta,\theta)\,\pi(d\theta).
$$
Conditioning on the observation and applying Tonelli gives
$$
r_\pi(\delta)=\mathbb E_{X}\left[
\mathbb E\{L(\delta(X),\theta)\mid X\}
\right].
$$
Thus choosing, for every observed $x$, an action minimizing the posterior expected loss minimizes each integrand and hence minimizes the Bayes risk.

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= Solution
{parent=c}

Under the uniform prior, the posterior density after $X=x$ is proportional to $\theta^x(1-\theta)^{n-x}$. For $0<x<n$, differentiating posterior risk gives the unique stationary point
$$
a=
\frac{\mathbb E[(1-\theta)^{-1}\mid x]}
     {\mathbb E[(\theta(1-\theta))^{-1}\mid x]}
=\frac{B(x+1,n-x)}{B(x,n-x)}=\frac xn.
$$
Strict convexity makes it the unique minimum. For $x=0$, every $a\ne0$ gives infinite posterior risk at zero and $a=0$ is optimal; symmetrically $a=1$ is optimal for $x=n$. Hence the unique Bayes rule is $\delta(x)=x/n$.

Solved by gpt-5.6-sol high.

= d
{parent=28l}
{scope}

= Solution
{parent=d}

The constant risk from part (a) gives
$$
\sup_\theta R(\delta_{\rm MLE},\theta)=\frac1n.
$$
Every estimator has maximum risk at least its uniform-prior Bayes risk, and the minimum possible Bayes risk is $1/n$ by part (c). Hence no estimator has smaller maximum risk, so the MLE is minimax.

An estimator is admissible if no other estimator has risk no larger at every parameter and strictly smaller somewhere. If an estimator dominated the MLE, continuity of its binomial risk would make the inequality strict on a set of positive prior measure, lowering its uniform-prior Bayes risk below $1/n$. This contradicts Bayes optimality. Equivalently, uniqueness of the Bayes action at every $x$ rules out equality for a distinct estimator. Thus the MLE is admissible.

Solved by gpt-5.6-sol high.

= 29L
{parent=Paper 3}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=29l}
{scope}

= Solution
{parent=a}

Under a risk-neutral measure,
$$
\mathbb E_Q\xi_n=1+r.
$$
Writing $q=Q(\xi_n=1+b)$ gives
$$
q(1+b)+(1-q)(1+a)=1+r,
\qquad q=\frac{r-a}{b-a}.
$$
Equivalence requires $0<q<1$, hence $a<r<b$. At every node the risk-neutral condition determines this same conditional probability $q$, so the multipliers remain independent and identically distributed under $Q$.

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= Solution
{parent=b}

Let
$$
c=q\log(1+b)+(1-q)\log(1+a),
\qquad m=N-n.
$$
Risk-neutral valuation or direct backward induction gives
$$
V(n,s)=(1+r)^{-m}\bigl(\log s+mc\bigr).
$$
Thus
$$
A_n=(1+r)^{-(N-n)},
\qquad
B_n=(N-n)c(1+r)^{-(N-n)}.
$$
Substitution into the recursion verifies the terminal condition and the induction step.

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= Solution
{parent=c}

At a node with time $n-1$ price $s$, choose the stock holding
$$
\Delta_n(s)=\frac{V(n,s(1+b))-V(n,s(1+a))}{s(b-a)}.
$$
The two possible values of the stock position differ by exactly the difference between the two continuation claims. Choose the bank holding so that total wealth is $V(n-1,s)$. The pricing recursion makes its next value equal to the appropriate continuation value in both states. Backward induction from $V(N,s)=g(s)$ therefore replicates $g(S_N)$ from initial capital $V(0,S_0)$.

Solved by gpt-5.6-sol high.

= d
{parent=29l}
{scope}

= Solution
{parent=d}

Before the barrier has been hit, the two next-step continuation values are
$$
U_u=U(n,s(1+b))\mathbf1_{\{s(1+b)<B\}},
\qquad U_d=U(n,s(1+a)).
$$
The replicating stock holding on $(n-1,n]$ is therefore
$$
\Delta_n^{\rm UO}(s)=\frac{U_u-U_d}{s(b-a)}.
$$
The bank holding is chosen from current wealth $U(n-1,s)$ as before. If the barrier has already been hit, the claim and both holdings are zero. Backward induction proves replication of the up-and-out payout.

Solved by gpt-5.6-sol high.

= 30C
{parent=Paper 3}
{scope}
{title2=Asymptotic Methods}

= a
{parent=30c}
{scope}

= Solution
{parent=a}

<Watson lemma> states that if, near zero,
$$
f(t)\sim\sum_{n=0}^\infty a_nt^{n+\lambda-1},
\qquad \lambda>0,
$$
and the tail is suitably controlled, then
$$
\int_0^b e^{-xt}f(t)dt
\sim\sum_{n=0}^\infty a_n\Gamma(n+\lambda)x^{-n-\lambda}
$$
as $x\to+\infty$.

Solved by gpt-5.6-sol high.

= b
{parent=30c}
{scope}

= Solution
{parent=b}

The endpoint $t=0$ is treated by rotating onto its <method of steepest descent> ray. It contributes
$$
\Gamma(3/2)(-ix)^{-3/2}=\frac{\sqrt\pi}{2}(-ix)^{-3/2}.
$$
At $t=1$, put $s=1-t$ and expand
$$
(1-s)^{1/2}=\sum_{n=0}^\infty(-1)^n\binom{1/2}{n}s^n.
$$
Termwise endpoint integration, using $\int_0^\infty s^ne^{-ixs}ds=n!(ix)^{-n-1}$ on the rotated ray, gives
$$
I(x)\sim\frac{\sqrt\pi}{2}(-ix)^{-3/2}
-e^{ix}\sum_{n=0}^\infty a_n(-ix)^{-n-1},
$$
where
$$
a_n=n!\binom{1/2}{n}.
$$
The two displayed contributions come from the two endpoints; the remaining deformed contour is exponentially or algebraically smaller after any fixed truncation.

Solved by gpt-5.6-sol high.

= 31A
{parent=Paper 3}
{scope}
{title2=Dynamical Systems}

= a
{parent=31a}
{scope}

= Solution
{parent=a}

Linearization about $X(t)$ gives
$$
\dot\Phi(t)=Df(X(t))\Phi(t),\qquad \Phi(0)=I.
$$
One <floquet multiplier> of $\Phi(T)$ is $1$, corresponding to a phase shift along the orbit. Liouville's formula gives
$$
\det\Phi(T)=\exp\left(\int_0^T\nabla\cdot f(X(t))dt\right).
$$
In the plane this <determinant> is the other, transverse multiplier. The periodic orbit is transversely asymptotically stable when the displayed quantity is less than one, equivalently when the divergence <integral> is negative, and unstable when it is greater than one.

Solved by gpt-5.6-sol high.

= b
{parent=31a}
{scope}

= Solution
{parent=b}

For the unperturbed oscillator take
$$
x=R\cos t,\qquad y=-R\sin t,
$$
and energy $E=(x^2+y^2)/2$. Its change over one nearly circular orbit is
$$
\Delta E=\varepsilon\int_0^{2\pi}(1-x^2+ay^2)y^2dt
=\varepsilon\pi\left(R^2+\frac{3a-1}{4}R^4\right).
$$
A nonzero balance therefore requires
$$
R^2=\frac4{1-3a},
$$
which exists exactly when $a<1/3$. The averaged energy drift is positive below this radius and negative above it, so the resulting <limit> cycle is stable.

Solved by gpt-5.6-sol high.

= 32C
{parent=Paper 3}
{scope}
{title2=Integrable Systems}

= Solution
{parent=32C}

A <vector> field $V=V_1\partial_x+\phi\partial_u$ is a Lie symmetry when
$$
\operatorname{pr}^{(N)}V(\Delta)=0
$$
on the equation manifold $\Delta=0$. With total <derivative> $D_x$, the prolongation coefficients obey
$$
\phi_0=\phi,\qquad
\phi_{j+1}=D_x\phi_j-u^{(j+1)}D_xV_1.
$$
For $V_1=f(x)$, direct iteration gives
$$
\begin{aligned}
\phi_3={}&\phi_{xxx}+u'(3\phi_{xxu}-f''')+3(u')^2\phi_{xuu}
 +(u')^3\phi_{uuu}\\
&+3(\phi_{xu}-f''+u'\phi_{uu})u''+(\phi_u-3f')u'''.
\end{aligned}
$$
Thus $\alpha=\beta=3$.

For $u'''=u^{-3}$, invariance requires $\phi_3+3u^{-4}\phi=0$ after substituting $u'''=u^{-3}$. Equating coefficients of the independent jet variables gives
$$
\phi_{uu}=0,\qquad \phi_{xu}=f'',\qquad f'''=0,
$$
and the remaining terms force $\phi=(3a/4)u$ and $f=ax+b$. Hence the Lie algebra is spanned by
$$
\partial_x,
\qquad
4x\partial_x+3u\partial_u.
$$

Solved by gpt-5.6-sol high.

= 33B
{parent=Paper 3}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=33b}
{scope}

= Solution
{parent=a}

The states are defined by
$$
J^2|j,m\rangle=\hbar^2j(j+1)|j,m\rangle,
\qquad J_3|j,m\rangle=\hbar m|j,m\rangle,
$$
where $j=0,\tfrac12,1,\ldots$ and $m=-j,-j+1,\ldots,j$. From
$$
J_\mp J_\pm=J^2-J_3^2\mp\hbar J_3
$$
and positivity of the norm,
$$
C_\pm(j,m)=\hbar\sqrt{j(j+1)-m(m\pm1)}.
$$
Thus $\lambda_1^+=\lambda_1^-=-1$, $\lambda_2^+=1$, and $\lambda_2^-=-1$.

Solved by gpt-5.6-sol high.

= i
{parent=a}
{scope}

= Solution
{parent=i}

Since $O$ commutes with every <angular momentum operator>, it commutes with $J^2$. Taking a <matrix> element of $[O,J^2]=0$ gives
$$
\bigl(j'(j'+1)-j(j+1)\bigr)\langle j,m|O|j',m'\rangle=0.
$$
The allowed nonnegative values of $j$ make $j(j+1)$ injective, so the <matrix> element vanishes unless $j=j'$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Taking a <matrix> element of $[O,J_3]=0$ gives
$$
\hbar(m'-m)\langle j,m|O|j',m'\rangle=0.
$$
Thus the <matrix> element vanishes unless $m=m'$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

For $-j\leq m<j$, the <matrix> element of $[O,J_+]$ between $|j,m+1\rangle$ and $|j,m\rangle$ gives
$$
C_+(j,m)\left(\langle j,m+1|O|j,m+1\rangle
-\langle j,m|O|j,m\rangle\right)=0.
$$
The ladder coefficient is nonzero, so adjacent diagonal entries agree. Induction makes the diagonal entry independent of $m$.

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= Solution
{parent=b}

The coupled <basis> consists of the triplet
$$
|1,1\rangle=|++\rangle,
\quad |1,0\rangle=\frac{|+-\rangle+|-+\rangle}{\sqrt2},
\quad |1,-1\rangle=|--\rangle,
$$
and the singlet
$$
|0,0\rangle=\frac{|+-\rangle-|-+\rangle}{\sqrt2}.
$$
These equations list all nonzero Clebsch--Gordan coefficients, each coefficient being $1$, $1/\sqrt2$, or $-1/\sqrt2$ as displayed, and
$$
I=\sum_{m=-1}^{1}|1,m\rangle\langle1,m|+|0,0\rangle\langle0,0|.
$$
Because each irreducible total-spin representation occurs once, an operator commuting with total <angular momentum> is
$$
O=\alpha P_{j=1}+\beta P_{j=0}.
$$
In the product <basis>, the nonzero <matrix> elements are
$$
\langle++|O|++\rangle=\langle--|O|--\rangle=\alpha,
$$
$$
\langle+-|O|+-\rangle=\langle-+|O|-+\rangle=\frac{\alpha+\beta}{2},
$$
and
$$
\langle+-|O|-+\rangle=\langle-+|O|+-\rangle=\frac{\alpha-\beta}{2}.
$$

Solved by gpt-5.6-sol high.

= 34E
{parent=Paper 3}
{scope}
{title2=Applications of Quantum Mechanics}

= a
{parent=34e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

For two solutions of the <Schrodinger equation>, their Wronskian is constant because the equation has no first-derivative term. The <determinant> of the Floquet <matrix> is the ratio of the Wronskian at $x+a$ to that at $x$, hence
$$
\det F(E)=1.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For real $E$ and real $V$, choose a real fundamental pair of solutions. The corresponding Floquet <matrix> has real entries, so $\operatorname{tr}F(E)$ is real. The trace is <basis> independent, so this holds for every fundamental pair.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The Floquet multipliers satisfy
$$
\lambda^2-(\operatorname{tr}F)\lambda+1=0.
$$
If $(\operatorname{tr}F)^2<4$, they are complex conjugates on the unit circle, $e^{\pm ika}$, and bounded Bloch waves exist: $E$ lies in an allowed band. If $(\operatorname{tr}F)^2>4$, the multipliers are real reciprocal numbers, one growing and one decaying, so there is no extended bounded Bloch state and $E$ lies in a gap.

Solved by gpt-5.6-sol high.

= b
{parent=34e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Translation preserves the alternating on-site term only when $(-1)^{j+\ell}=(-1)^j$, so
$$
[H_t,T_\ell]=0\quad\Longleftrightarrow\quad \ell\in2\mathbb Z.
$$
The primitive lattice spacing is therefore $2a$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Using one even and one odd site per unit cell, <Bloch theorem> reduces the <Hamiltonian> to
$$
H(k)=t\begin{pmatrix}
1&1+e^{-2ika}\\
1+e^{2ika}&-1
\end{pmatrix}.
$$
Its <eigenvalues> are
$$
E_\pm(k)=\pm t\sqrt{1+|1+e^{2ika}|^2}
=\pm t\sqrt{1+4\cos^2(ka)}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For lattice spacing $2a$, the first Brillouin zone is
$$
-\frac{\pi}{2a}\leq k\leq\frac{\pi}{2a}.
$$
At its boundaries $E_\pm=\pm t$, while at $k=0$ the energies are $E_\pm=\pm\sqrt5,t$. Thus the lower band runs from $-\sqrt5,t$ to $-t$, the upper from $t$ to $\sqrt5,t$, and each has bandwidth
$$
(\sqrt5-1)t.
$$
The requested plot consists of these two even branches, separated by a gap of size $2t$ at the zone boundaries.

Solved by gpt-5.6-sol high.

= 35B
{parent=Paper 3}
{scope}
{title2=Statistical Physics}

= a
{parent=35b}
{scope}

= Solution
{parent=a}

A grand <canonical ensemble> describes a system exchanging both energy and particles with a reservoir at fixed $T$, volume, and <chemical potential> $\mu$. For a noninteracting fermion level $r$, occupation is $0$ or $1$, so
$$
\mathcal Z_r=1+e^{-\beta(E_r-\mu)}.
$$
Factorization over levels gives $\mathcal Z=\prod_r\mathcal Z_r$, and <differentiation> with respect to $\beta\mu$ yields the <Fermi-Dirac distribution>
$$
\langle n_r\rangle=\frac{1}{e^{\beta(E_r-\mu)}+1}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35b}
{scope}

= Solution
{parent=b}

The number of <momentum> states in a shell is
$$
\frac{g_sV}{(2\pi\hbar)^d}S_{d-1}p^{d-1}dp.
$$
With $E=p^2/(2m)$ this gives
$$
g(E)=g_sBVE^{(d-2)/2},
\qquad
B=\frac{S_{d-1}m(2m)^{(d-2)/2}}{(2\pi\hbar)^d}.
$$
At zero <temperature> all states up to $E_F$ are occupied, hence
$$
N=\frac{2g_sBV}{d}E_F^{d/2},
\qquad
E_F=\left(\frac{dN}{2g_sBV}\right)^{2/d}.
$$
The energy is $U=dNE_F/(d+2)$. Since a nonrelativistic <ideal> gas in $d$ dimensions obeys $pV=2U/d$,
$$
pV=\frac{2}{d+2}NE_F.
$$

Solved by gpt-5.6-sol high.

= c
{parent=35b}
{scope}

= Solution
{parent=c}

The number flux through the surface is
$$
J_z=\frac{g_s}{(2\pi\hbar)^3}
\int_{p_z>p_0}\frac{p_z}{m}
\frac{d^3p}{e^{\beta(p^2/2m-E_F)}+1},
\qquad p_0=\sqrt{2m(E_F+V_0)}.
$$
Throughout this tail the occupation is Boltzmann to leading order. The two transverse Gaussian <integrals> contribute $2\pi m/\beta$, while
$$
\int_{p_0}^\infty\frac{p_z}{m}
 e^{-\beta(p_z^2/2m-E_F)}dp_z
=\frac1\beta e^{-\beta V_0}.
$$
Therefore
$$
J_z=\frac{g_sm}{4\pi^2\hbar^3}(k_BT)^2e^{-V_0/(k_BT)}.
$$

Solved by gpt-5.6-sol high.

= 36D
{parent=Paper 3}
{scope}
{title2=Electrodynamics}

= a
{parent=36d}
{scope}

= Solution
{parent=a}

Differentiating $\mathbf g=\epsilon_0\mathbf E\times\mathbf B$, substituting <Maxwell equations>, and using the stated <vector> identity gives the <Maxwell stress tensor> conservation law
$$
\partial_tg_j
=\partial_i\left[\epsilon_0\left(E_iE_j-\frac12E^2\delta_{ij}\right)
+\frac1{\mu_0}\left(B_iB_j-\frac12B^2\delta_{ij}\right)\right]
-(\rho\mathbf E+\mathbf J\times\mathbf B)_j.
$$
Moving the divergence to the left identifies
$$
\widetilde a=-\epsilon_0,
\quad \widetilde b=\frac{\epsilon_0}{2},
\quad \widetilde c=-\frac1{\mu_0},
\quad \widetilde d=\frac1{2\mu_0}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=36d}
{scope}

= Solution
{parent=b}

The tensor $\sigma_{ij}$ is the flux of the $j$th component of electromagnetic <momentum> across a surface normal to the $i$ direction, with the sign convention used in the conservation law. Its diagonal components are normal <momentum> fluxes, interpreted as electromagnetic <pressures> or tensions. Its off-diagonal components are tangential <momentum> fluxes, interpreted as shear stresses.

Solved by gpt-5.6-sol high.

= c
{parent=36d}
{scope}

= Solution
{parent=c}

Since $\mathbf B=0$,
$$
\mathcal E(r)=\frac{\epsilon_0E(r)^2}{2}
=\begin{cases}
\displaystyle\frac{q^2r^2}{32\pi^2\epsilon_0R^6},&r\leq R,\\[4pt]
\displaystyle\frac{q^2}{32\pi^2\epsilon_0r^4},&r>R.
\end{cases}
$$
Radial integration gives
$$
W_{\rm in}=\frac{q^2}{40\pi\epsilon_0R},
\qquad
W_{\rm out}=\frac{q^2}{8\pi\epsilon_0R},
$$
and therefore
$$
W_{\rm em}=\frac{3q^2}{20\pi\epsilon_0R}.
$$

To first order in $v/c$, $\mathbf E'=\mathbf E$ and $\mathbf B'=-\mathbf v\times\mathbf E/c^2$. Thus
$$
g'_z=\frac{\epsilon_0v}{c^2}E^2\sin^2\theta.
$$
Using the angular average $\langle\sin^2\theta\rangle=2/3$ and $\int\epsilon_0E^2dV=2W_{\rm em}$,
$$
P_z=\frac{4}{3}\frac{W_{\rm em}}{c^2}v
=\frac{q^2v}{5\pi\epsilon_0Rc^2}.
$$

Solved by gpt-5.6-sol high.

= 37B
{parent=Paper 3}
{scope}
{title2=General Relativity}

= a
{parent=37b}
{scope}

= Solution
{parent=a}

Metric compatibility and the ordinary chain rule give
$$
\frac d{d\lambda}(g(U,W))
=T^\alpha\nabla_\alpha(g(U,W))
=g(\nabla_TU,W)+g(U,\nabla_TW),
$$
which in components is the required identity.

Solved by gpt-5.6-sol high.

= b
{parent=37b}
{scope}

= Solution
{parent=b}

The <vectors> $T=\partial_\tau x$ and $S=\partial_\sigma x$ are coordinate <vector> fields on the parameter surface, so their ordinary commutator vanishes. Torsion-freeness then gives
$$
[T,S]=\nabla_TS-\nabla_ST=0.
$$
Since each $\tau$-curve is geodesic, $\nabla_TT=0$. Apply the Ricci identity to $V=T$ and use $\nabla_ST=\nabla_TS$:
$$
(\nabla_T\nabla_TS)^\alpha
=R^\alpha{}_{\beta\mu\nu}T^\beta T^\mu S^\nu.
$$
Thus the desired form holds with
$$
E^\alpha{}_{\nu}=-R^\alpha{}_{\beta\mu\nu}T^\beta T^\mu.
$$

Solved by gpt-5.6-sol high.

= c
{parent=37b}
{scope}

= Solution
{parent=c}

First,
$$
\nabla_T(T\cdot S)=(\nabla_TT)\cdot S+T\cdot\nabla_TS
=T\cdot\nabla_ST=\frac12\nabla_S(T\cdot T)=0.
$$
Hence orthogonality at one point persists. Substituting the constant-curvature tensor into geodesic deviation gives
$$
R^\alpha{}_{\beta\mu\nu}T^\beta T^\mu S^\nu
=K\{T^\alpha(T\cdot S)-S^\alpha(T\cdot T)\}=KS^\alpha,
$$
because $T\cdot S=0$ and $T\cdot T=-1$. Therefore
$$
(\nabla_T\nabla_TS)^\alpha-KS^\alpha=0
$$
along the whole geodesic.

Solved by gpt-5.6-sol high.

= 38C
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=38c}
{scope}

= Solution
{parent=a}

With streamwise and normal <velocities> $u,v$, zero exterior <pressure> <gradient> gives the steady boundary-layer equations
$$
u_x+v_y=0,
\qquad uu_x+vu_y=\nu u_{yy}.
$$
At the organism,
$$
v(x,0)=0,
\qquad \rho\nu u_y(x,0)=Sx^{-1/2},
$$
and matching to the quiescent exterior requires $u(x,y)\to0$ as $y\to\infty$.

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

Let $U$ be the streamwise speed and $\delta$ the layer width. Continuity gives $v\sim U\delta/x$, while inertia--viscosity balance gives
$$
\frac{U^2}{x}\sim\frac{\nu U}{\delta^2}.
$$
The imposed stress gives $\rho\nu U/\delta\sim Sx^{-1/2}$. Eliminating $U$ yields
$$
\delta\sim\left(\frac{\rho\nu^2}{S}\right)^{1/3}x^{1/2},
$$
and
$$
U\sim\left(\frac{S^2}{\rho^2\nu}\right)^{1/3},
$$
which is independent of $x$.

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= Solution
{parent=c}

Set
$$
A=\left(\frac{\rho\nu^2}{S}\right)^{1/3},
\qquad U_0=\frac{\nu}{A^2},
\qquad \eta=\frac{y}{A\sqrt x},
$$
and choose stream <function>
$$
\psi=U_0A\sqrt x,f(\eta).
$$
Then
$$
u=U_0f'(\eta),
\qquad
v=-\frac{U_0A}{2\sqrt x}\{f-\eta f'\}.
$$
Substitution gives
$$
f'''+\frac12ff''=0,
$$
with
$$
f(0)=0,
\qquad f''(0)=1,
\qquad f'(\infty)=0.
$$
Treat $f'(0)$ as a shooting parameter, integrate the initial-value problem numerically, and adjust it until the far-field condition holds. The adjacent-fluid speed is then $u(x,0)=U_0f'(0)$.

Solved by gpt-5.6-sol high.

= 39D
{parent=Paper 3}
{scope}
{title2=Waves}

= a
{parent=39d}
{scope}

= Solution
{parent=a}

The linearized horizontal and vertical <momentum>, density, and incompressibility equations are
$$
\rho_0u_t=-p_x,
\qquad \rho_0w_t=-p_z-g\rho',
\qquad \rho'_t+w\rho_0'=0,
\qquad u_x+w_z=0.
$$
Differentiate the <momentum> equations in time, take $\partial_x$ of the vertical equation minus $\partial_z$ of the horizontal one, and use incompressibility to eliminate $u,p,$ and $\rho'$. Neglecting the stated slow-background term gives
$$
\nabla^2w_{tt}+N^2(z)w_{xx}=0,
\qquad
N^2(z)=-\frac g{\rho_0}\frac{d\rho_0}{dz}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=39d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

In coordinates moving with the background fluid, $x'=x-Ut$. The stationary hill profile becomes
$$
h=h_0\sin(kx'+kUt),
$$
so in the convention $e^{i(kx'-\omega t)}$ its frequency is $\omega=-kU$; its physical angular frequency has magnitude $kU$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The <boundary condition> gives complex amplitude $w_0=Ukh_0$. Substitution of the <internal gravity wave> form
$$
w=w_0e^{i(kx+mz-\omega t)}
$$
into the internal-wave equation gives
$$
\omega^2=\frac{N^2k^2}{k^2+m^2}.
$$
Since $|\omega|=Uk$,
$$
m^2=\frac{N^2}{U^2}-k^2.
$$
For $k<N/U$, choose the sign of $m$ that gives upward energy propagation; with $\omega=-Uk$, this is $m>0$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Horizontal <momentum> and incompressibility give
$$
\widehat p=\rho_0\frac{\omega}{k}\widehat u
=-\rho_0\frac{\omega m}{k^2}\widehat w.
$$
Therefore
$$
\langle\overline I_z\rangle_0
=\frac12\operatorname{Re}(\widehat p\widehat w^*)
=-\frac{\rho_0\omega m}{2k^2}w_0^2.
$$
For upward radiation take $\omega=-Uk$ and $m=\sqrt{N^2/U^2-k^2}>0$, obtaining
$$
\langle\overline I_z\rangle_0
=\frac12\rho_0U^2kh_0^2\sqrt{N^2-U^2k^2}>0.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

If $k>N/U$, then $m=i\alpha$ with $\alpha>0$ chosen so that the disturbance decays as $z\to\infty$. The <pressure> amplitude is then ninety degrees out of phase with $w$, since $\widehat p/\widehat w$ is purely imaginary. Hence
$$
\langle\overline I_z\rangle_0
=\frac12\operatorname{Re}(\widehat p\widehat w^*)=0.
$$

Solved by gpt-5.6-sol high.

= v
{parent=b}
{scope}

= Solution
{parent=v}

The <mountain-wave cutoff> follows because the hills force intrinsic frequency $Uk$, whereas internal gravity waves require $|\omega|\leq N$. Thus $k_c=N/U$ separates long hills, which radiate vertically propagating waves and carry energy upward, from short hills, whose response is evanescent and carries no mean vertical energy flux.

Solved by gpt-5.6-sol high.

= 40A
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}

= a
{parent=40a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Writing the error as $e_k=x_k-x^*$ gives
$$
e_{k+1}=He_k
$$
provided the fixed-point consistency condition
$$
x^*=Hx^*+v
$$
holds. Convergence for every starting <vector> is therefore equivalent to $H^k\to0$, which in finite dimensions is equivalent to
$$
\rho(H)<1.
$$
These two conditions are necessary and sufficient.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Let $D=\operatorname{diag}(A)$. The <Jacobi method> is
$$
x_{k+1}=(I-D^{-1}A)x_k+D^{-1}b,
$$
so $H=I-D^{-1}A$ and $v=D^{-1}b$. If $A$ is strictly diagonally dominant, every Gershgorin disc of $H$ is centered at zero and has radius
$$
\sum_{j\ne i}\frac{|a_{ij}|}{|a_{ii}|}<1.
$$
Every <eigenvalue> therefore has <modulus> below one, so $\rho(H)<1$ and part (i) proves convergence.

Solved by gpt-5.6-sol high.

= b
{parent=40a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The <gradient> is $\nabla f(x)=Ax-b=A(x-x^*)$. Completing the square gives
$$
f(x)-f(x^*)=\frac12(x-x^*)^TA(x-x^*)>0
$$
for $x\ne x^*$, so $x^*$ is the unique global minimizer. Since $x-x^*=A^{-1}\nabla f(x)$,
$$
f(x)-f(x^*)=\frac12\nabla f(x)^TA^{-1}\nabla f(x).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Since $\nabla f(x_k)=-r_k$, exact line search minimizes $f(x_k+tr_k)$. <Differentiation> gives
$$
0=r_k^T(Ax_k-b)+t,r_k^TAr_k
=-r_k^Tr_k+t,r_k^TAr_k,
$$
so
$$
t_k=\frac{r_k^Tr_k}{r_k^TAr_k}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The <exact line search for a positive-definite quadratic> gives
$$
f(x_{k+1})-f(x_k)
=-\frac12\frac{(r_k^Tr_k)^2}{r_k^TAr_k}.
$$
Using
$$
f(x_k)-f(x^*)=\frac12r_k^TA^{-1}r_k
$$
produces the stated ratio formula. If $l$ and $L$ are the extreme <eigenvalues>,
$$
r_k^TAr_k\leq Lr_k^Tr_k,
\qquad
r_k^TA^{-1}r_k\leq l^{-1}r_k^Tr_k.
$$
Hence the fraction in the ratio is at least $l/L$, and iteration yields
$$
f(x_k)-f(x^*)\leq(1-l/L)^k\{f(x_0)-f(x^*)\}.
$$
When $n=1$, $l=L$, so the right side is zero after one iteration: exact line search reaches the solution in a single step.

Solved by gpt-5.6-sol high.