= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperii_4_2023.pdf

= 1G
{parent=Paper 4}
{scope}
{title2=Number Theory √}

= Solution
{parent=1G}

Apply the <continued-fraction algorithm>. Since $3<\sqrt{11}<4$,
$$
x_0=\sqrt{11},\qquad a_0=3.
$$
The successive complete quotients are
$$
x_1=\frac1{\sqrt{11}-3}
=\frac{\sqrt{11}+3}{2},
\qquad a_1=3,
$$
and
$$
x_2=\frac1{x_1-3}
=\sqrt{11}+3,
\qquad a_2=6.
$$
Taking one more reciprocal returns to $x_1$, so the <continued fraction of the square root of eleven> is
$$
\boxed{\sqrt{11}=[3;\overline{3,6}].}
$$

Write
$$
z_n=p_n+q_n\sqrt{11}.
$$
The first convergents are $p_0/q_0=3$ and $p_1/q_1=10/3$, whence
$$
z_1=10+3\sqrt{11}
=\frac{3+\sqrt{11}}2(3+\sqrt{11})
=\frac{3+\sqrt{11}}2z_0.
$$
The <continued fraction convergent> recurrence is
$$
z_{n+1}=a_{n+1}z_n+z_{n-1},
$$
where $a_{n+1}=3$ for even $n$ and $a_{n+1}=6$ for odd $n$.

Set
$$
\alpha=3+\sqrt{11},
\qquad
\beta=\frac{3+\sqrt{11}}2.
$$
If $n$ is even and $z_{n+1}=\beta z_n$, then
$$
z_{n+2}=6z_{n+1}+z_n
=\left(6+\frac1\beta\right)z_{n+1}
=(3+\sqrt{11})z_{n+1}
=\alpha z_{n+1},
$$
because $\beta^{-1}=\sqrt{11}-3$. If $n$ is odd and $z_{n+1}=\alpha z_n$, then
$$
z_{n+2}=3z_{n+1}+z_n
=\left(3+\frac1\alpha\right)z_{n+1}
=\frac{3+\sqrt{11}}2z_{n+1}
=\beta z_{n+1},
$$
because $\alpha^{-1}=(\sqrt{11}-3)/2$. The base case and induction prove the <alternating multiplier recurrence for convergents of the square root of eleven>:
$$
\boxed{
p_{n+1}+q_{n+1}\sqrt{11}
=\begin{cases}
(3+\sqrt{11})(p_n+q_n\sqrt{11}),&n\text{ odd},\\[2mm]
\dfrac{3+\sqrt{11}}2(p_n+q_n\sqrt{11}),&n\text{ even}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 4}
{scope}
{title2=Topics In Analysis}

= i
{parent=2f}
{scope}

= Solution
{parent=i}

Let
$$
h(x)=f(x)-x.
$$
The function $h$ is <continuous function>[continuous], with
$$
h(0)=f(0)\geq0,
\qquad
h(1)=f(1)-1\leq0.
$$
The <intermediate value theorem> gives $c\in[0,1]$ such that $h(c)=0$, and hence $f(c)=c$. This proves the <fixed-point property of a closed interval>.

Now suppose $0,1\in f([0,1])$. For any $y\in[0,1]$, choose $a,b\in[0,1]$ with $f(a)=0$ and $f(b)=1$. Applying the intermediate value theorem to $f$ on the interval with endpoints $a,b$ gives a point $x$ with $f(x)=y$. Thus $f$ is a <surjective function>.

Solved by gpt-5.6-sol high.

= ii
{parent=2f}
{scope}

= Solution
{parent=ii}

Let $h:A\to B$ be a <homeomorphism>, and let $f:B\to B$ be continuous. The conjugate map
$$
g=h^{-1}\circ f\circ h:A\to A
$$
is continuous. By hypothesis it has a fixed point $a\in A$. Therefore
$$
h^{-1}(f(h(a)))=a,
$$
and applying $h$ gives $f(h(a))=h(a)$. Hence every continuous self-map of $B$ has a fixed point. This is the <homeomorphism invariance of the fixed-point property>.

Solved by gpt-5.6-sol high.

= iii
{parent=2f}
{scope}

= Solution
{parent=iii}

The <Brouwer fixed-point theorem> for the closed unit disc $\overline D\subset\mathbb R^2$ states that every continuous map
$$
f:\overline D\to\overline D
$$
has a fixed point.

Solved by gpt-5.6-sol high.

= iv
{parent=2f}
{scope}

= Solution
{parent=iv}

The closed unit disc has the <fixed-point property> by the <Brouwer fixed-point theorem>. The annulus admits the <fixed-point-free rotation of an annulus>, for example
$$
R(x,y)=(-x,-y),
$$
which is continuous, maps the annulus to itself, and has no fixed point because the origin is not in the annulus. If the disc and annulus were <homeomorphic>, the <homeomorphism invariance of the fixed-point property> would give the annulus the fixed-point property, a contradiction. Therefore they are not homeomorphic.

Solved by gpt-5.6-sol high.

= v
{parent=2f}
{scope}

= Solution
{parent=v}

No. Take
$$
B=[0,1]\times\{0\}\subset\mathbb R^2.
$$
It contains more than one point and has the <fixed-point property of a closed interval>, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a <cut point>. Deleting one point from a two-dimensional closed disc leaves a <path-connected space>. Since being a cut point is preserved by <homeomorphism>[homeomorphisms], the two spaces cannot be homeomorphic. This gives the required <counterexample>.

Solved by gpt-5.6-sol high.

= 3I
{parent=Paper 4}
{scope}
{title2=Coding and Cryptography}

= a
{parent=3i}
{scope}

= Solution
{parent=a}

For the standard dot product on $\mathbb F_2^n$, the <dual code> is
$$
\boxed{
C^\perp=\{y\in\mathbb F_2^n:y\cdot c=0
\text{ for every }c\in C\}.}
$$
It is an intersection of kernels of <linear function>[linear functions], and is therefore a <linear code>.

Let $\sigma$ denote cyclic right shift. If $y\in C^\perp$ and $c\in C$, then
$$
(\sigma y)\cdot c=y\cdot(\sigma^{-1}c).
$$
A <cyclic code> is closed under both $\sigma$ and $\sigma^{-1}$, so $\sigma^{-1}c\in C$ and the right-hand side vanishes. Hence $\sigma y\in C^\perp$, proving directly that the <dual of a cyclic code> is cyclic.

Identify words with polynomials in $\mathbb F_2[X]/(X^n-1)$. If the <generator polynomial of a cyclic code> is the monic divisor $g(X)$ and
$$
g(X)h(X)=X^n-1,
$$
then the generator polynomial of $C^\perp$ is the monic reciprocal
$$
\boxed{g^\perp(X)=h^*(X)
=X^{\deg h}h(X^{-1})/h(0).}
$$
Equivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.

Solved by gpt-5.6-sol high.

= b
{parent=3i}
{scope}

= Solution
{parent=b}

Over $\mathbb F_2$, subtraction equals addition. The polynomial $X^7-1$ has the factorization
$$
\boxed{
X^7-1=(X+1)(X^3+X+1)(X^3+X^2+1).}
$$
The two cubic factors have no root in $\mathbb F_2$, so they are irreducible, and they are reciprocal to one another. Put
$$
f=X^3+X+1,
\qquad
f^*=X^3+X^2+1.
$$
Every cyclic code corresponds to one monic divisor of $X^7-1$. The complete list is
$$
\begin{array}{c|c|c}
g(X)&\text{parameters}&\text{code}\\ \hline
1 &[7,7,1]&\text{whole-space linear code}\\
X+1 &[7,6,2]&\text{even-weight binary code}\\
f &[7,4,3]&\text{Hamming code}\\
f^* &[7,4,3]&\text{reversed Hamming code}\\
(X+1)f &[7,3,4]&\text{binary simplex code}\\
(X+1)f^* &[7,3,4]&\text{reversed binary simplex code}\\
ff^*=1+X+\cdots+X^6 &[7,1,7]&\text{binary repetition code}\\
(X+1)ff^*=X^7-1 &[7,0]&\text{zero code}.
\end{array}
$$
Thus the two cubic generators give the two cyclic coordinate orientations of Hamming's $[7,4,3]$ code. Their duals have generators $(X+1)f$ and $(X+1)f^*$ and are the $[7,3,4]$ simplex codes. The remaining four are the familiar whole-space, even-weight, repetition, and zero codes. This is the <binary cyclic codes of length seven>[complete classification of binary cyclic codes of length seven].

Solved by gpt-5.6-sol high.

= 4I
{parent=Paper 4}
{scope}
{title2=Automata & Formal Languages}

= i
{parent=4i}
{scope}

= Solution
{parent=i}

A grammar $G=(\Sigma,V,P,S)$ is a right-linear <regular grammar> when every production has one of the forms
$$
\boxed{A\to wB\quad\text{or}\quad A\to w,}
$$
where $A,B\in V$ and $w\in\Sigma^*$. Thus a production has at most one variable on its right-hand side, and that variable is at the right end.

Solved by gpt-5.6-sol high.

= ii
{parent=4i}
{scope}

= Solution
{parent=ii}

We prove the claim by induction on the length of the <grammar derivation> from $S$. At length zero the sentential form is $S=\epsilon S$, which has the required form.

Suppose the current sentential form is $wA$. Because $w$ is terminal, the only symbol to which a production can apply is $A$. A production $A\to uB$ produces
$$
wA\Rightarrow wuB,
$$
again a terminal word followed by one variable. A production $A\to u$ produces the terminal word $wu$. Once a completely terminal word is reached, no further production can apply. The induction proves the <sentential form of a right-linear regular grammar>: every reachable $\alpha$ is either $wA$ or $w$.

Solved by gpt-5.6-sol high.

= iii
{parent=4i}
{scope}

= Solution
{parent=iii}

In every derivation beginning at $wA$, the terminal prefix $w$ remains unchanged because no production has a terminal on its left-hand side. By the sentential-form result, each step rewrites the sole variable at the right and appends terminals immediately before its successor variable. Hence
$$
wA\Rightarrow^*vB
$$
forces $v=wu$ for some $u\in\Sigma^*$. Removing the inert prefix $w$ from every sentential form in the derivation gives
$$
\boxed{A\Rightarrow^*uB.}
$$

Solved by gpt-5.6-sol high.

= iv
{parent=4i}
{scope}

= Solution
{parent=iv}

An accepting derivation in a right-linear grammar has the form
$$
S\Rightarrow w_1A_1\Rightarrow
w_1w_2A_2\Rightarrow\cdots
\Rightarrow w_1\cdots w_m,
$$
with one active variable until the final terminal production.

If no variable repeated on any accepting derivation, an accepting derivation could contain at most $|V|$ variable occurrences. Since the production set is finite, only finitely many such derivations and hence finitely many terminal words would exist. The hypothesis that $\mathcal L(G)$ is infinite therefore supplies an accepting derivation in which some variable $A$ occurs twice.

Split this derivation at the two occurrences:
$$
S\Rightarrow^*uA,\qquad
A\Rightarrow^*vA,\qquad
A\Rightarrow^*z,
$$
where $u,v,z\in\Sigma^*$. The first segment makes $A$ an <accessible variable of a regular grammar>, the middle segment makes it a <looping variable of a regular grammar>, and the last makes it a <terminable variable of a regular grammar>. Thus $A$ has all three properties, proving the <accessible looping terminable variable criterion>.

Solved by gpt-5.6-sol high.

= 5J
{parent=Paper 4}
{scope}
{title2=Statistical Modelling}

= i
{parent=5j}
{scope}

= Solution
{parent=i}

Use the <R linear-model formula> with only its implicit intercept:

`fit1 <- lm(log(price) ~ 1, data = cars)`

This models every log-price by one common mean plus an error term.

Solved by gpt-5.6-sol high.

= ii
{parent=5j}
{scope}

= Solution
{parent=ii}

The other quantitative columns are mpg, psngr, length, width, and weight, so use

`fit2 <- lm(log(price) ~ mpg + psngr + length + width + weight, data = cars)`

This is a <normal linear model> with an intercept and five slopes.

Solved by gpt-5.6-sol high.

= iii
{parent=5j}
{scope}

= Solution
{parent=iii}

Treat make as a categorical predictor, converting it to a factor explicitly if necessary:

`fit3 <- lm(log(price) ~ mpg + psngr + length + width + weight + factor(make), data = cars)`

The models are nested. The first <nested-model F-test> compares fit2 with the intercept-only fit1. Adding the five quantitative predictors reduces the residual sum of squares from $8584.0$ to $3349.1$, with
$$
F=69.7334,\qquad p<2.2\times10^{-16}.
$$
There is overwhelming evidence that these quantitative car properties jointly improve the model.

The second test compares fit3 with fit2. Adding make reduces the residual sum of squares further to $840.8$, with
$$
F=5.3891,\qquad p=2.541\times10^{-8}.
$$
Thus manufacturer has a highly significant effect even after adjusting for the five quantitative properties, and fit3 is preferred among these nested models.

There are $93$ observations because the intercept-only model has $92=93-1$ residual degrees of freedom. The make term uses $31$ additional degrees of freedom. By the <degrees of freedom of a factor predictor>, a factor with $m$ represented levels contributes $m-1$ degrees of freedom when an intercept is present. Hence
$$
m-1=31,
\qquad
\boxed{m=32}
$$
unique manufacturers occur in the dataset.

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 4}
{scope}
{title2=Mathematical Biology}

= Solution
{parent=6C}

Put $a=f'(0)>0$. Linearizing the <morphogen reaction-diffusion equation> at $C=0$ gives
$$
C_t=DC_{xx}+aC.
$$
Seek separated modes $C(x,t)=X(x)e^{\lambda t}$. Then
$$
DX''+aX=\lambda X,
\qquad
X(0)=0,\quad X'(L)=0.
$$
The <mixed Dirichlet-Neumann modes on an interval> are
$$
X_n(x)=\sin(k_nx),
\qquad
k_n=\frac{(n+1/2)\pi}{L},
\qquad n=0,1,2,\ldots.
$$
Indeed, the Dirichlet condition selects sine functions, while the Neumann condition requires $\cos(k_nL)=0$. Their growth rates are
$$
\lambda_n=a-Dk_n^2
=a-D\frac{(n+1/2)^2\pi^2}{L^2}.
$$
The largest growth rate is the lowest mode,
$$
\lambda_0=a-\frac{D\pi^2}{4L^2}.
$$
Linear stability requires $\lambda_0<0$, after which all higher modes also decay. Therefore the <critical length for a linearly growing morphogen> is
$$
\boxed{
L<\frac\pi2\sqrt{\frac{D}{f'(0)}}.}
$$
Equality gives a neutral lowest mode, while a larger domain is linearly unstable.

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 4}
{scope}
{title2=Further Complex Methods}

= Solution
{parent=7E}

The <Papperitz symbol>
$$
P\left\{
\begin{array}{ccccc}
0&1&\infty&&\\
0&0&a&z&\\
1-c&c-a-b&b&&
\end{array}\right\}
$$
specifies a second-order <Fuchsian differential equation>. The first row lists its three distinct regular singular points $0,1,\infty$; $z$ is the independent variable. The two entries below each singular point are its <characteristic exponent at a regular singular point>[characteristic exponents]. Thus local solutions have leading behaviors
$$
1,\ z^{1-c}\quad(z\to0),
$$
$$
1,\ (1-z)^{c-a-b}\quad(z\to1),
$$
and, with the usual convention at infinity,
$$
z^{-a},\ z^{-b}\quad(z\to\infty).
$$
When an exponent difference is an integer, a logarithmic second solution may replace the naive second Frobenius power. The entries obey the <Fuchs relation>
$$
0+(1-c)+0+(c-a-b)+a+b=1.
$$
For a second-order equation with exactly three regular singular points, these exponent data determine the equation up to multiplication by a nonzero function; there is no accessory parameter. The symbol is therefore the one for the <Gauss hypergeometric equation>. Its distinguished solution $F(a,b;c;z)$ is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes $c\notin\{0,-1,-2,\ldots\}$, with exceptional parameter values handled separately or by continuation.

Now put
$$
u=\frac{z}{z-1},
\qquad
Y(z)=F(a,c-b;c;u).
$$
The hypergeometric equation in $u$ has exponent pairs
$$
(0,1-c)\text{ at }u=0,\qquad
(0,b-a)\text{ at }u=1,\qquad
(a,c-b)\text{ at }u=\infty.
$$
The <Möbius transformation of a Papperitz symbol> sends
$$
u=0,1,\infty
\quad\longleftarrow\quad
z=0,\infty,1,
$$
so $Y$ has symbol
$$
P\left\{
\begin{array}{ccccc}
0&1&\infty&&\\
0&a&0&z&\\
1-c&c-b&b-a&&
\end{array}\right\}.
$$

On the other hand, $F(a,b;c;z)$ has exponent pairs
$$
(0,1-c),\qquad(0,c-a-b),\qquad(a,b)
$$
at $0,1,\infty$. Multiplication by $(1-z)^a$ applies the <dependent-variable rescaling of a Papperitz symbol>: it adds $a$ to both exponents at $z=1$ and subtracts $a$ from both at infinity. Hence
$$
(1-z)^aF(a,b;c;z)
$$
has exactly the same three exponent pairs as $Y$.

Both functions are analytic near $z=0$ and equal one at $z=0$, so uniqueness of the normalized exponent-zero hypergeometric solution gives the <Pfaff transformation>
$$
\boxed{
F\left(a,c-b;c;\frac{z}{z-1}\right)
=(1-z)^aF(a,b;c;z).}
$$
The identity first holds near zero with the branch of $(1-z)^a$ equal to one there, and then extends by <analytic continuation> on any domain where compatible branches are chosen.

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 4}
{scope}
{title2=Classical Dynamics}

= Solution
{parent=8D}

An adiabatic invariant remains constant to leading order when a system parameter changes on a timescale much longer than one period. For a periodic one-degree-of-freedom system, the <adiabatic invariance of the action> supplies such an invariant.

The momentum magnitude between collisions is $p_0=\sqrt{2mE}$. A complete orbit crosses the gap once in each direction, so the unnormalized action used in the question is
$$
\boxed{I=\oint p\,dq=2Lp_0=2L\sqrt{2mE}.}
$$
Consequently
$$
\frac{dI}{dE}=\frac{2Lm}{\sqrt{2mE}}
=\frac{2L}{\sqrt{2E/m}}=T,
$$
because the speed is $\sqrt{2E/m}$ and the round-trip distance is $2L$.

When $L(t)$ varies adiabatically, conservation of the <adiabatic particle between moving parallel walls> action gives
$$
\boxed{E(t)=E(0)\left(\frac{L(0)}{L(t)}\right)^2.}
$$
A moving wall does work at each collision: an approaching wall raises the particle's energy and a receding wall lowers it. Elasticity holds in the instantaneous rest frame of the wall.

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 4}
{scope}
{title2=Cosmology}

= Solution
{parent=9B}

The <flatness problem> is that the observed <cosmological density parameter> is close to $\Omega=1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The <cosmological perfect-fluid continuity equation> is
$$
\dot\rho+3H(\rho+P)=0.
$$
Consequently
$$
\frac d{dt}(\rho a^2)
=a^2\dot\rho+2a\dot a\rho
=-Ha^2(\rho+3P).
$$
For an expanding universe $H>0$. If
$$
\rho+3P<0,
$$
then $\rho a^2$ increases. The <Friedmann equation> implies
$$
\left|\Omega^{-1}-1\right|
=\frac{\text{constant}\times |k|}{\rho a^2}.
$$
It therefore decreases during this period, driving $\Omega$ toward one. By the <Friedmann acceleration equation>, the same condition gives $\ddot a>0$ when the cosmological constant is included in the effective fluid. This is the <inflationary solution of the flatness problem>.

For the scalar field, the <slow-roll approximation> means
$$
\dot\phi^2\ll V(\phi),
\qquad
|\ddot\phi|\ll3H|\dot\phi|.
$$
The equations reduce to
$$
3H^2\simeq m^2\phi^2,
\qquad
3H\dot\phi\simeq-2m^2\phi.
$$
Taking the positive-field branch gives $H\simeq m\phi/\sqrt3$, and therefore
$$
\dot\phi\simeq-\frac{2m}{\sqrt3}.
$$
With $\phi(0)=\phi_i$,
$$
\boxed{\phi(t)=\phi_i-\frac{2m}{\sqrt3}t.}
$$
Now integrate $H=\dot a/a\simeq m\phi/\sqrt3$:
$$
\log\frac{a(t)}{a_i}
=\int_0^t\frac m{\sqrt3}
\left(\phi_i-\frac{2m}{\sqrt3}s\right)ds
=\frac{m\phi_i}{\sqrt3}t-\frac{m^2t^2}{3}.
$$
Thus the <quadratic-potential slow-roll solution> is
$$
\boxed{
a(t)=a_i\exp\left[
\frac{m\phi_i}{\sqrt3}t-\frac{m^2t^2}{3}
\right].}
$$
Finally,
$$
\phi_i^2-\phi(t)^2
=\frac{4m\phi_i}{\sqrt3}t-\frac{4m^2}{3}t^2,
$$
so equivalently
$$
\boxed{
a(t)=a_i\exp\left[
\frac{\phi_i^2-\phi(t)^2}{4}
\right].}
$$

Solved by gpt-5.6-sol high.

= 10D
{parent=Paper 4}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10d}
{scope}

= Solution
{parent=a}

Let the unknown qubit be $|\alpha\rangle=a|0\rangle+b|1\rangle$. Alice holds this qubit and her half of the shared <Bell state> $|\Psi^+\rangle$, while Bob holds the other half.

Alice performs a measurement of her two qubits in the Bell basis and sends the two-bit outcome to Bob through the classical channel. For the shared state $|\Psi^+\rangle$, the <teleportation with the psi-plus Bell state> correction table is
$$
\begin{array}{c|c|c}
\text{Alice's outcome}&\text{Bob's state}&\text{Bob applies}\\ \hline
\Phi^+&X|\alpha\rangle&X\\
\Phi^-&XZ|\alpha\rangle&ZX\\
\Psi^+&|\alpha\rangle&I\\
\Psi^-&Z|\alpha\rangle&Z
\end{array}
$$
up to physically irrelevant global phases. Here $X$ and $Z$ are the <Pauli X gate> and <Pauli Z gate>. After the indicated correction, Bob's qubit is $|\alpha\rangle$. This is <quantum teleportation>; no physical copy of the unknown qubit is sent.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

Let $D$ denote Alice's unknown input qubit and $A$ her GHZ qubit. Expanding the four-qubit state in the <Bell state>[Bell basis] of $D,A$ gives
$$
|\alpha\rangle_D|\operatorname{GHZ}\rangle_{ABC}
=\frac12\Big[
|\Phi^+\rangle_{DA}(a|00\rangle+b|11\rangle)_{BC}
+|\Phi^-\rangle_{DA}(a|00\rangle-b|11\rangle)_{BC}
$$
$$
\qquad
+|\Psi^+\rangle_{DA}(b|00\rangle+a|11\rangle)_{BC}
+|\Psi^-\rangle_{DA}(b|00\rangle-a|11\rangle)_{BC}
\Big],
$$
where a global sign in the last term is immaterial.

For outcomes $\Phi^+$ and $\Phi^-$, Bob and Charlie already have
$$
|\phi_1\rangle=a|00\rangle+b|11\rangle,
\qquad
|\phi_2\rangle=a|00\rangle-b|11\rangle.
$$
For either $\Psi$ outcome, both apply the <Pauli X gate>. Since $X\otimes X$ exchanges $|00\rangle$ and $|11\rangle$, the $\Psi^+$ branch becomes $|\phi_1\rangle$ and the $\Psi^-$ branch becomes $|\phi_2\rangle$, up to a global phase. This proves the <Bell measurement on one qubit and one leg of a GHZ state> result.

Finally, tracing out Charlie gives
$$
\rho_B=|a|^2|0\rangle\langle0|+|b|^2|1\rangle\langle1|
$$
for either sign. This reduced state has rank two exactly when $a\ne0$ and $b\ne0$. By the <entanglement criterion for a two-term correlated state>,
$$
\boxed{|\phi_1\rangle\text{ and }|\phi_2\rangle
\text{ are entangled iff }a\ne0\text{ and }b\ne0.}
$$

Solved by gpt-5.6-sol high.

= 11G
{parent=Paper 4}
{scope}
{title2=Number Theory}

= a
{parent=11g}
{scope}

= Solution
{parent=a}

For an odd prime $p$, the <Legendre symbol> is
$$
\left(\frac ap\right)=
\begin{cases}
0,&p\mid a,\\
1,&a\not\equiv0\pmod p\text{ is a quadratic residue},\\
-1,&a\text{ is a quadratic nonresidue}.
\end{cases}
$$
<Euler criterion> states that
$$
\boxed{
a^{(p-1)/2}\equiv\left(\frac ap\right)\pmod p.}
$$
If $p\mid a$, both sides vanish. Otherwise choose a <primitive root> $g$ and write $a\equiv g^m$. Since $g^{(p-1)/2}\equiv-1$,
$$
a^{(p-1)/2}\equiv(-1)^m.
$$
The power $g^m$ is a square exactly when $m$ is even, proving the criterion. Taking $a=-1$ gives the <first supplementary law for quadratic reciprocity>
$$
\boxed{\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=11g}
{scope}

= Solution
{parent=b}

Because $I$ and $A$ commute, the matrix binomial theorem gives
$$
(I+A)^p=\sum_{j=0}^p\binom pjA^j.
$$
For prime $p$, every intermediate coefficient $\binom pj$ is divisible by $p$, proving the <prime-power binomial congruence for a matrix>
$$
(I+A)^p\equiv I+A^p\pmod p.
$$

For
$$
A=\begin{pmatrix}0&-1\\1&0\end{pmatrix},
\qquad A^2=-I,
$$
one has $(I+A)^2=2A$ and $(I+A)^4=-4I$. If $p=4k+1$, then
$$
(I+A)^p=(-4)^k(I+A)
\equiv I+A,
$$
so $(-4)^k\equiv1\pmod p$. If $p=4k-1$, then $A^p=-A$ and
$$
(I+A)^p=(-4)^k(I+A)^{-1}
=\frac{(-4)^k}{2}(I-A)
\equiv I-A,
$$
so $(-4)^k\equiv2\pmod p$.

By <Euler criterion>, $(\frac2p)\equiv2^{(p-1)/2}\pmod p$. In either case the preceding congruence gives $(\frac2p)=(-1)^k$. Since $(p^2-1)/8$ has the same parity as $k$ for $p=4k\pm1$, this is the <second supplementary law for quadratic reciprocity>
$$
\boxed{\left(\frac2p\right)=(-1)^{(p^2-1)/8}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=11g}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Write $\chi(a)=(\frac ap)$. When $p\equiv1\pmod4$, the <first supplementary law for quadratic reciprocity> gives $\chi(-1)=1$, so $\chi(p-a)=\chi(a)$. Pairing $a$ with $p-a$ gives
$$
T=p\sum_{a=1}^{(p-1)/2}\chi(a).
$$
The sum of the nontrivial character $\chi$ over all nonzero residue classes is zero. Because its two half-sums are equal when $\chi(-1)=1$,
$$
0=\sum_{a=1}^{p-1}\chi(a)
=2\sum_{a=1}^{(p-1)/2}\chi(a).
$$
Hence
$$
\boxed{T=0.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

By <Euler criterion>,
$$
T\equiv\sum_{a=1}^{p-1}a^{(p+1)/2}\pmod p.
$$
Choose a <primitive root> $g$. For any integer $m$ not divisible by $p-1$,
$$
\sum_{a=1}^{p-1}a^m
=\sum_{j=0}^{p-2}g^{jm}
=\frac{g^{m(p-1)}-1}{g^m-1}
\equiv0\pmod p.
$$
For $p>3$, the exponent $m=(p+1)/2$ satisfies $0<m<p-1$. Therefore
$$
\boxed{T\equiv0\pmod p.}
$$
Together with part (i), this is the <weighted complete Legendre-symbol sum>.

Solved by gpt-5.6-sol high.

= d
{parent=11g}
{scope}

= Solution
{parent=d}

Put $H=(p-1)/2$. Because $p\equiv7\pmod8$, the supplementary laws give
$$
\left(\frac{-1}{p}\right)=-1,
\qquad
\left(\frac2p\right)=1.
$$
Let $R$ and $N$ be the quadratic residues and nonresidues, respectively, among $1,\ldots,H$, and put
$$
S=\sum_{r\in R}r.
$$
Since $-1$ is a nonresidue, the complete set $Q$ of least positive quadratic residues is
$$
Q=R\cup\{p-n:n\in N\}.
$$
If $u=|N|$, its sum is
$$
\sum_{q\in Q}q
=S+up-\sum_{n\in N}n
=2S+up-\frac{H(H+1)}2.
$$

Since $2$ is a residue, multiplication by two permutes $Q$ modulo $p$. Exactly the $u$ upper-half residues $p-n$ cross $p$ when doubled. Equality of the sums before and after reduction therefore gives
$$
\sum_{q\in Q}q
=2\sum_{q\in Q}q-up,
\qquad
\sum_{q\in Q}q=up.
$$
Comparing the two formulas yields
$$
2S=\frac{H(H+1)}2.
$$
As $H=(p-1)/2$,
$$
\boxed{
S=\frac{H(H+1)}4=\frac{p^2-1}{16}.}
$$
This is the <lower-half quadratic-residue sum for primes congruent to seven modulo eight>.

Solved by gpt-5.6-sol high.

= 12F
{parent=Paper 4}
{scope}
{title2=Topics In Analysis}

= a
{parent=12f}
{scope}

= Solution
{parent=a}

For $\mathbf a,\mathbf b\in\mathbb R^{n+1}$, the <triangle inequality> and the <Cauchy-Schwarz inequality> give
$$
\begin{aligned}
|S(\mathbf a)-S(\mathbf b)|
&\leq \lVert T(\mathbf a-\mathbf b)\rVert_\infty\\
&\leq \sum_{r=0}^n|a_r-b_r|
\leq \sqrt{n+1}\,\lVert\mathbf a-\mathbf b\rVert_2.
\end{aligned}
$$
Thus $S$ is a <continuous function>. The <unit sphere> in the finite-dimensional <Euclidean normed vector space> $\mathbb R^{n+1}$ is <compact>. Moreover, $S(\mathbf a)=0$ implies that the <polynomial> $T\mathbf a$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the <extreme value theorem>, it has a positive minimum
$$
\delta=\min_{\lVert\mathbf a\rVert_2=1}S(\mathbf a)>0.
$$

The <supremum norm> is homogeneous. For $\mathbf a\ne0$,
$$
\lVert T\mathbf a\rVert_\infty
=\lVert\mathbf a\rVert_2
 S\left(\frac{\mathbf a}{\lVert\mathbf a\rVert_2}\right)
\geq\delta\lVert\mathbf a\rVert_2.
$$
It follows directly that $\lVert T\mathbf a\rVert_\infty\to\infty$ whenever $\lVert\mathbf a\rVert_2\to\infty$.

Solved by gpt-5.6-sol high.

= b
{parent=12f}
{scope}

= Solution
{parent=b}

Write each <polynomial> of <degree of a polynomial>[degree] at most $n$ as $T\mathbf a$ and define
$$
F(\mathbf a)=\lVert T\mathbf a-f\rVert_\infty.
$$
This is a <continuous function>, and the reverse <triangle inequality> together with part (a) gives
$$
F(\mathbf a)\geq \lVert T\mathbf a\rVert_\infty-\lVert f\rVert_\infty
\geq\delta\lVert\mathbf a\rVert_2-\lVert f\rVert_\infty.
$$
Consequently $F$ is a <coercive function>. Choose $R$ so large that $F(\mathbf a)>F(\mathbf0)$ outside the closed <Euclidean ball> of radius $R$. That ball is <compact>, so the <extreme value theorem> supplies a minimizer $\mathbf a_*$ there. Then $P=T\mathbf a_*$ is a <best uniform approximation> to $f$ among polynomials of degree at most $n$.

Solved by gpt-5.6-sol high.

= c
{parent=12f}
{scope}

= Solution
{parent=c}

The <Chebyshev alternation theorem> says that a <polynomial> $P$ of <degree of a polynomial>[degree] at most $n$ is a <best uniform approximation> to $f\in C([0,1])$ if and only if there are $n+2$ points
$$
0\leq t_0<t_1<\cdots<t_{n+1}\leq1
$$
and a sign $\varepsilon\in\{1,-1\}$ for which
$$
f(t_j)-P(t_j)=\varepsilon(-1)^j\lVert f-P\rVert_\infty.
$$
This is the equiripple criterion.

To prove sufficiency, put $E=\lVert f-P\rVert_\infty$ and suppose that another polynomial $Q$ of degree at most $n$ satisfies $\lVert f-Q\rVert_\infty<E$. At every $t_j$, the difference $R=Q-P$ must have the same sign as $f-P$; hence its signs alternate at the $n+2$ ordered points. The <intermediate value theorem> then gives at least one distinct root of $R$ in every interval $(t_j,t_{j+1})$, for at least $n+1$ roots in all. The <Lagrange root bound over a field> forces the nonzero polynomial $R$ to have degree at least $n+1$, contradicting $\deg R\leq n$. Therefore no $Q$ gives a smaller error, and $P$ is best.

Solved by gpt-5.6-sol high.

= d
{parent=12f}
{scope}

= Solution
{parent=d}

As printed, this part contains an indexing error. The displayed product ends at $m-1$, but the hypotheses contain $m+1$ alternating intervals indexed by $0,\ldots,m$. On the first $m$ intervals the printed polynomial has sign $(-1)^i$, while on the last interval it has sign $(-1)^{m-1}$ rather than the required $(-1)^m$. For example, when $m=1$ the printed product is the constant $Q=1$, which cannot reduce a negative extremum of $g$ on the final interval. The intended polynomial is
$$
Q(t)=(-1)^m\prod_{j=1}^{m}(t-w_j),
\qquad w_j=\frac{v_{j-1}+u_j}{2}.
$$
We prove the stated conclusion with this correction.

Each root $w_j$ lies in the gap between the $(j-1)$st and $j$th active intervals. Therefore $Q$ has sign $(-1)^i$ throughout $[u_i,v_i]$. In particular, wherever $|g|=M$, the numbers $Q$ and $g$ have the same sign, and $Q$ is nonzero.

Let
$$
K_+=\{t:g(t)=M\},\qquad K_-=\{t:g(t)=-M\}.
$$
These are <compact sets>. On their union, the <continuous function> $|Q|$ has a positive minimum. Hence, throughout some open neighbourhood $U$ of $K_+\cup K_-$, subtracting a sufficiently small positive multiple $\eta Q$ moves $g$ strictly towards zero and gives $|g-\eta Q|<M$. On the compact complement $[0,1]\setminus U$, continuity gives a uniform margin $|g|\leq M-\epsilon$ for some $\epsilon>0$. Taking also
$$
\eta\lVert Q\rVert_\infty<\epsilon
$$
shows that $|g-eta Q|<M$ there. Thus
$$
\lVert\eta Q-g\rVert_\infty<M
$$
for every sufficiently small $\eta>0$.

It remains to prove necessity in the <Chebyshev alternation theorem>. Let $P$ be a best approximation of degree at most $n$, put $g=f-P$, and let $M=\lVert g\rVert_\infty$. If $g$ has no sequence of $n+2$ extrema with alternating signs, its sets of positive and negative extrema can be collected, from left to right, into $m+1$ alternating compact groups with $m\leq n$. Choose disjoint intervals $[u_i,v_i]$ containing those groups and separated by regions on which $|g|<M$. After replacing $g$ by $-g$ if necessary, they satisfy the displayed hypotheses. The corrected construction produces a polynomial $Q$ of degree $m\leq n$ and a small $\eta>0$ such that
$$
\lVert f-(P+\eta Q)\rVert_\infty
=\lVert g-\eta Q\rVert_\infty<M.
$$
This contradicts the assumed optimality of $P$. Therefore every best polynomial has the required $n+2$ alternating extrema.

Solved by gpt-5.6-sol high.

= 13J
{parent=Paper 4}
{scope}
{title2=Statistical Modelling}

= i
{parent=13j}
{scope}

= Solution
{parent=i}

Let $Y_i$ be the first team's goal count in match $i$, let $x_i^T$ be row $i$ of the <design matrix>, and let $\beta$ be the <regression coefficient> vector. The fitted <Poisson regression> assumes that the $Y_i$ are <independent random variables> with
$$
Y_i\sim\operatorname{Pois}(\mu_i),
\qquad \log\mu_i=x_i^T\beta.
$$
The <likelihood function> maximized by `glm` is therefore
$$
L(\beta;y)
=\prod_{i=1}^{64}
 \frac{\mu_i^{y_i}e^{-\mu_i}}{y_i!}
=\prod_{i=1}^{64}
 \frac{\exp\{y_i x_i^T\beta-e^{x_i^T\beta}\}}{y_i!}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=13j}
{scope}

= Solution
{parent=ii}

None of the displayed one-coefficient <Wald tests> for `team1_form` rejects at the $5\%$ <significance level>. That does not establish that formation has no effect: the relevant <null hypothesis> sets all nine displayed `team1_form` coefficients to zero simultaneously, and several individually imprecise estimates may still be jointly significant. One should compare the fitted model with the nested model omitting `team1_form` by an <analysis of deviance for nested generalized linear models>, equivalently a <likelihood-ratio test> of those joint restrictions.

Solved by gpt-5.6-sol high.

= iii
{parent=13j}
{scope}

= Solution
{parent=iii}

Writing $Y_i$ for `team1_goal` and $x_i$ for `team1_xg`, the first hypothesis concerns only the <expected value>:
$$
H_1:\quad \mathbb E(Y_i\mid x_i)=x_i
\quad\text{for every }i.
$$
The second specifies the entire conditional <probability distribution>:
$$
H_2:\quad Y_i\mid x_i\sim\operatorname{Pois}(x_i)
\quad\text{for every }i,
$$
with independence between matches as assumed in the question. The <Poisson limit theorem>, often called the law of small numbers in this setting, suggests a <Poisson distribution> when a goal count is the sum of many approximately independent rare scoring opportunities.

Solved by gpt-5.6-sol high.

= iv
{parent=13j}
{scope}

= Solution
{parent=iv}

Under $H_1$, the slope in this through-origin <linear regression> is $1$. Its observed <Wald statistic> is
$$
t=\frac{1.15790-1}{0.08643}=1.83.
$$
The corresponding two-sided <p-value> is about $0.07$, so the fit does not give evidence against $H_1$ at the $5\%$ <significance level>.

The usual least-squares <standard error> assumes constant conditional <variance>. Goal counts are counts, and under the natural <Poisson distribution> model their variance equals their mean $x_i$, so observations with different expected goals are <heteroscedastic>. The reported homoscedastic standard error is therefore doubtful; a heteroscedasticity-robust calculation or a suitable count model would be more credible.

Solved by gpt-5.6-sol high.

= v
{parent=13j}
{scope}

= Solution
{parent=v}

The fitted <Poisson regression> has
$$
Y_i\mid x_i\sim\operatorname{Pois}(\mu_i),
\qquad
\log\mu_i=\beta_0+\beta_1\log x_i,
$$
so that $\mu_i=e^{\beta_0}x_i^{\beta_1}$. Hypothesis $H_2$ requires both restrictions
$$
\beta_0=0,\qquad \beta_1=1,
$$
as well as the adequacy of the assumed <Poisson distribution>. The displayed interval addresses only the single slope restriction $\beta_1=1$. Even if the separate interval for the intercept also contains zero, separate one-parameter intervals do not implement the relevant <joint hypothesis test>, because the two coefficient estimators can be correlated. One should test $(\beta_0,\beta_1)=(0,1)$ jointly, for example by a <likelihood-ratio test>, and separately assess the Poisson goodness of fit. Thus the slope interval alone cannot decide whether $H_2$ is rejected at the $5\%$ level.

Solved by gpt-5.6-sol high.

= 14C
{parent=Paper 4}
{scope}
{title2=Mathematical Biology}

= Solution
{parent=14C}

This is a <birth-death process> with birth and death rates
$$
b_n=\gamma+\beta n,
\qquad
d_n=\alpha n(n-1).
$$
Its transition diagram has the two arrows
$$
n\xrightarrow{\ \gamma+\beta n\ }n+1,
\qquad
n\xrightarrow{\ \alpha n(n-1)\ }n-1.
$$
The gain into state $n$ comes from a death in state $n+1$ or a birth in state $n-1$, while the loss is the sum of both rates out of state $n$. The <birth-death master equation> is consequently
$$
\frac{\partial P(n,t)}{\partial t}
=\alpha n(n+1)P(n+1,t)
-[\alpha n(n-1)+\gamma+\beta n]P(n,t)
+[\gamma+\beta(n-1)]P(n-1,t).
$$

Multiplying by $n$, summing over the nonnegative integers, and shifting the summation indices shows that each birth contributes $+1$ and each death contributes $-1$. Thus the <first-moment equation of a birth-death process> gives
$$
\begin{aligned}
\frac{d\langle n\rangle}{dt}
&=\langle b_n-d_n\rangle\\
&=-\alpha\langle n^2\rangle
 +(\alpha+\beta)\langle n\rangle+\gamma.
\end{aligned}
$$
At a stationary state, use the <variance> identity $\langle n^2\rangle=\langle n\rangle^2+(\Delta n)^2$. Solving the resulting <quadratic equation> gives
$$
\langle n\rangle
=\frac{\alpha+\beta}{2\alpha}
\mathbin{\pm}
\sqrt{\frac{(\alpha+\beta)^2}{4\alpha^2}
 +\frac\gamma\alpha-(\Delta n)^2}.
$$
The minus root is inadmissible whenever it is negative, namely when $(\Delta n)^2<\gamma/\alpha$. Equality would give zero mean, which is also incompatible with a positive immigration rate $\gamma$, so for $\gamma>0$ the minus branch requires $(\Delta n)^2>\gamma/\alpha$ even to be a possible mean.

For a continuum approximation, write $b(n)=\gamma+\beta n$ and $d(n)=\alpha n(n-1)$. Applying the <Kramers-Moyal expansion> to the two gain terms and retaining derivatives through second order gives the <Fokker-Planck equation>
$$
\frac{\partial P}{\partial t}
=\frac{\partial}{\partial n}[g(n)P]
 +\frac12\frac{\partial^2}{\partial n^2}[h(n)P],
$$
where the negative drift and infinitesimal jump variance are
$$
\begin{aligned}
g(n)&=d(n)-b(n)
=\alpha n^2-(\alpha+\beta)n-\gamma,\\
h(n)&=d(n)+b(n)
=\alpha n^2+(\beta-\alpha)n+\gamma.
\end{aligned}
$$
This truncation requires the typical population and the scale on which $P$ varies to be much larger than the unit jump size, with the rates varying smoothly across that scale.

The positive zero of $g$ is
$$
n_*
=\frac{\alpha+\beta+sqrt{(\alpha+\beta)^2+4\alpha\gamma}}{2\alpha}
=\frac\beta\alpha+1+\frac\gamma\beta+O(\alpha),
$$
so in particular $n_*\sim\beta/\alpha$ as $\alpha/\beta\to0$. Put $x=n-n_*$. The <linear noise approximation> uses
$$
g(n)=g_*'x+O(x^2),
\qquad h(n)=h_*+O(x),
$$
where
$$
g_*'=2\alpha n_*-(\alpha+\beta)
=\sqrt{(\alpha+\beta)^2+4\alpha\gamma}>0
$$
and, because $b(n_*)=d(n_*)$,
$$
h_*=h(n_*)=2b(n_*)=2(\gamma+\beta n_*).
$$
In a stationary state with zero <Fokker-Planck probability current>,
$$
g_*'xP+\frac{h_*}{2}\frac{dP}{dx}=0.
$$
Its normalized solution is the <normal distribution>
$$
P(x)\simeq
\sqrt{\frac{g_*'}{\pi h_*}}
\exp\left(-\frac{g_*'x^2}{h_*}\right).
$$
It follows that
$$
\langle n\rangle\simeq n_*\sim\frac\beta\alpha,
\qquad
(\Delta n)^2\simeq\frac{h_*}{2g_*'}
\sim\frac\beta\alpha.
$$

These estimates agree with the exact stationary first-moment relation on its plus branch to leading order: inserting $(\Delta n)^2\sim\beta/\alpha$ gives $\langle n\rangle\sim\beta/\alpha$. Moreover,
$$
\frac{\Delta n}{n_*}sim\sqrt{\frac\alpha\beta}\ll1,
\qquad
\Delta n\sim\sqrt{\frac\beta\alpha}\gg1.
$$
Thus the stationary mass lies far from the boundary $n=0$, is narrow relative to its mean, yet changes across many lattice sites. These are precisely the large-population and slow-variation conditions needed for the diffusion approximation.

Solved by gpt-5.6-sol high.

= 15D
{parent=Paper 4}
{scope}
{title2=Classical Dynamics}

= Solution
{parent=15D}

A transformation of <phase space> is <canonical transformation>[canonical] when it preserves the <symplectic form>, or equivalently all <Poisson brackets>. For one degree of freedom this is
$$
dQ\wedge dP=dq\wedge dp,
$$
equivalently $\{Q,P\}_{q,p}=1$.

Suppose a <type-two generating function for a canonical transformation> $F(q,P)$ defines
$$
p=F_q,
\qquad Q=F_P.
$$
Then
$$
\begin{aligned}
dq\wedge dp&=F_{qP}\,dq\wedge dP,\\
dQ\wedge dP&=F_{Pq}\,dq\wedge dP.
\end{aligned}
$$
Equality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.

For
$$
F_0(q,P)=\int_0^q\sqrt{2P-u^2}\,du,
$$
the <fundamental theorem of calculus> and differentiation under the integral sign give
$$
p=\sqrt{2P-q^2},
\qquad
Q=\int_0^q\frac{du}{\sqrt{2P-u^2}}
=\arcsin\frac{q}{\sqrt{2P}}.
$$
Thus
$$
q=\sqrt{2P}\sin Q,
\qquad
p=\sqrt{2P}\cos Q,
$$
and in particular $P=(p^2+q^2)/2$.

For the unit-frequency <simple harmonic motion>, the transformed <Hamiltonian> is simply $H=P$. <Hamilton's equations> become
$$
\dot Q=\frac{\partial H}{\partial P}=1,
\qquad
\dot P=-\frac{\partial H}{\partial Q}=0.
$$
Therefore
$$
P(t)=E,
\qquad Q(t)=t-t_0,
$$
and transformation back gives the familiar solution
$$
q(t)=\sqrt{2E}\sin(t-t_0),
\qquad
p(t)=\sqrt{2E}\cos(t-t_0).
$$

Now consider the weak <quartic oscillator>
$$
H(q,p)=\frac12(p^2+q^2)+\epsilon q^4.
$$
Choose the modified generating function
$$
F_\epsilon(q,P)
=\int_0^q\sqrt{2P-u^2-2\epsilon u^4}\,du.
$$
It gives
$$
p=\sqrt{2P-q^2-2\epsilon q^4},
$$
so the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the <Taylor series> gives
$$
F_\epsilon(q,P)
=F_0(q,P)
-\epsilon\int_0^q\frac{u^4}{\sqrt{2P-u^2}}\,du
+O(\epsilon^2).
$$
Differentiation with respect to $P$ therefore yields
$$
Q
=\arcsin\frac{q}{\sqrt{2P}}
+\epsilon I(q,P)+O(\epsilon^2),
$$
where
$$
I(x,y)=\int_0^x\frac{u^4}{(2y-u^2)^{3/2}}\,du.
$$

Set $\theta=t-t_0$ and
$$
q_0=\sqrt{2E}\sin\theta,
\qquad
p_0=\sqrt{2E}\cos\theta.
$$
Writing $q=q_0+\epsilon q_1+O(\epsilon^2)$ and expanding the <inverse sine> relation at $q_0$ gives
$$
0=\frac{q_1}{p_0}+I(q_0,E),
\qquad
q_1=-p_0I(q_0,E).
$$
Expanding $p^2=2E-q^2-2\epsilon q^4$ similarly gives
$$
p=p_0+\epsilon p_1+O(\epsilon^2),
\qquad
p_1=q_0I(q_0,E)-\frac{q_0^4}{p_0}.
$$
Consequently
$$
\begin{aligned}
\frac qp
&=\frac{q_0}{p_0}
 +\epsilon\left(\frac{q_1}{p_0}
 -\frac{q_0p_1}{p_0^2}\right)
 +O(\epsilon^2)\\
&=\tan\theta
 -\epsilon I(q_0,E)(1+\tan^2\theta)
 +\epsilon q_0^2\tan^3\theta
 +O(\epsilon^2).
\end{aligned}
$$
Since $\theta=t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.

Solved by gpt-5.6-sol high.

= 16H
{parent=Paper 4}
{scope}
{title2=Logic and Set Theory}

= a
{parent=16h}
{scope}

= Solution
{parent=a}

Assume the <axiom of choice>. By the <well-ordering theorem>, every set is bijective with an <ordinal>, and hence with an <initial ordinal>. Any two ordinals are comparable, so for any sets $K,L$ either $K$ injects into $L$ or $L$ injects into $K$. Transitivity is immediate, and antisymmetry on cardinalities follows from the <Cantor-Schröder-Bernstein theorem>. Thus cardinal comparison is a total order.

Conversely, assume the <cardinal comparability principle>. Given any set $X$, let $h(X)$ be its Hartogs ordinal. By <Hartogs theorem>, there is no injection $h(X)\to X$. Comparability therefore supplies an injection $X\to h(X)$. Pulling the ordinal order back to $X$ gives a well-order of $X$. Hence every set can be well-ordered, and the <well-ordering theorem> implies the axiom of choice.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

Under the <axiom of choice>, identify every infinite cardinal $\kappa$ with its <initial ordinal>. We prove $\kappa^2=\kappa$ by <transfinite induction> on infinite cardinals. The countable case follows from the usual diagonal enumeration of $\mathbb N\times\mathbb N$.

For the induction step, well-order the <Cartesian product> $\kappa\times\kappa$ by first comparing
$$
\max\{\alpha,\beta\}
$$
and then using lexicographic order among pairs with the same maximum. The predecessors of $(\alpha,\beta)$ lie in $(\eta+1)\times(\eta+1)$, where $\eta=\max\{\alpha,\beta\}<\kappa$. Put $\mu=|\eta+1|<\kappa$. If $\mu$ is infinite, the induction hypothesis makes this predecessor set have cardinal at most $\mu^2=\mu<\kappa$; if $\mu$ is finite, the same conclusion is immediate.

Let $\theta$ be the order type of this well-order. Every proper initial segment of $\theta$ therefore has cardinal less than $\kappa$. Hence $\theta<\kappa^+$, since otherwise its initial segment of order type $\kappa$ would have cardinal $\kappa$. Thus $|\kappa\times\kappa|=|\theta|\leq\kappa$. The map $\alpha\mapsto(\alpha,0)$ gives the reverse injection, and the <Cantor-Schröder-Bernstein theorem> yields
$$
\kappa^2=\kappa.
$$

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= Solution
{parent=c}

Let
$$
i:K\times L\longrightarrow K\sqcup L
$$
be the injection witnessing $\kappa\lambda\leq\kappa+\lambda$, and fix $(k_*,l_*)\in K\times L$. Extend the inverse of $i$ to a surjection
$$
p:K\sqcup L\longrightarrow K\times L
$$
by setting
$$
p(z)=
\begin{cases}
i^{-1}(z),&z\in i[K\times L],\\
(k_*,l_*),&z\notin i[K\times L].
\end{cases}
$$
This construction uses only the fixed pair, not the <axiom of choice>.

Restrict $p$ to the $K$-summand and take its second coordinate:
$$
r:K\longrightarrow L,
\qquad r(k)=\pi_L(p(k,0)).
$$
If $r$ is onto, it is the required <surjective function> from $K$ to $L$.

Otherwise choose one $l_0\in L\setminus r[K]$. For each $k\in K$, surjectivity of $p$ gives a preimage of $(k,l_0)$. No such preimage lies in the $K$-summand, by the choice of $l_0$, so it lies in the $L$-summand. Unless $(k,l_0)=(k_*,l_*)$, this preimage is exactly $i(k,l_0)$ and is unique: all points outside the range of $i$ map only to the default pair. Distinct $k$ give distinct preimages because $i$ is injective.

If $l_0\ne l_*$, these unique preimages therefore define an <injective function> $K\to L$. The only exceptional case is $l_0=l_*$, where they initially define an injection
$$
j:K\setminus\{k_*\}\longrightarrow L.
$$
If $j$ is onto, extend it arbitrarily at $k_*$ to obtain a surjection $K\to L$. If it is not onto, choose one $l'\in L\setminus j[K\setminus\{k_*\}]$ and set $j(k_*)=l'$; this gives an injection $K\to L$. These cases prove the <product-sum comparison lemma> entirely in ZF.

Solved by gpt-5.6-sol high.

= d
{parent=16h}
{scope}

= Solution
{parent=d}

Part (b) proves one implication. Conversely, assume in ZF that
$$
\kappa^2=\kappa
$$
for every infinite cardinal $\kappa$. Let $X$ be any set and let $\gamma=h(X)$ be its Hartogs ordinal. Finite $X$ is already well-orderable, so suppose $X$ is infinite and form the <disjoint union>
$$
A=X\sqcup\gamma.
$$
By hypothesis, $A\times A$ is bijective with $A$. Since $\gamma\times X$ injects into $A\times A$, we obtain
$$
|\gamma|\,|X|\leq|\gamma|+|X|.
$$

Apply part (c), with $K=\gamma$ and $L=X$. It gives either an injection $\gamma\to X$ or a surjection $s:\gamma\to X$. The first alternative contradicts <Hartogs theorem>. In the second, every fiber $s^{-1}(\{x\})$ is a nonempty set of ordinals and therefore has a least member. The map
$$
x\longmapsto\min s^{-1}(\{x\})
$$
is an injection of $X$ into the ordinal $\gamma$ and pulls its well-order back to $X$. Thus every set is well-orderable, so the <well-ordering theorem> gives the <axiom of choice>. This proves the <Tarski cardinal-square theorem>.

Solved by gpt-5.6-sol high.

= 17H
{parent=Paper 4}
{scope}
{title2=Graph Theory}

= a
{parent=17h}
{scope}

= Solution
{parent=a}

The vertex form of <Menger theorem> states that if $x$ and $y$ are distinct nonadjacent vertices of a finite graph, then the maximum number of pairwise <internally vertex-disjoint paths> from $x$ to $y$ equals the minimum cardinality of an $x$-$y$ <vertex separator>.

Solved by gpt-5.6-sol high.

= b
{parent=17h}
{scope}

= Solution
{parent=b}

The <Hall marriage theorem> says that a <bipartite graph> with classes $A,B$ has a <matching in a graph>[matching] saturating $A$ if and only if
$$
|N(S)|\geq|S|
\qquad\text{for every }S\subseteq A,
$$
where $N(S)$ is the <graph neighbourhood> of $S$. Necessity follows because the matching sends the vertices of $S$ to distinct vertices of $N(S)$.

For sufficiency, add vertices $x,y$, join $x$ to every vertex of $A$, and join every vertex of $B$ to $y$. Let $W$ be an $x$-$y$ vertex separator, and write $W_A=W\cap A$ and $W_B=W\cap B$. If an edge joined a vertex of $A\setminus W_A$ to a vertex of $B\setminus W_B$, it would give an $x$-$y$ path avoiding $W$. Hence
$$
N(A\setminus W_A)\subseteq W_B.
$$
Hall's condition gives
$$
|W_B|\geq|N(A\setminus W_A)|
\geq|A\setminus W_A|,
$$
and therefore $|W|\geq|A|$. By <Menger theorem>, there are $|A|$ internally vertex-disjoint $x$-$y$ paths. Each has the form $xaby$ with $a\in A$ and $b\in B$; disjointness makes all the $a$ and all the $b$ distinct. Their middle edges form a matching saturating $A$.

Solved by gpt-5.6-sol high.

= c
{parent=17h}
{scope}

= Solution
{parent=c}

Construct a <bipartite graph> with left vertices $A_1,\ldots,A_k$, right vertices $B_1,\ldots,B_k$, and an edge $A_iB_j$ exactly when $A_i\cap B_j\ne\varnothing$. For $S\subseteq[k]$, every point of $\bigcup_{i\in S}A_i$ lies in some $B_j$ adjacent to one of those $A_i$. Thus
$$
\bigcup_{i\in S}A_i
\subseteq
\bigcup_{j\in N(S)}B_j.
$$
Using finite additivity of <Lebesgue measure> and the equal-volume hypotheses,
$$
\frac{|S|}{k}
=\operatorname{vol}\left(\bigcup_{i\in S}A_i\right)
\leq
\operatorname{vol}\left(\bigcup_{j\in N(S)}B_j\right)
=\frac{|N(S)|}{k}.
$$
Hence $|N(S)|\geq|S|$ for every $S$. The <Hall marriage theorem> supplies a <perfect matching>, which has the form $A_iB_{\sigma(i)}$ for a permutation $\sigma$ of $[k]$. Every matched edge means precisely that $A_i\cap B_{\sigma(i)}\ne\varnothing$.

Solved by gpt-5.6-sol high.

= d
{parent=17h}
{scope}

= Solution
{parent=d}

Let
$$
T=\{x_1,\ldots,x_k,y_1,\ldots,y_k\}
$$
be the terminal set, and let $C$ be the vertex set of the given <complete graph> $K_{2k}$. Any vertex set $S$ meeting every $T$-$C$ path has size at least $2k$: if $|S|<2k$, choose $t\in T\setminus S$ and $c\in C\setminus S$. Since $G$ is $2k$-connected, $G-S$ is connected, so it contains a $t$-$c$ path avoiding $S$, a contradiction.

The <set version of Menger theorem> therefore gives $2k$ pairwise vertex-disjoint $T$-$C$ paths. Truncate them at their first and last vertices in $T\cup C$. Since both sets have $2k$ vertices, every terminal is the endpoint of one path and the other endpoints are distinct vertices of $C$. Denote the path from a terminal $z$ to its clique endpoint by $R_z$, and call that endpoint $c_z$.

For each $i$, the clique contains the edge $c_{x_i}c_{y_i}$. Concatenating
$$
R_{x_i},\quad c_{x_i}c_{y_i},\quad R_{y_i}
$$
gives an $x_i$-$y_i$ <path in a graph>. The linkage paths are mutually vertex-disjoint, their clique endpoints are all distinct, and the joining clique edges pair those endpoints without introducing a new vertex. The resulting paths $P_1,\ldots,P_k$ are therefore mutually vertex-disjoint. In particular, under the stated clique hypothesis, the $2k$-connected graph is <linked graph>[$k$-linked] for these terminals.

Solved by gpt-5.6-sol high.

= 18I
{parent=Paper 4}
{scope}
{title2=Galois Theory}

= a
{parent=18i}
{scope}

= Solution
{parent=a}

If a monic <polynomial> $f$ has roots $\alpha_1,\ldots,\alpha_n$ in a <splitting field>, its <polynomial discriminant> is
$$
\operatorname{disc}(f)
=\prod_{1\leq i<j\leq n}(\alpha_i-\alpha_j)^2.
$$
It is a symmetric polynomial in the roots, so it belongs to the base <field>; it is nonzero exactly when $f$ is a <separable polynomial>.

Put
$$
D=\prod_{i<j}(\alpha_i-\alpha_j),
$$
the <Vandermonde determinant>. A root permutation $\sigma$ sends $D$ to $\operatorname{sgn}(\sigma)D$. Hence the <Galois group of a polynomial> is contained in the <alternating group> $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\operatorname{disc}(f)=D^2$ is a square in $K$. Conversely, if $\operatorname{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the <discriminant criterion for an alternating Galois group>.

Now let
$$
f(T)=T^3-2T+2.
$$
The <rational root theorem> shows that none of $\pm1,\pm2$ is a root, so this cubic is irreducible over $\mathbb Q$. The <discriminant of a depressed cubic> is
$$
\operatorname{disc}(f)
=-4(-2)^3-27(2)^2
=32-108=-76=-4\cdot19.
$$
This is not a square in $\mathbb Q$, so the <Galois group of an irreducible cubic> is $S_3$ over $\mathbb Q$.

Over $K=\mathbb Q(\sqrt{-19})$, the discriminant is the square
$$
-76=(2\sqrt{-19})^2.
$$
The cubic remains irreducible: a root in the degree-two extension $K/\mathbb Q$ would generate over $\mathbb Q$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the <tower law for field extensions>. Its Galois group over $K$ is consequently $A_3\cong C_3$.

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

In <characteristic of a field>[characteristic] two, the <formal derivative in positive characteristic> of
$$
f(T)=T^3+aT+b
$$
is
$$
f'(T)=T^2+a.
$$
A <polynomial> is separable exactly when it and its derivative have no common root. If $r$ is a root of $f'$, then
$$
f(r)=r^3+ar+b=r(r^2+a)+b=b.
$$
Thus $f$ and $f'$ have a common root exactly when $b=0$. Therefore $f$ is a <separable polynomial> if and only if $b\ne0$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $x_1,x_2,x_3\in L$ be the distinct roots of $f$, and define the two cyclic sums
$$
r=x_1^2x_2+x_2^2x_3+x_3^2x_1,
\qquad
s=x_2^2x_1+x_3^2x_2+x_1^2x_3.
$$
Vieta's relations in characteristic two are
$$
x_1+x_2+x_3=0,
\qquad
x_1x_2+x_2x_3+x_3x_1=a,
\qquad
x_1x_2x_3=b.
$$
A direct symmetric expansion using these identities gives
$$
r+s=b,
\qquad
rs=a^3+b^2.
$$
Consequently $r$ and $s$ are the roots of the <characteristic-two cubic resolvent>
$$
g(T)=T^2+bT+a^3+b^2.
$$
They are distinct because $r+s=b\ne0$. Thus $g$ splits into distinct linear factors in $L[T]$.

A three-cycle of the roots fixes each cyclic sum, whereas any transposition interchanges $r$ and $s$. The induced action on $\{r,s\}$ is therefore exactly the sign action of the <symmetric group> $S_3$ on $S_3/A_3$. If the Galois group $G$ is contained in $A_3$, it fixes $r$ and $s$, so both lie in the fixed field $K$ and $g$ splits over $K$. Conversely, if $g$ splits in $K[T]$, its distinct roots $r,s$ belong to $K$ and every element of $G$ fixes them individually. No element of $G$ can then act as an odd root permutation, so $G\subseteq A_3$.

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 4}
{scope}
{title2=Representation Theory}

= Solution
{parent=19H}

<Schur lemma> says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A <continuous representation of a topological group> is a continuous homomorphism
$$
\rho:G\longrightarrow\operatorname{GL}(V)
$$
for a finite-dimensional complex <vector space> $V$. It is a <unitary representation> if $V$ has a positive-definite Hermitian <inner product> for which
$$
\langle\rho(g)v,\rho(g)w\rangle
=\langle v,w\rangle
$$
for every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized <Haar measure>:
$$
\langle v,w\rangle_{S^1}
=\int_{S^1}
 \langle\rho(z)v,\rho(z)w\rangle\,dz.
$$
Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving <unitarization of a compact-group representation>[unitarity]. Since $S^1$ is abelian, the operators $\rho(z)$ commute; since they are unitary, they are <normal matrix>[normal]. <Simultaneous diagonalization> therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every <representation of the circle group> is a direct sum of one-dimensional representations.

Write
$$
g(x,y,z)=
\begin{pmatrix}
1&x&z\\0&1&y\\0&0&1
\end{pmatrix}.
$$
The group law and inverse are
$$
\begin{aligned}
g(x,y,z)g(x',y',z')
 &=g(x+x',y+y',z+z'+xy'),\\
g(x,y,z)^{-1}
 &=g(-x,-y,-z+xy).
\end{aligned}
$$
A calculation gives
$$
g(x,y,z)^{-1}g(x',y',z')^{-1}g(x,y,z)g(x',y',z')
=g(0,0,xy'-x'y).
$$
Every element of the <center of a group>[centre] $Z$ occurs by taking, for example, $x'=y=0$ and $y'=1$. Hence the <commutator subgroup> is exactly $Z$. The image of a one-dimensional representation is abelian, so the <one-dimensional representation kills the commutator subgroup> and its kernel contains $Z$.

Now let $(\rho,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroup
$$
Z/Z_0\cong\mathbb R/\mathbb Z\cong S^1
$$
is a representation of the circle group. Decompose it into its distinct weight spaces:
$$
V=V_1\oplus\cdots\oplus V_d,
\qquad
\rho(z)|_{V_i}=\theta_i(z)\operatorname{id}_{V_i}.
$$
Because $Z/Z_0$ is central, every $\rho(g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the <central circle weight-space decomposition>.

Let $n_i=\dim V_i$. The map
$$
g\longmapsto\det(\rho(g)|_{V_i})
$$
is a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value is
$$
\det(\theta_i(z)I_{n_i})=\theta_i(z)^{n_i},
$$
so $\theta_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every $\theta_i$ is trivial. Since the $\theta_i$ were distinct, $d=1$ and $\theta_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is <faithful representation>[faithful]. This is the <real Heisenberg quotient has no faithful finite-dimensional representation>.

Solved by gpt-5.6-sol high.

= 20H
{parent=Paper 4}
{scope}
{title2=Number Fields}

= a
{parent=20h}
{scope}

= Solution
{parent=a}

<Minkowski convex body theorem>[Minkowski's lemma] states that if $\Lambda$ is a full lattice in $\mathbb R^n$ and $C\subset\mathbb R^n$ is measurable, <convex set>[convex], and centrally symmetric, then
$$
\operatorname{vol}(C)>2^n\operatorname{covol}(\Lambda)
$$
implies that $C$ contains a nonzero point of $\Lambda$.

Let $\sigma_1,\sigma_2:K\to\mathbb R$ be the two real embeddings. The <Minkowski embedding of a real quadratic field> identifies $\mathcal O_K$ with a lattice
$$
\Lambda=\{(\sigma_1(\alpha),\sigma_2(\alpha)):\alpha\in\mathcal O_K\}
$$
of covolume $\sqrt{|D_K|}$. Choose a constant $c$ satisfying
$$
2\sqrt{|D_K|}<c<4\sqrt{|D_K|}.
$$
For $R>0$, consider the closed diamond
$$
C_R=
\left\{(x,y):\frac{|x|}{R}+\frac{|y|}{c/R}\leq1\right\}.
$$
It is convex and centrally symmetric, and its area is
$$
\operatorname{vol}(C_R)=2R\frac cR=2c>4\sqrt{|D_K|}.
$$
Minkowski's lemma supplies a nonzero $\alpha_R\in\mathcal O_K$ whose two embeddings lie in $C_R$.

For $(x,y)\in C_R$, the <arithmetic-geometric mean inequality> gives
$$
|xy|leq\frac14R\frac cR=\frac c4<\sqrt{|D_K|}.
$$
Since the coordinate product is the <field norm>,
$$
|N_{K/\mathbb Q}(\alpha_R)|<\sqrt{|D_K|}.
$$
Finally, $|\sigma_2(\alpha_R)|\leq c/R\to0$ as $R\to\infty$. No fixed nonzero <algebraic integer> can occur for arbitrarily large $R$, because its second embedding is nonzero. Consequently the elements $\alpha_R$ obtained along an unbounded sequence of $R$ contain infinitely many distinct values. Here $N(\alpha)$ denotes the absolute norm, as usual in this inequality.

Solved by gpt-5.6-sol high.

= b
{parent=20h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For $K=\mathbb Q(\sqrt{10})$, the <ring of integers of a quadratic field> is
$$
\mathcal O_K=\mathbb Z[\sqrt{10}].
$$
The element
$$
\varepsilon=3+\sqrt{10}
$$
has <field norm> $9-10=-1$, so it is a <unit from norm one or minus one>[unit]. We claim that
$$
\mathcal O_K^\times
=\{\mathord\pm\varepsilon^n:n\in\mathbb Z\}.
$$

Let $u$ be any unit. Its norm is $\pm1$. After changing its sign and replacing it by its inverse if necessary, its first real embedding satisfies $u\geq1$. Choose $n\in\mathbb Z$ so that
$$
1\leq v=u\varepsilon^{-n}<\varepsilon.
$$
Write $v=a+b\sqrt{10}$ with $a,b\in\mathbb Z$. Its conjugate is $v'=\pm v^{-1}$, so $|v'|\leq1$. Therefore
$$
|a|=\frac{|v+v'|}{2}<\frac{\varepsilon+1}{2}<4,
\qquad
|b|=\frac{|v-v'|}{2\sqrt{10}}
<\frac{\varepsilon+1}{2\sqrt{10}}<2.
$$
Thus $|a|\leq3$ and $|b|\leq1$. The equation $a^2-10b^2=\pm1$ now shows directly that the only value in $[1,\varepsilon)$ is $v=1$: if $b=0$ then $a=\pm1$, while if $|b|=1$ then $a=\pm3$ and the positive possibilities are either below $1$ or equal to $\varepsilon$. Hence $u=\varepsilon^n$ after normalization, proving the claimed description of the <units of Q of square root ten>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $K=\mathbb Q(\sqrt{-3})$, put
$$
\omega=\frac{-1+\sqrt{-3}}2.
$$
Then the <ring of integers of a quadratic field> is $\mathcal O_K=\mathbb Z[\omega]$, and
$$
N(a+b\omega)=a^2-ab+b^2.
$$
A unit must have norm $1$. Multiplying by four gives
$$
(2a-b)^2+3b^2=4,
$$
so $b\in\{0,\pm1\}$, and checking these cases gives exactly
$$
\mathcal O_K^\times
=\{\mathord\pm1,\mathord\pm\omega,\mathord\pm\omega^2\}
\cong C_6.
$$
These are precisely the sixth <roots of unity>, proving the description of the <units of Q of square root minus three>.

Solved by gpt-5.6-sol high.

= c
{parent=20h}
{scope}

= Solution
{parent=c}

Assume that $\zeta$ is a <primitive root of unity>[primitive] fifth root of unity. The <cyclotomic field> $K=\mathbb Q(\zeta)$ has degree four and signature $(r_1,r_2)=(0,2)$. The <Dirichlet unit theorem> therefore gives free rank
$$
r_1+r_2-1=1.
$$
The <roots of unity in a rational cyclotomic field> are the ten elements $\mathord\pm\zeta^j$, so
$$
\mathcal O_K^\times\cong C_{10}\times\mathbb Z.
$$

For explicit units, $-\zeta$ has order ten, while
$$
N_{K/\mathbb Q}(1+\zeta)
=\prod_{j=1}^4(1+\zeta^j)
=\Phi_5(-1)=1.
$$
Hence $1+\zeta$ is a unit. It has infinite order because in the standard complex embedding
$$
|1+\zeta|=2\cos(\pi/5)>1,
$$
whereas every root of unity has modulus one. Its image in the free $\mathbb Z$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequently
$$
H=\langle-\zeta,1+\zeta\rangle
$$
has finite index in $\mathcal O_K^\times$, as described by the <unit group of the fifth cyclotomic field>.

Solved by gpt-5.6-sol high.

= 21G
{parent=Paper 4}
{scope}
{title2=Algebraic Topology}

= Solution
{parent=21G}

Let $f:(C,d)\to(C',d')$ be a <chain map>, so
$$
d'_if_i=f_{i-1}d_i.
$$
If $x\in C_i$ is a cycle, then $d'f_i(x)=f_{i-1}d(x)=0$, so $f_i(x)$ is a cycle. If $x=d_{i+1}y$ is a boundary, then
$$
f_i(x)=f_id_{i+1}y=d'_{i+1}f_{i+1}y
$$
is a boundary. Hence
$$
f_*:H_i(C)\longrightarrow H_i(C'),
\qquad [x]\longmapsto[f_i(x)]
$$
is a well-defined <induced map on homology>.

The maps $f$ and $g$ are <chain homotopy>[chain homotopic] if there are homomorphisms
$$
h_i:C_i\longrightarrow C'_{i+1}
$$
such that
$$
f_i-g_i=d'_{i+1}h_i+h_{i-1}d_i.
$$
For a cycle $x$, this gives
$$
f_i(x)-g_i(x)=d'_{i+1}h_i(x),
$$
which is a boundary. Thus $f_*[x]=g_*[x]$ and $f_*=g_*$ on <homology>.

Now consider the proposed <mapping cone> $M(f)$. Applying its differential twice gives
$$
(d_f)_{i-1}(d_f)_i
=
\begin{pmatrix}
d_{i-2}d_{i-1}&0\\
(-1)^{i-1}f_{i-2}d_{i-1}+(-1)^id'_{i-1}f_{i-1}
&d'_{i-1}d'_i
\end{pmatrix}.
$$
The diagonal entries vanish because $C$ and $C'$ are <chain complexes>, and the lower-left entry is
$$
(-1)^{i-1}(f_{i-2}d_{i-1}-d'_{i-1}f_{i-1})=0
$$
by the chain-map identity. Therefore $d_f^2=0$ and $M(f)$ is a chain complex.

There is a <short exact sequence of chain complexes>
$$
0\longrightarrow C'
\xrightarrow{\ j\ }M(f)
\xrightarrow{\ p\ }C[-1]
\longrightarrow0,
$$
where
$$
j_i(y)=(0,y),
\qquad p_i(x,y)=x,
\qquad C[-1]_i=C_{i-1}.
$$
Its <long exact sequence in homology> is
$$
\cdots\longrightarrow H_i(C')
\longrightarrow H_i(M(f))
\longrightarrow H_{i-1}(C)
\xrightarrow{\ \delta_i\ }H_{i-1}(C')
\longrightarrow\cdots.
$$
To identify the connecting map, represent a class in $H_{i-1}(C)$ by a cycle $x$ and lift it to $(x,0)\in M(f)_i$. Then
$$
(d_f)_i(x,0)
=(0,(-1)^if_{i-1}x),
$$
so
$$
\delta_i=(-1)^if_*.
$$
At the preceding occurrence the degree is $i+1$, giving $(-1)^{i+1}f_*$. Hence the sequence is exactly
$$
\cdots\longrightarrow H_i(C)
\xrightarrow{(-1)^{i+1}f_*}H_i(C')
\longrightarrow H_i(M(f))
\longrightarrow H_{i-1}(C)
\xrightarrow{(-1)^if_*}H_{i-1}(C')
\longrightarrow\cdots.
$$

Finally suppose $f-g=d'h+hd$. Define
$$
\Phi_i:M(f)_i\longrightarrow M(g)_i,
\qquad
\Phi_i(x,y)=\bigl(x,y+(-1)^ih_{i-1}x\bigr).
$$
A direct calculation gives
$$
\begin{aligned}
(d_g)_i\Phi_i(x,y)
&=\left(d x,(-1)^igx+d'y+(-1)^id'hx\right),\\
\Phi_{i-1}(d_f)_i(x,y)
&=\left(d x,(-1)^ifx+d'y+(-1)^{i-1}hdx\right).
\end{aligned}
$$
These are equal precisely because $f-g=d'h+hd$. Thus $\Phi$ is a chain map. Replacing the plus sign in its definition by a minus sign gives its inverse, so $M(f)$ and $M(g)$ are isomorphic as chain complexes.

Solved by gpt-5.6-sol high.

= 22F
{parent=Paper 4}
{scope}
{title2=Linear Analysis}

= a
{parent=22f}
{scope}

= Solution
{parent=a}

For each $n$, define the bounded linear functional
$$
T_n:H\longrightarrow\mathbb C,
\qquad T_n(y)=\langle x_n,y\rangle.
$$
The hypothesis is precisely that $x_n$ converges <weak convergence in a Hilbert space>[weakly] to $x$. In particular, for each fixed $y\in H$, the scalar sequence $T_n(y)$ converges and is therefore bounded. Thus the family $\{T_n\}$ is pointwise bounded.

The <Uniform boundedness principle> gives
$$
\sup_n\lVert T_n\rVert<\infty.
$$
By the <Riesz representation theorem>, $\lVert T_n\rVert=\lVert x_n\rVert_H$. Hence
$$
\sup_n\lVert x_n\rVert_H<\infty,
$$
so the sequence is bounded.

Solved by gpt-5.6-sol high.

= b
{parent=22f}
{scope}

= Solution
{parent=b}

Put $z_n=x_n-x$, so $z_n\rightharpoonup0$. We show that the norm closure of the <convex hull> of every tail
$$
A_k=\{z_n:n\geq k\}
$$
contains zero. If it did not, the <Hahn-Banach separation theorem> would give a bounded linear functional $\phi$, a real number $c>0$, and, after multiplying $\phi$ by a complex scalar if necessary,
$$
\operatorname{Re}\phi(z)\geq c
$$
for every $z$ in that closed convex hull. In particular $\operatorname{Re}\phi(z_n)\geq c$ for all $n\geq k$. By the <Riesz representation theorem>, $\phi(z)=\langle z,y\rangle$ for some $y\in H$, contradicting $z_n\rightharpoonup0$.

Consequently, for every $k$ there is a finite <convex combination>
$$
w_k=\sum_{n=k}^{N_k}\lambda_{k,n}z_n,
\qquad
\lambda_{k,n}\geq0,
\qquad
\sum_{n=k}^{N_k}\lambda_{k,n}=1,
$$
with $\lVert w_k\rVert_H<1/k$. Define
$$
\widetilde x_k=x+w_k
=\sum_{n=k}^{N_k}\lambda_{k,n}x_n.
$$
Then every $\widetilde x_k$ is a convex combination of terms of the original sequence and
$$
\lVert\widetilde x_k-x\rVert_H=\lVert w_k\rVert_H<\frac1k\longrightarrow0.
$$
This is <Mazur lemma> in the present Hilbert-space setting.

Solved by gpt-5.6-sol high.

= c
{parent=22f}
{scope}

= Solution
{parent=c}

Apply part (b). Since every $x_n$ belongs to the <convex set> $C$, every convex combination $\widetilde x_k$ also belongs to $C$. Since $C$ is norm closed and $\widetilde x_k\to x$ in norm, it follows that $x\in C$. Thus a <norm-closed convex set is weakly closed>.

Solved by gpt-5.6-sol high.

= d
{parent=22f}
{scope}

= Solution
{parent=d}

No. Take the <Hilbert space> $H=\ell^2$ and its standard <orthonormal sequence> $(e_n)$. Let
$$
C=\{u\in\ell^2:\lVert u\rVert_2=1\},
$$
the unit sphere. This set is norm closed but is not convex. By the <Bessel inequality>, for every $y=(y_j)\in\ell^2$,
$$
\langle e_n,y\rangle=\overline{y_n}\longrightarrow0,
$$
so $e_n\rightharpoonup0$; equivalently, this is a <weakly null orthonormal sequence>. Every $e_n$ lies in $C$, but the weak limit $0$ does not. Hence convexity is essential.

Solved by gpt-5.6-sol high.

= 23F
{parent=Paper 4}
{scope}
{title2=Analysis of Functions}

= a
{parent=23f}
{scope}

= Solution
{parent=a}

Choose a nonzero $\phi\in C_c^\infty(\mathbb R^n)$ supported in a ball of radius $1$, and let
$$
\phi_k(x)=\phi(x-3ke_1).
$$
Translation preserves both the $L^2$ norm and the weak-derivative norms, so
$$
\lVert\phi_k\rVert_{H^1}=\lVert\phi\rVert_{H^1}
$$
for every $k$. Thus $(\phi_k)$ is bounded in the <Sobolev space> $H^1(\mathbb R^n)$.

Distinct translates have disjoint supports. Hence, for $j\ne k$,
$$
\lVert\phi_j-\phi_k\rVert_2^2
=\lVert\phi_j\rVert_2^2+\lVert\phi_k\rVert_2^2
=2\lVert\phi\rVert_2^2.
$$
No subsequence is Cauchy in $L^2$, so no subsequence converges there. The embedding $H^1(\mathbb R^n)\hookrightarrow L^2(\mathbb R^n)$ is therefore not a <compact operator>. This is the <noncompactness of a Sobolev embedding by translation> caused by <loss of compactness at infinity>.

Solved by gpt-5.6-sol high.

= b
{parent=23f}
{scope}

= Solution
{parent=b}

Regard the <space of continuous functions vanishing at infinity> $C_0(\mathbb R^n)$ as a subspace of $L^\infty(\mathbb R^n)$. Point evaluation at the origin,
$$
\ell_0(f)=f(0),
$$
is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the <Hahn-Banach theorem>, $\ell_0$ extends to a bounded linear functional
$$
\ell:L^\infty(\mathbb R^n)\longrightarrow\mathbb C.
$$

Suppose that some $g\in L^1(\mathbb R^n)$ represented this extension:
$$
\ell(f)=\int_{\mathbb R^n}f(x)g(x)\,dx.
$$
Choose $\varphi\in C_c(\mathbb R^n)$ with $0\leq\varphi\leq1$, $\varphi(0)=1$, and support in the unit ball, and set $\varphi_k(x)=\varphi(kx)$. Then $\ell(\varphi_k)=1$ for every $k$. On the other hand, $\varphi_k(x)\to0$ for almost every $x$ and $|\varphi_k g|\leq|g|$. The <dominated convergence theorem> gives
$$
\int\varphi_k(x)g(x)\,dx\longrightarrow0,
$$
a contradiction. Thus $\ell$ is a <singular functional on L infinity> and has no $L^1$ density.

Solved by gpt-5.6-sol high.

= c
{parent=23f}
{scope}

= Solution
{parent=c}

We prove the estimate first for $u$ in the <Schwartz space>. By the <Fourier inversion theorem> and the <Cauchy-Schwarz inequality>,
$$
\begin{aligned}
|u(x)|
&\leq C\int_{\mathbb R^n}|\widehat u(\xi)|\,d\xi\\
&\leq C
\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-n}\,d\xi\right)^{1/2}
\lVert u\rVert_{H^n}.
\end{aligned}
$$
The integral is finite because $n>n/2$.

For $0<\alpha\leq1$, the elementary bound
$$
|e^{it}-1|\leq C_\alpha|t|^\alpha
$$
gives
$$
\begin{aligned}
|u(x)-u(y)|
&\leq C|x-y|^\alpha
 \int_{\mathbb R^n}|\xi|^\alpha|\widehat u(\xi)|\,d\xi\\
&\leq C|x-y|^\alpha
\left(\int_{\mathbb R^n}
 |\xi|^{2\alpha}(1+|\xi|^2)^{-n}\,d\xi\right)^{1/2}
\lVert u\rVert_{H^n}.
\end{aligned}
$$
The last integral is finite near zero for every $\alpha>0$ and at infinity exactly when $2\alpha<n$. Choose, for example, $\alpha=1/4$, which works for every $n\geq1$. We obtain
$$
\lVert u\rVert_{C^{0,1/4}}
\leq C_n\lVert u\rVert_{H^n}.
$$
Density of the Schwartz space in $H^n$ extends the estimate and supplies a unique Hölder-continuous representative. Therefore
$$
H^n(\mathbb R^n)\hookrightarrow C^{0,1/4}(\mathbb R^n)
$$
continuously. This is the <Fourier proof of Hölder regularity from a Sobolev norm>.

Solved by gpt-5.6-sol high.

= d
{parent=23f}
{scope}

= Solution
{parent=d}

Use the centered <Hardy-Littlewood maximal function>
$$
M\theta(x)=\sup_{r>0}\frac1{2r}
\int_{x-r}^{x+r}\theta(t)\,dt.
$$
If $x>0$, choosing $0<r<x$ makes the whole interval lie in the positive half-line, so $M\theta(x)=1$. If $x=0$, every centered interval has exactly half its length in the positive half-line, so $M\theta(0)=1/2$.

If $x<0$, the average is zero for $r\leq-x$. For $r>-x$, it is
$$
\frac{x+r}{2r}
=\frac12+\frac{x}{2r}<\frac12,
$$
and these values tend to $1/2$ as $r\to\infty$. Hence
$$
M\theta(x)=
\begin{cases}
1,&x>0,\\
\frac12,&x\leq0.
\end{cases}
$$
For the uncentered maximal-function convention, intervals extending arbitrarily far to the right make the supremum equal to $1$ at every $x$.

Solved by gpt-5.6-sol high.

= 24G
{parent=Paper 4}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=24G}

Two irreducible <algebraic variety>[varieties] $X$ and $Y$ are <birational variety>[birational] if they contain nonempty <Zariski-open set>[Zariski-open subsets] $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their <function field of an algebraic variety>[function fields] are isomorphic:
$$
k(X)\cong k(Y).
$$
For an irreducible variety,
$$
\dim X=\operatorname{trdeg}_k k(X).
$$
An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the <dimension from the function field>.

Now let $K/\mathbb C$ be a finitely generated field extension. Choose field generators $a_1,\ldots,a_m$ and set
$$
A=\mathbb C[a_1,\ldots,a_m]\subseteq K.
$$
Then $A$ is a finitely generated integral $\mathbb C$-algebra and $\operatorname{Frac}(A)=K$. The affine irreducible variety
$$
X_0=\operatorname{Spec}A
$$
has function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space $\mathbb A^N$, identify that affine space with the standard open chart of <projective space> $\mathbb P^N$, and take the projective closure $X$ of $X_0$. The closure of an <irreducible topological space> is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a <projective variety> and
$$
k(X)=k(X_0)=K.
$$
This constructs the <projective model of a finitely generated field>.

For the curve
$$
X=V\bigl(y^2-x(x-1)^2\bigr),
$$
put
$$
t=\frac y{x-1}.
$$
In its function field, the defining equation gives
$$
t^2=x,
\qquad
y=t(x-1)=t(t^2-1).
$$
Hence $\mathbb C(X)=\mathbb C(t)$, so $X$ is rational. Equivalently, this is the <normalization of y squared equals x times x minus one squared>, whose smooth projective normalization is $\mathbb P^1$ and has <geometric genus> zero.

A <smooth projective curve> in the <projective plane> of degree $d$ has, by the <genus of a smooth plane curve> formula,
$$
g=\frac{(d-1)(d-2)}2.
$$
Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb C$ are rational, so each is birational to $X$. Therefore the required degrees are exactly
$$
\boxed{d=1\text{ or }d=2}.
$$

Finally, in $\mathbb A^3_{x,y,z}$ take
$$
S=V(F),
\qquad
F(x,y,z)=y^2-x(x-1)^2.
$$
The polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1)^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible <affine hypersurface> of dimension two. Its function field is
$$
\mathbb C(S)=\mathbb C(t,z),
\qquad
x=t^2,
\qquad
y=t(t^2-1),
$$
so $S$ is birational to $\mathbb A^2$.

The partial derivatives are
$$
F_x=-(x-1)(3x-1),
\qquad F_y=2y,
\qquad F_z=0.
$$
Solving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the <Zariski tangent space>[Jacobian criterion],
$$
\operatorname{Sing}(S)=\{(1,0,z):z\in\mathbb C\}\cong\mathbb A^1.
$$
This is an irreducible subvariety of dimension one, giving the requested <singular cylinder over a nodal curve>.

Solved by gpt-5.6-sol high.

= 25G
{parent=Paper 4}
{scope}
{title2=Differential Geometry}

= a
{parent=25g}
{scope}

= Solution
{parent=a}

For a positively oriented local parametrization $X(u,v)$ with <first fundamental form>
$$
I=E\,du^2+2F\,du\,dv+G\,dv^2,
$$
the <area element of a surface> is
$$
dA=\sqrt{EG-F^2}\,du\wedge dv.
$$
Under an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The <Euler characteristic> may be defined from any finite triangulation by
$$
\chi(S)=V-E+F,
$$
or equivalently by the alternating sum of the dimensions of the rational <homology> groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by <arc length>. If $T$ is its unit tangent and $N$ is the chosen <unit normal> to the surface, its signed <geodesic curvature> is
$$
k_g=\langle D_sT,N\times T\rangle,
$$
where $D_s$ is the <surface covariant derivative>. The <Gauss-Bonnet theorem> for a compact oriented surface with smooth boundary and no corners is
$$
\int_S K\,dA+\int_{\partial S}k_g\,ds=2\pi\chi(S),
$$
where $K$ is the <Gaussian curvature>.

Solved by gpt-5.6-sol high.

= b
{parent=25g}
{scope}

= Solution
{parent=b}

Because $S_1$ and $S_2$ are <isometry>[isometric],
$$
\chi(S_1)=\chi(S_2),
\qquad
\int_{S_1}K\,dA=\int_{S_2}K\,dA.
$$
The two induced orientations on their common boundary $\gamma$ are opposite. If
$$
B=\int_\gamma k_g\,ds
$$
is computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the <Gauss-Bonnet theorem> to the two halves gives
$$
\int_{S_1}K\,dA+B=2\pi\chi(S_1),
\qquad
\int_{S_2}K\,dA-B=2\pi\chi(S_2).
$$
The equalities above imply $B=0$. This is the <isometric halves give zero total boundary geodesic curvature> argument.

Every ambient isometry $\phi_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the <transitive curve symmetry makes geodesic-curvature magnitude constant>. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv0$. A curve has zero geodesic curvature exactly when it is a <geodesic>, so $\gamma$ is geodesic.

Solved by gpt-5.6-sol high.

= c
{parent=25g}
{scope}

= Solution
{parent=c}

No. On the round <unit sphere>[sphere], take a latitude circle other than the equator. Rotations about the polar axis preserve the circle and act transitively on its points, so all the isometries $\phi_{x,y}$ exist. The latitude separates the sphere into two caps of unequal area, so those caps are not isometric. It is not a great circle and therefore is not a geodesic; equivalently, its geodesic curvature is nonzero. This is a <homogeneous nongeodesic latitude>.

Solved by gpt-5.6-sol high.

= d
{parent=25g}
{scope}

= Solution
{parent=d}

Again, no. On the unit round sphere and for a sufficiently small nonzero $\varepsilon$, consider
$$
\gamma(\theta)=
\frac{(\cos\theta,\sin\theta,\varepsilon\sin3\theta)}
{\sqrt{1+\varepsilon^2\sin^23\theta}},
\qquad0\leq\theta\leq2\pi.
$$
This is a smooth embedded closed curve and may be reparametrized by arc length. It obeys
$$
\gamma(\theta+\pi)=-\gamma(\theta),
$$
so the antipodal <isometry> preserves the curve and exchanges the two complementary discs. Its restriction therefore gives an isometry $S_1\to S_2$.

The curve is not contained in any plane through the origin: its third-harmonic height cannot satisfy a nontrivial linear relation with $\cos\theta$ and $\sin\theta$. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles, $\gamma$ is not geodesic. This <antipodally symmetric nongeodesic spherical curve> shows that the point-transitivity assumption cannot be omitted.

Solved by gpt-5.6-sol high.

= 26K
{parent=Paper 4}
{scope}
{title2=Probability and Measure}

= a
{parent=26k}
{scope}

= Solution
{parent=a}

Write $S_n=\sum_{j=1}^nY_j$ for the <partial sum>. Fix an integer $M\geq1$. The <tail-sum formula for expectation>, applied to the nonnegative <random variable> $|Y_1|$, gives
$$
\sum_{n=1}^{\infty}\mathbb P(|Y_n|>Mn)
=\sum_{n=1}^{\infty}\mathbb P(|Y_1|>Mn)
=\infty,
$$
because $\mathbb E|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are <independent random variables>[independent], so the second <Borel-Cantelli lemmas>[Borel--Cantelli lemma] shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in\mathbb N$ yields
$$
\limsup_{n\to\infty}\frac{|Y_n|}{n}=\infty
$$
<almost sure convergence>[almost surely].

Now $Y_n=S_n-S_{n-1}$, and the <triangle inequality> gives
$$
\frac{|Y_n|}{n}
\leq \frac{|S_n|}{n}
+\frac{n-1}{n}\frac{|S_{n-1}|}{n-1}.
$$
If the <limit superior> of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Therefore
$$
\limsup_{n\to\infty}\frac{|Y_1+\cdots+Y_n|}{n}=\infty
$$
almost surely.

Solved by gpt-5.6-sol high.

= b
{parent=26k}
{scope}

= Solution
{parent=b}

Let $(Y_n)$ be <independent and identically distributed random variables> with the standard <Cauchy distribution>, whose <probability density function> is
$$
f(y)=\frac{1}{\pi(1+y^2)}.
$$
Its first absolute moment diverges since
$$
\mathbb E|Y_1|=\frac{2}{\pi}\int_0^\infty\frac{y}{1+y^2}\,dy=\infty.
$$
On the other hand, its <characteristic function> is $\varphi(t)=e^{-|t|}$. The <characteristic function of a sum of independent variables> therefore gives
$$
\mathbb E\exp\left(it\frac{S_n}{n}\right)
=\varphi(t/n)^n=e^{-|t|}.
$$
Thus the <stability of the Cauchy distribution> implies that $S_n/n$ has the standard Cauchy distribution for every $n$. In particular,
$$
\frac{Y_1+\cdots+Y_n}{n}\xrightarrow{d}Z,
$$
where $Z$ is standard Cauchy and $\xrightarrow d$ denotes <convergence in distribution>.

Solved by gpt-5.6-sol high.

= 27J
{parent=Paper 4}
{scope}
{title2=Applied Probability}

= a
{parent=27j}
{scope}

= Solution
{parent=a}

A <Non-homogeneous Poisson point process> with <intensity function of a point process>[intensity function] $\lambda$ is a random counting measure $N$ such that, for every finite family of pairwise disjoint bounded Borel sets $A_1,\ldots,A_k$, the counts $N(A_1),\ldots,N(A_k)$ are <independent random variables> and
$$
N(A_i)\sim\operatorname{Poisson}\left(\int_{A_i}\lambda(x)\,dx\right).
$$
The assumed local integrability makes each <Poisson distribution>[Poisson] parameter finite. This count law and independence characterize the process.

Solved by gpt-5.6-sol high.

= b
{parent=27j}
{scope}

= Solution
{parent=b}

Let $N$ be the homogeneous <Poisson point process> in $\mathbb R^3$ with intensity measure $\lambda\,dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq\mathbb R_+$, the <pushforward measure> of the intensity is
$$
\lambda\,\operatorname{vol}\{x:a<|x|\leq b\}
=\frac{4\pi\lambda}{3}(b^3-a^3).
$$
The <Mapping theorem for Poisson point processes> therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function
$$
\rho(r)=\frac{d}{dr}\left(\frac{4\pi\lambda r^3}{3}\right)=4\pi\lambda r^2,
\qquad r>0.
$$

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a <Poisson distribution> gives
$$
\mathbb P(R>r)=\exp\left(-\frac{4\pi\lambda r^3}{3}\right).
$$
Differentiating its <cumulative distribution function> yields the <probability density function>
$$
f_R(r)=4\pi\lambda r^2
\exp\left(-\frac{4\pi\lambda r^3}{3}\right),
\qquad r\geq0.
$$

Solved by gpt-5.6-sol high.

= c
{parent=27j}
{scope}

= Solution
{parent=c}

For one visitor, let
$$
T_0=0,
\qquad
T_j=Z_1+\cdots+Z_j,
$$
and define
$$
p_j(u)=\mathbb P(T_{j-1}\leq u<T_j),
\qquad 1\leq j\leq10.
$$
A visitor arriving at time $s\leq t$ is in room $j$ at time $t$ exactly when its independent service-time mark satisfies $T_{j-1}\leq t-s<T_j$. Thus it receives mark $j$ with probability $p_j(t-s)$; a further mark records that it has already left the gallery.

The <Independent marking theorem for Poisson point processes> makes the point processes carrying the different room marks independent. Consequently $V_1(t),\ldots,V_{10}(t)$ are independent, and <Poisson thinning theorem>[Poisson thinning] gives
$$
V_j(t)\sim\operatorname{Poisson}(m_j(t)),
\qquad
m_j(t)=\int_0^t\lambda(s)p_j(t-s)\,ds.
$$
Equivalently,
$$
m_j(t)=\int_0^t\lambda(s)
\mathbb P\left(
Z_1+\cdots+Z_{j-1}\leq t-s
<Z_1+\cdots+Z_j
\right)ds.
$$

Solved by gpt-5.6-sol high.

= 28K
{parent=Paper 4}
{scope}
{title2=Principles of Statistics}

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Assume that $f(x)=0$ wherever $h(x)=0$ on the part of the space relevant to $g$. The <importance sampling> algorithm draws <independent and identically distributed random variables> $Y_1,\ldots,Y_m$ from $h$, forms the weights
$$
w(Y_i)=\frac{f(Y_i)}{h(Y_i)},
$$
and returns
$$
\widehat\mu_m=\frac1m\sum_{i=1}^m g(Y_i)w(Y_i).
$$
A <change of measure> gives
$$
\mathbb E_h[g(Y)w(Y)]
=\int_{\mathcal X}g(y)\frac{f(y)}{h(y)}h(y)\,dy
=\mathbb E_f[g(X)].
$$
Hence the estimator is unbiased whenever the expectation exists, and, if $\mathbb E_h|g(Y)w(Y)|<\infty$, the <strong law of large numbers> gives $\widehat\mu_m\to\mathbb E_f[g(X)]$ almost surely.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

Conditionally on $Y_1,\ldots,Y_m$, the <cumulative distribution function> of $X^*$ is the weighted empirical distribution
$$
F_m^*(t)=\mathbb P(X^*\leq t\mid Y_1,\ldots,Y_m)
=\frac{m^{-1}\sum_{i=1}^m w(Y_i)\mathbf1_{\{Y_i\leq t\}}}
{m^{-1}\sum_{i=1}^m w(Y_i)},
\qquad w(y)=\frac{f(y)}{h(y)}.
$$
For fixed $t$, the <strong law of large numbers> gives, almost surely,
$$
\frac1m\sum_{i=1}^m w(Y_i)\mathbf1_{\{Y_i\leq t\}}
\longrightarrow
\int_{(-\infty,t]}f(y)\,dy=F(t),
$$
while
$$
\frac1m\sum_{i=1}^m w(Y_i)\longrightarrow\int f(y)\,dy=1.
$$
The ratio therefore converges almost surely to $F(t)$. This is precisely the claimed <convergence in distribution> of the <self-normalized importance sampling> resample.

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

Apply part (b) simultaneously at every rational $t$; the intersection of these probability-one events still has probability one. On that event, monotonicity and continuity extend the convergence from the rationals to every real $t$.

More directly, fix $\varepsilon>0$. Since $F$ is a continuous <cumulative distribution function>, choose rational points
$$
t_0<t_1<\cdots<t_k
$$
so that the two tails have $F(t_0)<\varepsilon$ and $1-F(t_k)<\varepsilon$, while every increment $F(t_i)-F(t_{i-1})<\varepsilon$. Part (b) makes $|F_m^*(t_i)-F(t_i)|<\varepsilon$ at all these finitely many points for all sufficiently large $m$. If $t_{i-1}\leq t\leq t_i$, monotonicity gives
$$
F_m^*(t_{i-1})\leq F_m^*(t)\leq F_m^*(t_i),
$$
so $|F_m^*(t)-F(t)|<2\varepsilon$; the same estimate follows in the tails. This proves the <uniform convergence of distribution functions to a continuous limit>:
$$
\sup_{t\in\mathbb R}|F_m^*(t)-F(t)|\longrightarrow0
$$
almost surely.

Solved by gpt-5.6-sol high.

= 29K
{parent=Paper 4}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=29k}
{scope}

= Solution
{parent=a}

The stock tree has $S_0=5$, first-period values $6,4$, and terminal successors $7,5$ from $6$ and $5,3$ from $4$. When $r=1/6$, the risk-free growth factor is $R=7/6$. At the upper time-one node, investing $6$ in the risk-free asset produces $7$ at time two, while one share bought for $6$ produces only $7$ or $5$.

Use the following predictable <self-financing portfolio>. Hold nothing initially. If $S_1=6$, short one share and invest the proceeds $6$ in the risk-free asset; if $S_1=4$, continue to hold nothing. Its terminal payoff is
$$
7-S_2=
\begin{cases}
0,&S_2=7,\\
2,&S_2=5
\end{cases}
$$
on the upper branch, and zero on the lower branch. It costs zero, is never negative, and is positive with positive probability, so it is an <arbitrage>.

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

Now $R=1+r=9/8$. The local <risk-neutral probability in a binomial market>[risk-neutral up probabilities] solve $RS=qS_u+(1-q)S_d$. They are
$$
q_0=\frac{(9/8)5-4}{6-4}=\frac{13}{16},
\qquad
q_u=\frac{(9/8)6-5}{7-5}=\frac78,
\qquad
q_d=\frac{(9/8)4-3}{5-3}=\frac34.
$$
The terminal payoff of the <European put option> with strike $5$ is $0,0,2$ at stock prices $7,5,3$. <Backward option pricing> gives time-one values
$$
V_u=0,
\qquad
V_d=R^{-1}\left(\frac14\cdot2\right)=\frac49,
$$
and hence
$$
V_0=R^{-1}\left(\frac{3}{16}\frac49\right)=\boxed{\frac{2}{27}}.
$$
The first-period stock holding in the <replicating portfolio in a binomial market> is
$$
\Delta_0=\frac{V_u-V_d}{6-4}=-\frac29.
$$
Thus the hedge initially shorts $2/9$ of a share.

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

The <European call option> with strike $5$ pays $2,0,0$ at terminal stock prices $7,5,3$. Its lower-node value is zero, while its upper-node value is
$$
V_u=R^{-1}\left(\frac78\cdot2\right)=\frac{14}{9}.
$$
A second step of <backward option pricing> gives
$$
V_0=R^{-1}\frac{13}{16}\frac{14}{9}
=\boxed{\frac{91}{81}}.
$$
By the <fundamental theorem of asset pricing>, this discounted expectation under the unique risk-neutral probabilities is the unique no-arbitrage price.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

For the <American put option>, compare exercise with continuation at time one. At $S_1=6$, both values are zero. At $S_1=4$, immediate exercise pays $1$, whereas the continuation value is the European value $4/9$, so exercise is optimal. At time zero, immediate exercise pays zero and continuation is
$$
R^{-1}\left(\frac{13}{16}\cdot0+\frac{3}{16}\cdot1\right)
=\boxed{\frac16}.
$$
The optimal policy is therefore to continue at time zero, exercise at time one if $S_1=4$, and otherwise continue to maturity. This nodewise maximum is the <Snell envelope> value.

Solved by gpt-5.6-sol high.

= 30J
{parent=Paper 4}
{scope}
{title2=Mathematics of Machine Learning}

= a
{parent=30j}
{scope}

= Solution
{parent=a}

A <regression tree> chooses a threshold $s$ and constant predictions $c_L,c_R$ to minimize the squared-error criterion
$$
\min_{s,c_L,c_R}
\left\{
\sum_{i:X_i\leq s}(Y_i-c_L)^2
+\sum_{i:X_i>s}(Y_i-c_R)^2
\right\},
$$
where $s$ ranges over thresholds producing two nonempty groups. For a fixed split, the <ordinary least squares estimators>[least-squares] minimizers are the two sample means,
$$
c_L=\frac{1}{n_L}\sum_{X_i\leq s}Y_i,
\qquad
c_R=\frac{1}{n_R}\sum_{X_i>s}Y_i.
$$
Thus the first split is the candidate threshold with the smallest sum of the two within-node residual sums of squares.

Solved by gpt-5.6-sol high.

= b
{parent=30j}
{scope}

= Solution
{parent=b}

With $X_1<\cdots<X_n$, it suffices to put the threshold between $X_m$ and $X_{m+1}$ for $m=1,\ldots,n-1$. Form the <prefix sum>[prefix sums]
$$
A_m=\sum_{i=1}^mY_i,
\qquad
B_m=\sum_{i=1}^mY_i^2.
$$
They take $O(n)$ operations to compute. The minimized squared errors for the two children are
$$
L_m=B_m-\frac{A_m^2}{m},
$$
and
$$
R_m=(B_n-B_m)-\frac{(A_n-A_m)^2}{n-m}.
$$
Each candidate value $L_m+R_m$ therefore takes constant time once the prefix sums are available. Scanning all $n-1$ candidates takes $O(n)$ operations in the sense of <Big O notation>, and retaining the minimizing $m$ gives the optimal split.

Solved by gpt-5.6-sol high.

= c
{parent=30j}
{scope}

= Solution
{parent=c}

Write
$$
\bar X_m=\frac1m\sum_{i=1}^mX_i,
\qquad
\bar Y_m=\frac1m\sum_{i=1}^mY_i.
$$
The <ordinary least squares estimators> give
$$
\widehat f_m(x)=\widehat\alpha_m+\widehat\beta_mx
=\bar Y_m+\widehat\beta_m(x-\bar X_m),
$$
where
$$
\widehat\beta_m
=\frac{\sum_{i=1}^m(X_i-\bar X_m)(Y_i-\bar Y_m)}
{\sum_{i=1}^m(X_i-\bar X_m)^2}
=\frac{\sum_{i=1}^mX_iY_i-m\bar X_m\bar Y_m}
{\sum_{i=1}^mX_i^2-m\bar X_m^2}.
$$
The denominator is positive because the sorted inputs are distinct and $m\geq2$.

Solved by gpt-5.6-sol high.

= d
{parent=30j}
{scope}

= Solution
{parent=d}

For each $m$, the displayed objective is the sum of the minimized linear-regression residual sums of squares on the prefix $1{:}m$ and suffix $(m+1){:}n$. Precompute prefix sums of
$$
X_i,quad Y_i,quad X_i^2,quad X_iY_i,quad Y_i^2.
$$
This requires $O(n)$ operations. The corresponding five sums over a suffix are obtained by subtracting a prefix sum from the total.

For any block of $k$ observations, these raw sums give in constant time
$$
S_{xx}=\sum x_i^2-\frac{(\sum x_i)^2}{k},
\quad
S_{xy}=\sum x_iy_i-\frac{(\sum x_i)(\sum y_i)}{k},
\quad
S_{yy}=\sum y_i^2-\frac{(\sum y_i)^2}{k}.
$$
By the formula for the <residual sum of squares in simple linear regression>, its minimized loss is
$$
S_{yy}-\frac{S_{xy}^2}{S_{xx}}.
$$
Hence both losses for a proposed $m$ take constant time. Evaluating their sum for $m=2,\ldots,n-2$ and keeping the minimum is another $O(n)$ scan, so the complete minimization uses $O(n)$ computations.

Solved by gpt-5.6-sol high.

= 31E
{parent=Paper 4}
{scope}
{title2=Asymptotic Methods}

= Solution
{parent=31E}

Write the phase and amplitude as
$$
\phi(z)=\cosh z,
\qquad
f(z)=\frac1{z^2+16}.
$$
The <saddle point> relevant to the contour is $z=0$, since $\phi'(z)=\sinh z$ and
$$
\phi(0)=1,
\qquad
\phi''(0)=1.
$$
The local <method of steepest descent>[steepest-descent] direction is therefore vertical: with $z=iy$,
$$
\phi(iy)=\cos y=1-\frac{y^2}{2}+O(y^4).
$$

The given contour approaches the lines $\operatorname{Im}z=-\pi$ at its left end and $\operatorname{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm4i$, outside the closed strip $|\operatorname{Im}z|\leq\pi$. The <Cauchy integral theorem> therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty-i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty+i\pi$. On either horizontal ray,
$$
\cosh(p\pm i\pi)=-\cosh p,
$$
so those two integrals are exponentially small, of order $e^{-x}x^{-1/2}$.

The vertical segment, oriented upward, contributes
$$
i\int_{-\pi}^{\pi}\frac{e^{x\cos y}}{16-y^2}\,dy.
$$
Its real phase has a unique maximum at $y=0$. Since
$$
\frac1{16-y^2}=\frac1{16}+O(y^2),
$$
the <simple-saddle contribution in steepest descent>, equivalently the local <Gaussian integral>, gives
$$
I(x)
\sim \frac{i e^x}{16}
\int_{-\infty}^{\infty}e^{-xy^2/2}\,dy
=\boxed{\frac{i e^x}{16}\sqrt{\frac{2\pi}{x}}}
\qquad(x\to\infty).
$$
The factor $i$ comes from the upward tangent $dz=i\,dy$ of the deformed contour.

Solved by gpt-5.6-sol high.

= 32A
{parent=Paper 4}
{scope}
{title2=Dynamical Systems}

= i
{parent=32a}
{scope}

= Solution
{parent=i}

For $x\in[0,1]$, the nonnegative function $x(1-x^2)$ has derivative $1-3x^2$, so its maximum occurs at $x=1/\sqrt3$ and equals
$$
\frac1{\sqrt3}\left(1-\frac13\right)=\frac{2}{3\sqrt3}.
$$
The map preserves the interval exactly when this maximum is at most one:
$$
\lambda\frac{2}{3\sqrt3}\leq1.
$$
Thus the upper limit is
$$
\boxed{\lambda_{\max}=\frac{3\sqrt3}{2}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=32a}
{scope}

= Solution
{parent=ii}

A <fixed point stability for an iteration>[fixed point] satisfies
$$
x=\lambda x(1-x^2),
$$
so either $x=0$ or $x^2=1-1/\lambda$. When $0<\lambda<1$, the latter equation has no real solution, and the origin is the only fixed point. Since
$$
F_x(x,\lambda)=\lambda(1-3x^2),
\qquad
F_x(0,\lambda)=\lambda,
$$
the criterion for <fixed point stability for an iteration> shows that the origin is stable for $\lambda<1$.

Solved by gpt-5.6-sol high.

= iii
{parent=32a}
{scope}

= Solution
{parent=iii}

For $\lambda>1$, the new fixed point in $[0,1]$ is
$$
x^*=\sqrt{1-\frac1\lambda}.
$$
The origin has multiplier $F_x(0,\lambda)=\lambda>1$ and is therefore unstable. At the new fixed point,
$$
F_x(x^*,\lambda)
=\lambda\left(1-3\left(1-\frac1\lambda\right)\right)
=3-2\lambda.
$$
Hence
$$
|F_x(x^*,\lambda)|<1
\quad\Longleftrightarrow\quad
1<\lambda<2.
$$
The point $x^*$ is stable in that interval and unstable for $\lambda>2$. At $\lambda=2$ its multiplier is $-1$, the threshold for a <period-doubling bifurcation>.

Solved by gpt-5.6-sol high.

= iv
{parent=32a}
{scope}

= Solution
{parent=iv}

Let $x^*(\lambda)=\sqrt{1-1/\lambda}$ and center the map at this parameter-dependent fixed point:
$$
G(X,\mu)=F(x^*(2+\mu)+X,2+\mu)-x^*(2+\mu).
$$
A <Taylor polynomial> in $X$, with the coefficients expanded at $(x,\lambda)=(1/\sqrt2,2)$, gives
$$
G(X,\mu)=-X+\alpha\mu X+\beta X^2+\gamma X^3+O(\mu^2)
$$
under the period-doubling scaling $X=O(\sqrt\mu)$. The coefficients are
$$
\alpha=\left.\frac{d}{d\lambda}F_x(x^*(\lambda),\lambda)\right|_{\lambda=2}=-2,
$$
$$
\beta=\frac12F_{xx}\left(\frac1{\sqrt2},2\right)=-3\sqrt2,
\qquad
\gamma=\frac16F_{xxx}\left(\frac1{\sqrt2},2\right)=-2.
$$
Here we used $F_x(x^*(\lambda),\lambda)=3-2\lambda$, $F_{xx}=-6\lambda x$, and $F_{xxx}=-6\lambda$.

Composing the local map with itself and retaining the terms of order $\mu X$ and $X^3$ gives
$$
G(G(X,\mu),\mu)-X
=-2\alpha\mu X-2(\gamma+\beta^2)X^3
+o(\mu X+X^3).
$$
Besides the fixed solution $X=0$, the two leading solutions therefore satisfy
$$
X^2=-\frac{\alpha\mu}{\gamma+\beta^2}.
$$
Thus the period-two points born at $\lambda=2$ are
$$
\boxed{x_\pm=x^*\pm
\sqrt{\frac{-\alpha\mu}{\gamma+\beta^2}}}
$$
to leading order. With the coefficients above, the radicand is positive for $\mu>0$.

Solved by gpt-5.6-sol high.

= v
{parent=32a}
{scope}

= Solution
{parent=v}

The exact second-iterate equation factors as
$$
F(F(x))-x
=x(\lambda x^2-\lambda-1)(\lambda x^2-\lambda+1)
(\lambda^2x^4-\lambda^2x^2+1).
$$
The positive nonfixed period-two points come from the final factor. If $y=x^2$, their squared values are
$$
y_\pm=\frac{1\pm\sqrt{1-4/\lambda^2}}2,
\qquad
y_++y_-=1,
\qquad
y_+y_-=\frac1{\lambda^2}.
$$
They are real and distinct for $\lambda>2$.

The <multiplier of a periodic orbit of an iteration> is
$$
F'(x_+)F'(x_-)
=\lambda^2(1-3y_+)(1-3y_-)
=\lambda^2\left(1-3(y_++y_-)+9y_+y_-\right)
=9-2\lambda^2.
$$
For $\lambda=2+\mu$ with $0<\mu\ll1$,
$$
9-2\lambda^2=1-8\mu+O(\mu^2),
$$
whose modulus is less than one. The new two-cycle is therefore locally asymptotically stable.

Solved by gpt-5.6-sol high.

= 33B
{parent=Paper 4}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=33b}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

On the tensor-product Hilbert space, <addition of angular momentum> gives
$$
\mathbf J=\mathbf J_1\otimes I_2+I_1\otimes\mathbf J_2.
$$
Suppressing the identity operators, its components and <spin ladder operator>[ladder operators] are
$$
J_x=J_{1x}+J_{2x},
\quad
J_y=J_{1y}+J_{2y},
\quad
J_z=J_{1z}+J_{2z},
$$
$$
J_\pm=J_x\pm iJ_y=J_{1\pm}+J_{2\pm}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The <Clebsch-Gordan decomposition> permits
$$
j=|j_1-j_2|,|j_1-j_2|+1,\ldots,j_1+j_2.
$$
Thus a total-angular-momentum singlet $j=0$ occurs exactly when $j_1=j_2$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Put $J=j_1+j_2$. The only state with $m=J$ is the maximal <highest-weight states in angular momentum addition>[highest-weight state]
$$
|J,J\rangle=|j_1,j_1\rangle_1|j_2,j_2\rangle_2.
$$
Applying $J_-=J_{1-}+J_{2-}$ and normalizing gives
$$
|J,J-1\rangle
=\sqrt{\frac{j_1}{J}},
|j_1,j_1-1\rangle_1|j_2,j_2\rangle_2
+\sqrt{\frac{j_2}{J}},
|j_1,j_1\rangle_1|j_2,j_2-1\rangle_2.
$$
The orthonormal vector in the same two-dimensional $m=J-1$ subspace is
$$
|J-1,J-1\rangle
=\sqrt{\frac{j_2}{J}},
|j_1,j_1-1\rangle_1|j_2,j_2\rangle_2
-\sqrt{\frac{j_1}{J}},
|j_1,j_1\rangle_1|j_2,j_2-1\rangle_2.
$$
It is annihilated by $J_+$, so it is the highest-weight state of the $j=J-1$ multiplet. These are all states with $m\geq J-1$. The last state exists when both subsystem spins are nonzero.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

A <product state> factors as $|\psi_1\rangle\otimes|\psi_2\rangle$; an <entangled state> admits no such factorization. The state $|J,J\rangle$ is a product state. When $j_1,j_2>0$, each state with $m=J-1$ displayed in part (iii) has two nonzero terms with orthogonal states in both subsystems, so its Schmidt rank is two and it is entangled. If one subsystem has spin zero, only the $j=J$ multiplet remains and its states factor through the unique state of that subsystem.

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= Solution
{parent=b}

For $|J,J\rangle$, tracing out subsystem 2 gives the <reduced density matrix>
$$
\rho_1=|j_1,j_1\rangle\langle j_1,j_1|,
$$
which is pure, so its <entanglement entropy> is zero.

For $|J,J-1\rangle$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and gives
$$
\rho_1=
\frac{j_1}{J}|j_1,j_1-1\rangle\langle j_1,j_1-1|
+\frac{j_2}{J}|j_1,j_1\rangle\langle j_1,j_1|.
$$
Its nonzero <eigenvalue>[eigenvalues] are $j_1/J$ and $j_2/J$, so the <Von Neumann entropy> is
$$
S=-\frac{j_1}{J}\log\frac{j_1}{J}
-\frac{j_2}{J}\log\frac{j_2}{J}.
$$
When $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log2$, the maximal entropy for a rank-two reduced state.

Solved by gpt-5.6-sol high.

= c
{parent=33b}
{scope}

= Solution
{parent=c}

A $j=0$ state exists only when $j_1=j_2=:j$. Since its total magnetic quantum number is zero, every uncoupled basis term must have $m_1+m_2=0$, so
$$
|0,0\rangle=\sum_{m=-j}^j\alpha_m
|j,m\rangle_1|j,-m\rangle_2.
$$
Apply the total raising operator $J_+=J_{1+}+J_{2+}$. The coefficient of
$|j,m+1\rangle_1|j,-m\rangle_2$ is
$$
\sqrt{(j-m)(j+m+1)}\,(\alpha_m+\alpha_{m+1}).
$$
The <angular momentum singlet state> is annihilated by $J_+$, hence
$$
\alpha_{m+1}=-\alpha_m,
\qquad -j\leq m<j.
$$
All coefficients consequently have the same modulus and alternating signs. Normalization fixes that modulus to $(2j+1)^{-1/2}$, and an arbitrary overall phase may be chosen so that
$$
\boxed{
\alpha_m=\frac{(-1)^{j-m}}{\sqrt{2j+1}}
}.
$$
Thus
$$
|0,0\rangle=\frac1{\sqrt{2j+1}}
\sum_{m=-j}^j(-1)^{j-m}|j,m\rangle_1|j,-m\rangle_2.
$$

Solved by gpt-5.6-sol high.

= 34D
{parent=Paper 4}
{scope}
{title2=Applications of Quantum Mechanics}

= a
{parent=34d}
{scope}

= Solution
{parent=a}

Take $B>0$ and use the <Landau gauge> $\mathbf A=(0,Bx,0)$. With zero $z$-momentum, the Hamiltonian becomes
$$
H=\frac1{2m}\left[p_x^2+(p_y+eBx)^2\right].
$$
For a state $e^{ik_y y}\chi(x)$, this is a <quantum harmonic oscillator> in $x$, centred at
$$
x_0=-\frac{\hbar k_y}{eB},
$$
with <cyclotron frequency>
$$
\omega_c=\frac{|eB|}{m}.
$$
The <Landau level>[Landau levels] are therefore
$$
\boxed{E_n=\hbar\omega_c\left(n+\frac12\right)},
\qquad n=0,1,2,\ldots.
$$

Periodicity in $y$ quantizes $k_y=2\pi\ell/R_y$. Requiring the corresponding oscillator centre to lie in an $x$ interval of length $R_x$ leaves
$$
\boxed{D=\frac{|eB|R_xR_y}{2\pi\hbar}}
$$
allowed centres for each $n$. This is the <degeneracy of a Landau level>, equal to the magnetic flux through the rectangle divided by the <magnetic flux quantum> $h/|e|$. Consistent magnetic periodic boundary conditions make $D$ an integer.

Solved by gpt-5.6-sol high.

= b
{parent=34d}
{scope}

= Solution
{parent=b}

Assume $eB>0$; reversing the sign only exchanges the spin labels. Since $\mathbf B\mathbin{\cdot}\boldsymbol\sigma=B\sigma_3$, the Pauli term shifts the spinless energies by $\pm\hbar\omega_c/2$. Thus
$$
E_{n,\sigma}
=\hbar\omega_c\left(n+\frac12+\frac\sigma2\right),
\qquad \sigma=\pm1.
$$
After collecting equal energies, the <spin splitting of Landau levels> is
$$
\boxed{E_\ell=\ell\hbar\omega_c},
\qquad \ell=0,1,2,\ldots.
$$
The level $\ell=0$ contains only $(n,sigma)=(0,-1)$ and has degeneracy $D$. Every level $\ell\geq1$ contains $(\ell,-1)$ and $(\ell-1,+1)$ and has degeneracy $2D$.

For $N$ noninteracting electrons, the <Pauli exclusion principle> fills these states from the bottom. If $0\leq N\leq D$, every electron fits in the zero-energy level and
$$
E_{\mathrm{gs}}(N)=0.
$$
For $N>D$, write
$$
N-D=2Dk+r,
\qquad k\in\mathbb Z_{\geq0},
\qquad 0\leq r<2D.
$$
Then the first $k$ positive levels are full and $r$ states in level $k+1$ are occupied. The <ground state>[ground-state] energy is
$$
\boxed{
E_{\mathrm{gs}}(N)
=\hbar\omega_c\left[Dk(k+1)+(k+1)r\right].
}
$$
Equivalently, on
$$
D(2k+1)\leq N\leq D(2k+3),
$$
$$
E_{\mathrm{gs}}(N)
=D\hbar\omega_c k(k+1)
+(k+1)\hbar\omega_c\,[N-D(2k+1)].
$$
The graph is continuous and piecewise linear: it is flat up to $N=D$, then has slopes $\hbar\omega_c,2\hbar\omega_c,3\hbar\omega_c,\ldots$, with kinks at
$$
N=D,3D,5D,\ldots.
$$

Solved by gpt-5.6-sol high.

= 35A
{parent=Paper 4}
{scope}
{title2=Statistical Physics}

= a
{parent=35a}
{scope}

= Solution
{parent=a}

The <Clausius statement of the second law> says that no cyclic device can have as its sole effect the transfer of heat from a colder reservoir to a hotter one. The <Kelvin-Planck statement of the second law> says that no cyclic device can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work.

Suppose first that the Clausius statement fails. Use the forbidden device to return to a hot reservoir the heat $Q_C$ rejected by an ordinary heat engine. The combined device extracts $Q_H-Q_C$ from the hot reservoir and converts all of it into work, violating the Kelvin--Planck statement.

Conversely, suppose the Kelvin--Planck statement fails. Use the work produced by the forbidden one-reservoir engine to drive an ordinary refrigerator. The engine's heat withdrawal and the refrigerator's heat discharge cancel by the work-energy balance, leaving only a transfer of heat from the cold reservoir to the hot reservoir with no external work. This violates the Clausius statement. Each statement therefore implies the other, so the two forms of the <Second law of thermodynamics> are equivalent.

Solved by gpt-5.6-sol high.

= b
{parent=35a}
{scope}

= Solution
{parent=b}

For $N$ particles, the <ideal gas> equation of state is
$$
\boxed{pV=Nk_BT}.
$$
The <internal energy of an ideal gas> depends only on temperature and obeys
$$
dE=C_V\,dT.
$$
For constant heat capacity and a convenient choice of energy zero,
$$
\boxed{E=C_VT}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=35a}
{scope}

= Solution
{parent=c}

An <adiabatic process> has no heat transfer, $\delta Q=0$. With work done by the gas taken as positive, the <first law of thermodynamics> gives
$$
dE=\delta Q-p\,dV=-p\,dV.
$$
Using $dE=C_VdT$ and $p=Nk_BT/V$,
$$
C_V\frac{dT}{T}=-Nk_B\frac{dV}{V}.
$$
Integration, followed by the <Mayer relation> $Nk_B/C_V=\gamma-1$, yields
$$
TV^{\gamma-1}=\text{constant}.
$$
Eliminating $T$ with the equation of state gives the <reversible ideal-gas adiabat>
$$
\boxed{pV^\gamma=\text{constant}},
\qquad \gamma=\frac{C_p}{C_V}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=35a}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

This is the ideal <Diesel cycle>. In the $(p,V)$ plane, $A\to B$ is an adiabatic curve running up and left, $B\to C$ is a horizontal constant-pressure segment running right, $C\to D$ is an adiabatic curve running down and right, and $D\to A$ is a vertical constant-volume segment running down. The traversal is clockwise, so the enclosed area is positive work output.

In the $(T,S)$ plane, the two reversible adiabats are vertical because <entropy> is constant: $A\to B$ rises at $S_A=S_B$, and $C\to D$ falls at $S_C=S_D$. Along $B\to C$, the heat increment at constant pressure is
$$
\delta Q=C_p\,dT=T\,dS,
$$
so
$$
\boxed{S-S_B=C_p\log\frac{T}{T_B}},
\qquad
\boxed{T=T_Be^{(S-S_B)/C_p}}.
$$
Along the constant-volume curve $D\to A$,
$$
\delta Q=C_V\,dT=T\,dS,
$$
which gives
$$
\boxed{S-S_A=C_V\log\frac{T}{T_A}},
\qquad
\boxed{T=T_Ae^{(S-S_A)/C_V}}.
$$
Thus $B\to C$ runs up and right and $D\to A$ runs down and left between the two vertical adiabats.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

Heat enters only during the constant-pressure step, so the <heat capacity at constant pressure> gives
$$
Q_1=C_p(T_C-T_B).
$$
The positive magnitude of the heat rejected during constant-volume cooling is
$$
Q_2=C_V(T_D-T_A).
$$
Because the working gas returns to its initial state, the <thermodynamic cycle> has zero net change in internal energy and $W=Q_1-Q_2$. Its <thermal efficiency> is therefore
$$
\boxed{
\eta=\frac{W}{Q_1}
=1-\frac{C_V(T_D-T_A)}{C_p(T_C-T_B)}
=1-\frac1\gamma\frac{T_D-T_A}{T_C-T_B}
}.
$$

Solved by gpt-5.6-sol high.

= 36A
{parent=Paper 4}
{scope}
{title2=Electrodynamics}

= a
{parent=36a}
{scope}

= Solution
{parent=a}

The macroscopic <Maxwell equations in matter> are
$$
\boxed{\nabla\cdot\mathbf D=\rho},
\qquad
\boxed{\nabla\cdot\mathbf B=0},
$$
$$
\boxed{\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}},
\qquad
\boxed{\nabla\times\mathbf H=mathbf J+\frac{\partial\mathbf D}{\partial t}}.
$$
Here $\rho$ and $\mathbf J$ are the free charge and current densities.

Solved by gpt-5.6-sol high.

= b
{parent=36a}
{scope}

= Solution
{parent=b}

Dot the <Ampère-Maxwell equation> with $\mathbf E$ and <Faraday's law> with $\mathbf H$:
$$
\mathbf E\cdot\frac{\partial\mathbf D}{\partial t}
=\mathbf E\cdot(\nabla\times\mathbf H)-\mathbf E\cdot\mathbf J,
$$
$$
\mathbf H\cdot\frac{\partial\mathbf B}{\partial t}
=-\mathbf H\cdot(\nabla\times\mathbf E).
$$
Adding and using the vector identity
$$
\nabla\cdot(\mathbf E\times\mathbf H)
=\mathbf H\cdot(\nabla\times\mathbf E)
-\mathbf E\cdot(\nabla\times\mathbf H)
$$
gives
$$
\boxed{
\mathbf E\cdot\frac{\partial\mathbf D}{\partial t}
+\mathbf H\cdot\frac{\partial\mathbf B}{\partial t}
+\nabla\cdot(\mathbf E\times\mathbf H)
=-\mathbf E\cdot\mathbf J}.
$$
This is the local <Poynting theorem> before specializing the constitutive relations.

Solved by gpt-5.6-sol high.

= c
{parent=36a}
{scope}

= Solution
{parent=c}

Because $\varepsilon_{ij}$ and $\mu_{ij}$ are symmetric and time-independent, the <linear anisotropic dielectric> relations give
$$
E_i\partial_tD_i
=\varepsilon_{ij}E_i\partial_tE_j
=\frac12\partial_t(\varepsilon_{ij}E_iE_j),
$$
and similarly
$$
H_i\partial_tB_i
=\frac12\partial_t(\mu_{ij}H_iH_j).
$$
Integrate the identity from part (b) over $V$ and apply the <divergence theorem> to obtain
$$
\boxed{
\frac{\partial}{\partial t}\int_V
\frac12(\varepsilon_{ij}E_iE_j+\mu_{ij}H_iH_j)\,dV
+\int_S(\mathbf E\times\mathbf H)\cdot d\mathbf S
=-\int_V\mathbf E\cdot\mathbf J\,dV}.
$$
When $\mathbf J=0$, this is conservation of electromagnetic energy. The <Poynting theorem in a linear anisotropic medium> identifies
$$
\boxed{u=\frac12(\varepsilon_{ij}E_iE_j+\mu_{ij}H_iH_j)}
$$
as the energy density and
$$
\boxed{\mathbf N=\mathbf E\times\mathbf H}
$$
as the energy flux, or <Poynting vector>.

Solved by gpt-5.6-sol high.

= d
{parent=36a}
{scope}

= Solution
{parent=d}

Substitution of the sinusoidal <plane electromagnetic wave>[plane waves] into <Faraday's law> gives
$$
\mathbf k\times\mathbf e=\omega\mathbf b.
$$
The source-free <Ampère-Maxwell equation>, with $\mathbf H=\mathbf B/\mu$ and $D_i=\varepsilon_{ij}E_j$, gives
$$
\mathbf k\times\frac{\mathbf b}{\mu}
=-\omega\boldsymbol\varepsilon\mathbf e.
$$
Eliminating $\mathbf b=(\mathbf k\times\mathbf e)/\omega$ yields the wave equation for a <plane wave in an anisotropic dielectric>:
$$
\boxed{
[\mathbf k\times(\mathbf k\times\mathbf e)]_i
+\omega^2\mu\varepsilon_{ij}e_j=0}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=36a}
{scope}

= Solution
{parent=e}

Suppose $\varepsilon_{ij}e_j=\varepsilon_e e_i$. Taking the scalar product of the equation in part (d) with $\mathbf k$ gives $\mathbf k\cdot\mathbf e=0$. The <vector triple product> then reduces the equation to
$$
(-k^2+\omega^2\mu\varepsilon_e)\mathbf e=0,
$$
so $k^2=\omega^2\mu\varepsilon_e$. From part (d),
$$
\mathbf H=\frac{\mathbf B}{\mu}
=\frac{\mathbf k\times\mathbf e}{\mu\omega}
\sin(\mathbf k\cdot\mathbf x-\omega t).
$$
Consequently
$$
\mathbf N=\mathbf E\times\mathbf H
=\frac{\sin^2(\mathbf k\cdot\mathbf x-\omega t)}{\mu\omega}
\mathbf e\times(\mathbf k\times\mathbf e)
=\frac{|\mathbf e|^2\sin^2(\mathbf k\cdot\mathbf x-\omega t)}{\mu\omega}\mathbf k.
$$
Thus the instantaneous flux, except at its zeros, and its time average both point in the wave-propagation direction $\mathbf k$.

Solved by gpt-5.6-sol high.

= 37B
{parent=Paper 4}
{scope}
{title2=General Relativity}

= a
{parent=37b}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <Lorenz gauge in linearized gravity> gives
$$
0=\partial^\mu\bar h_{\mu\nu}
=ik^\mu H_{\mu\nu}e^{ik\cdot x},
$$
so the symmetric amplitude is transverse:
$$
\boxed{k^\mu H_{\mu\nu}=0}.
$$
Under this condition the supplied linearized Einstein tensor reduces to
$$
G_{\mu\nu}=-\frac12\Box\bar h_{\mu\nu}.
$$
The vacuum <Einstein field equations> and the plane-wave ansatz then imply
$$
\boxed{k^\rho k_\rho=0}.
$$
Thus the wavevector is null and the perturbation is a <plane gravitational wave in linearized gravity>. Lorenz gauge alone does not require the trace of $H_{\mu\nu}$ to vanish; that is a further gauge choice.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Under an infinitesimal coordinate transformation generated by $\xi_\mu$, the <trace-reversed metric perturbation> changes by
$$
\delta\bar h_{\mu\nu}
=-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu
+\eta_{\mu\nu}\partial_\rho\xi^\rho.
$$
Taking a divergence gives
$$
\partial^\mu\delta\bar h_{\mu\nu}=-\Box\xi_\nu.
$$
The transformation therefore preserves Lorenz gauge precisely when $\Box\xi_\nu=0$. For a plane-wave parameter $\xi_\mu=a_\mu e^{ik\cdot x}$, this holds because $k^2=0$. Absorbing a factor of $i$ into an arbitrary constant vector $\epsilon_\mu$, the amplitude transformation is
$$
\boxed{
H_{\mu\nu}\mapsto H_{\mu\nu}
+k_\mu\epsilon_\nu+k_\nu\epsilon_\mu
-\eta_{\mu\nu}k\cdot\epsilon}.
$$
This four-parameter freedom is the <residual gauge symmetry of linearized gravity>.

Solved by gpt-5.6-sol high.

= b
{parent=37b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Use the total mass
$$
M\simeq60M_\odot\simeq1.2\times10^{32}\ \mathrm{kg}.
$$
With the stated last-orbit estimate,
$$
r=\frac{6GM}{c^2}\simeq5\times10^5\ \mathrm m.
$$
The <Kepler third law> then gives
$$
T=2\pi\sqrt{\frac{r^3}{GM}}
=12\pi\sqrt6\,\frac{GM}{c^3}
\simeq3\times10^{-2}\ \mathrm s.
$$
A binary's dominant quadrupole radiation has two cycles per orbit, so the <last-orbit frequency of a compact binary> is
$$
\boxed{f_{\rm GW}\simeq\frac2T\simeq7\times10^1\ \mathrm{Hz}}.
$$
Only one significant figure is meaningful under the approximations.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

A suitable <spatially flat FLRW metric> is
$$
\boxed{ds^2=-c^2dt^2+a^2(t)(dx^2+dy^2+dz^2)}.
$$
Both source and detector are comoving, so <cosmological redshift> stretches the wave period in proportion to the <scale factor>. If $f_e$ is the local source frequency at emission and $f_o$ is the frequency measured in Earth's local frame, then
$$
\boxed{
f_o=f_e\frac{a(t_e)}{a(t_o)}
=\frac{f_e}{1+z}},
\qquad
1+z=\frac{a(t_o)}{a(t_e)}.
$$
The same relation applies to gravitational waves propagating on the FLRW background in the geometric-optics regime.

Solved by gpt-5.6-sol high.

= 38C
{parent=Paper 4}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=38c}
{scope}

= Solution
{parent=a}

The unperturbed <fluid wake> is the horizontal strip $-h<y<h$: the fluid is stationary there and moves to the right with <velocity> $U\mathbf e_x$ above and below it. Its two <shear layers> are the straight lines $y=\pm h$. After the symmetric disturbance, draw the upper layer at $y=h+\eta(x,t)$ and the lower one as its reflection at $y=-h-\eta(x,t)$, so the local wake width is $2(h+\eta)$.

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

Write the <velocity potentials> in the upper, middle and lower domains as $\phi_+$, $\phi_0$ and $\phi_-$. <Irrotational flow> and <incompressible flow> imply the <Laplace equation>
$$
\nabla^2\phi_j=0,\qquad j\in\{+,0,-\}.
$$
The <far-field boundary conditions> are
$$
\nabla\phi_+\longrightarrow U\mathbf e_x\quad(y\to+\infty),
\qquad
\nabla\phi_-\longrightarrow U\mathbf e_x\quad(y\to-\infty).
$$

Every shear layer is a material interface. The exact <kinematic boundary condition> at $y=h+\eta$ is
$$
\phi_{+,y}=\eta_t+\phi_{+,x}\eta_x,
\qquad
\phi_{0,y}=\eta_t+\phi_{0,x}\eta_x,
$$
and at $y=-h-\eta$ it is
$$
\phi_{-,y}=-\eta_t-\phi_{-,x}\eta_x,
\qquad
\phi_{0,y}=-\eta_t-\phi_{0,x}\eta_x.
$$
Here each equation is evaluated on the indicated moving interface.

Finally, <pressure continuity> is the <dynamic boundary condition for an inviscid interface>. The <Unsteady Bernoulli equation> gives
$$
\phi_{+,t}+\frac12|\nabla\phi_+|^2
=\phi_{0,t}+\frac12|\nabla\phi_0|^2
\quad(y=h+\eta),
$$
$$
\phi_{-,t}+\frac12|\nabla\phi_-|^2
=\phi_{0,t}+\frac12|\nabla\phi_0|^2
\quad(y=-h-\eta),
$$
after absorbing spatially constant functions of time into the potentials.

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= Solution
{parent=c}

The displacement must have small relative amplitude and small slope,
$$
\frac{|\eta|}{h}\ll1,\qquad |\eta_x|\ll1.
$$
Write the potentials as perturbations of the base flow,
$$
\phi_\pm=Ux+\varphi_\pm,\qquad \phi_0=\varphi_0,
$$
with $|\nabla\varphi_\pm|/U$ and $|\nabla\varphi_0|/U$ small. A <linearization> evaluates the conditions at the unperturbed lines $y=\pm h$ and discards products of perturbations. At $y=h$ this gives
$$
\varphi_{+,y}=\eta_t+U\eta_x,\qquad
\varphi_{0,y}=\eta_t,\qquad
\varphi_{+,t}+U\varphi_{+,x}=\varphi_{0,t},
$$
whereas at $y=-h$ it gives
$$
\varphi_{-,y}=-\eta_t-U\eta_x,\qquad
\varphi_{0,y}=-\eta_t,\qquad
\varphi_{-,t}+U\varphi_{-,x}=\varphi_{0,t}.
$$
The perturbation potentials remain <harmonic functions>, and their gradients decay at the corresponding far fields.

Solved by gpt-5.6-sol high.

= d
{parent=38c}
{scope}

= Solution
{parent=d}

The linearized equations are homogeneous and have constant coefficients in $x$ and $t$. Consequently, spatial <Fourier modes> evolve independently, and the <superposition principle> reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the <normal modes>
$$
\eta(x,t)=\operatorname{Re}\{\eta_0e^{ikx+\sigma t}\}.
$$
Here the real $k$ is the <wavenumber>, $\sigma$ is the possibly complex temporal exponent, and taking the <real part> produces a physical displacement.

Solved by gpt-5.6-sol high.

= e
{parent=38c}
{scope}

= Solution
{parent=e}

Put $q=|k|$. Decay at infinity and the symmetry of the disturbance suggest the mode potentials
$$
\varphi_+=Ae^{-q(y-h)}e^{ikx+\sigma t},\qquad
\varphi_-=Ae^{q(y+h)}e^{ikx+\sigma t},
$$
$$
\varphi_0=B\cosh(qy)e^{ikx+\sigma t}.
$$
The middle potential is an <even function> of $y$ because its vertical derivative must take opposite values at the two interfaces. The upper kinematic conditions give
$$
-qA=(\sigma+ikU)\eta_0,\qquad
qB\sinh(qh)=\sigma\eta_0.
$$
The lower conditions give the same two relations. The upper dynamic condition is
$$
(\sigma+ikU)A=\sigma B\cosh(qh).
$$
Eliminating $A$, $B$ and $\eta_0$ yields the <dispersion relation>
$$
\boxed{(\sigma+ikU)^2+\sigma^2\coth(qh)=0}.
$$
Writing $C=\coth(qh)$, its two roots are
$$
\boxed{\sigma=\frac{-ikU\mathbin{\pm}qU\sqrt C}{1+C}}.
$$
For every real $k\ne0$, $C>0$ and one root has the positive <growth rate>
$$
\operatorname{Re}\sigma=\frac{|k|U\sqrt{\coth(|k|h)}}{1+\coth(|k|h)}>0.
$$
Thus every nonzero wavenumber is linearly unstable in this inviscid <vortex-sheet> model; $k=0$ is neutral.

Solved by gpt-5.6-sol high.

= f
{parent=38c}
{scope}

= Solution
{parent=f}

When $|k|h\gg1$, the two shear layers are separated by many disturbance decay lengths. Since $\coth(|k|h)\to1$, the roots reduce to
$$
\boxed{\sigma\sim-\frac{ikU}{2}\mathbin{\pm}\frac{|k|U}{2}}.
$$
This is exactly the <Kelvin-Helmholtz instability> of one isolated <vortex sheet> separating equal-density streams of speeds $U$ and $0$: the pattern travels at their mean speed $U/2$, while its unstable amplitude grows at rate $|k|U/2$. The exponentially weak interaction between the two layers disappears in this short-wavelength <limit>.

Solved by gpt-5.6-sol high.

= 39C
{parent=Paper 4}
{scope}
{title2=Waves}

= a
{parent=39c}
{scope}

= Solution
{parent=a}

The one-dimensional <mass flux> and <momentum flux> are respectively
$$
J_m=\rho u,\qquad J_p=p+\rho u^2.
$$
In a steady flow, the <conservation laws> make both the mass flux and the given <energy flux>
$$
J_E=u\left(p+W+\frac12\rho u^2\right)
$$
constant. Wherever $J_m\ne0$, their ratio is therefore constant:
$$
\frac{J_E}{J_m}
=\frac{p+W}{\rho}+\frac12u^2
=\frac{\gamma}{\gamma-1}\frac p\rho+\frac12u^2
=C,
$$
where $W=p/(\gamma-1)$ is the <internal energy> per unit volume of the <ideal gas>. This is the required steady-flow energy integral.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= Solution
{parent=b}

Work in the <shock frame>, and let the positive upstream and downstream speeds be $w_0$ and $w_1$. The <Rankine-Hugoniot conditions for a perfect gas> give conservation of mass and momentum:
$$
\rho_0w_0=\rho_1w_1,
\qquad
p_0+\rho_0w_0^2=p_1+\rho_1w_1^2.
$$
Put
$$
r=\frac{\rho_1}{\rho_0},\qquad
\frac{p_1}{p_0}=1+\beta.
$$
Mass conservation gives $w_1=w_0/r$, and momentum conservation then gives
$$
w_0^2=\frac{p_0}{\rho_0}\frac{\beta r}{r-1}. \qquad (1)
$$

Applying the energy integral from part (a) on the two sides gives
$$
\frac{\gamma}{\gamma-1}\frac{p_0}{\rho_0}+\frac12w_0^2
=\frac{\gamma}{\gamma-1}\frac{p_0(1+\beta)}{\rho_0r}
+\frac12\frac{w_0^2}{r^2}. \qquad (2)
$$
Substitution of (1) into (2), followed by cancellation of $p_0/\rho_0$, yields
$$
\frac{\beta(r+1)}2
=\frac{\gamma}{\gamma-1}(1+\beta-r).
$$
Solving this linear equation for the density ratio gives
$$
\boxed{\frac{\rho_1}{\rho_0}=r
=\frac{2\gamma+(\gamma+1)\beta}
{2\gamma+(\gamma-1)\beta}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=39c}
{scope}

= Solution
{parent=c}

The <Taylor series> of the shock density ratio at $\beta=0$ is
$$
\frac{\rho_1}{\rho_0}
=1+\frac{\beta}{\gamma}
-\frac{\gamma-1}{2\gamma^2}\beta^2
+O(\beta^3). \qquad (1)
$$
For a reversible <adiabatic process>, $p/p_0=(\rho/\rho_0)^\gamma$, so the density ratio predicted by the ideal-gas adiabat is
$$
\left(\frac{p_1}{p_0}\right)^{1/\gamma}
=(1+\beta)^{1/\gamma}
=1+\frac{\beta}{\gamma}
-\frac{\gamma-1}{2\gamma^2}\beta^2
+O(\beta^3). \qquad (2)
$$
Equations (1) and (2) agree through order $\beta^2$. Thus a <weak shock> satisfies the adiabatic pressure-density relation with an error of order $O(\beta^3)$, which in particular establishes the requested accuracy through $O(\beta^2)$.

Solved by gpt-5.6-sol high.

= 40C
{parent=Paper 4}
{scope}
{title2=Numerical Analysis}

= a
{parent=40c}
{scope}

= Solution
{parent=a}

For a nonzero real vector $x$, the <Rayleigh quotient> of the <matrix> $A$ is
$$
\boxed{R_A(x)=\frac{x^TAx}{x^Tx}}.
$$
The <Rayleigh quotient iteration> starts from a unit vector $v_0$. Given $v_k$, compute the shift
$$
\mu_k=R_A(v_k),
$$
solve the <linear system>
$$
(A-\mu_kI)w_k=v_k,
$$
and normalize in the <Euclidean norm>:
$$
v_{k+1}=\frac{w_k}{\|w_k\|_2}.
$$
The next eigenvalue estimate is $\mu_{k+1}=R_A(v_{k+1})$. The process stops when the <eigenpair residual> is sufficiently small.

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= Solution
{parent=b}

Differentiate the <eigenvalue> equation
$$
A(t)v(t)=\lambda(t)v(t)
$$
at $t=0$, where $A'(0)=E$. Suppressing the argument $0$ gives
$$
Ev+Av'=\lambda'v+\lambda v'.
$$
Left-multiplication by the transpose of the unit <left eigenvector> $u$ and use of $u^TA=\lambda u^T$ cancel the terms involving $v'$:
$$
u^TEv=\lambda' u^Tv.
$$
Because the eigenvalue is <simple eigenvalue>[simple], $u^Tv\ne0$, and hence the <first-order perturbation of a simple eigenvalue> is
$$
\lambda'(0)=\frac{u(0)^TEv(0)}{u(0)^Tv(0)}.
$$
The definition of the <operator norm>, followed by the <Cauchy-Schwarz inequality>, gives
$$
|u^TEv|\leq\|u\|_2\|E\|_2\|v\|_2=\|E\|_2.
$$
Therefore
$$
\boxed{|\lambda'(0)|
\leq\frac{\|E\|_2}{|u(0)^Tv(0)|}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=40c}
{scope}

= Solution
{parent=c}

If $A$ is a real <symmetric matrix>, the finite-dimensional spectral theorem lets us choose the normalized <left eigenvector> and <right eigenvector> to be the same vector. Thus
$$
|u^Tv|=1,\qquad \boxed{s(\lambda)=1}.
$$

More generally, the <unitary diagonalization of a normal matrix> shows that a real <normal matrix> also has coincident normalized left and right eigenspaces for each real eigenvalue. Since the question assumes $\lambda\in\mathbb R$, one can again choose $u=v$, and every such simple eigenvalue has
$$
\boxed{s(\lambda)=1}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=40c}
{scope}

= Solution
{parent=d}

The displayed matrix is an <upper bidiagonal matrix>. For $\lambda_1=1$, the <right eigenvector> equation $(A-I)v=0$ successively gives
$$
v_2=v_3=\cdots=v_n=0.
$$
We may therefore take the unit right eigenvector to be $v=e_1$.

For a <left eigenvector>, $(A^T-I)u=0$ gives
$$
u_{i-1}+(\lambda_i-1)u_i=0
\qquad(2\leq i\leq n).
$$
Since $\lambda_i-1=-1/i$,
$$
u_i=i\,u_{i-1}.
$$
Choosing the initial scale $u_1=1$ produces
$$
u=(1,2!,3!,\ldots,n!)^T.
$$
After normalization,
$$
\widehat u=
\frac{(1,2!,3!,\ldots,n!)^T}
{\sqrt{1+(2!)^2+\cdots+(n!)^2}}.
$$
The <eigenvalue sensitivity> is therefore
$$
s(1)=\frac1{|\widehat u^Tv|}
=\sqrt{1+(2!)^2+\cdots+(n!)^2}
\geq n!,
$$
where $n!$ is the <factorial>. Hence
$$
\boxed{s(\lambda_1)\geq n!}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=40c}
{scope}

= Solution
{parent=e}

Let the computed <eigenpair residual> be
$$
r=A\widetilde v-\widetilde\lambda\widetilde v,
\qquad \|r\|_2=\epsilon,
\qquad \|\widetilde v\|_2=1.
$$
Define the <rank-one matrix>
$$
E=-r\widetilde v^T.
$$
Then
$$
(A+E)\widetilde v
=A\widetilde v-r(\widetilde v^T\widetilde v)
=\widetilde\lambda\widetilde v,
$$
so $(\widetilde\lambda,\widetilde v)$ is an exact eigenpair of the perturbed matrix $A+E$. Moreover,
$$
\|E\|_2=\|r\|_2\|\widetilde v\|_2=\epsilon,
$$
which is the <backward error of an approximate eigenpair>.

Follow the simple eigenvalue branch from $\lambda(0)=\lambda$ for $A(t)=A+tE$. The <first-order perturbation of a simple eigenvalue> and part (b) give
$$
|\widetilde\lambda-\lambda|
=|\lambda(1)-\lambda(0)|
\approx|\lambda'(0)|
\leq\|E\|_2s(\lambda).
$$
Since the residual is at the <machine precision> scale, this proves
$$
\boxed{|\widetilde\lambda-\lambda|\lesssim\epsilon s(\lambda)}.
$$

Solved by gpt-5.6-sol high.