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

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperii_4_2022.pdf

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

= Solution
{parent=1I}

The <continued fraction>[simple continued fraction] algorithm for $\sqrt{29}$ begins with $a_0=5$. Its complete quotients repeat after
$$
\sqrt{29}
=5+\cfrac1{
2+\cfrac1{
1+\cfrac1{
1+\cfrac1{
2+\cfrac1{
10+\cdots}}}}}.
$$
Thus
$$
\boxed{\sqrt{29}=[5;\overline{2,1,1,2,10}]}.
$$

The convergents through one period are
$$
5,\quad \frac{11}{2},\quad\frac{16}{3},
\quad\frac{27}{5},\quad\frac{70}{13}.
$$
The last one gives the <Pell-type norm from a periodic continued fraction>[negative Pell equation] solution
$$
70^2-29\cdot13^2=4900-4901=-1.
$$
Therefore
$$
\boxed{x=70,\qquad y=13}.
$$

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 4}
{scope}
{title2=Topics in Analysis}

= Solution
{parent=2G}

For $t<1/2$, the number $f(t)$ is negative real, so the proposed identity would force
$$
\phi(t)\in(2\mathbb Z+1)\pi.
$$
Continuity on the connected interval $[0,1/2)$ makes $\phi$ equal to one fixed odd multiple of $\pi$ there. For $t>1/2$, the number $f(t)$ is positive real, so $\phi$ must instead equal one fixed even multiple of $\pi$. These two constants cannot agree, and their one-sided limits at $1/2$ therefore contradict continuity. The fact that $f(1/2)=0$ removes the phase condition at that one point but does not repair the discontinuity.

Now let $g:[0,1]\to\mathbb C\setminus\{0\}$ and put $h=g/|g|$. This is a continuous path in the <unit circle>. The exponential map $\theta\mapsto e^{i\theta}$ is a <covering space>[covering map], so the <path lifting theorem> supplies a continuous lift $\theta:[0,1]\to\mathbb R$ after one value of $\theta(0)$ is chosen. Hence
$$
g(t)=|g(t)|e^{i\theta(t)}.
$$
One can obtain the same lift directly from the allowed special case: by uniform continuity, subdivide $[0,1]$ so that on each subinterval $h(t)/h(t_j)$ lies in the right half-plane, choose its continuous local argument there, and add a multiple of $2\pi$ to match the preceding endpoint.

If $\theta$ and $\widetilde\theta$ are two such phases, then
$$
\frac{\theta(t)-\widetilde\theta(t)}{2\pi}\in\mathbb Z.
$$
This integer-valued function is continuous and hence constant. Therefore
$$
r(g)=\theta(1)-\theta(0)
$$
does not depend on the chosen lift.

For $u(t)=g(t^2)$, the phase $\theta(t^2)$ gives
$$
\boxed{r(u)=r(g)}.
$$
For $v(t)=g(t)^2$, the phase $2\theta(t)$ gives
$$
\boxed{r(v)=2r(g)}.
$$

Finally, take
$$
g_1(t)=1,
\qquad
g_2(t)=e^{2\pi it}.
$$
They have the same two endpoints, but $r(g_1)=0$ and $r(g_2)=2\pi$. This endpoint discrepancy records the <winding number> of the closed path.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3K}

A general binary <feedback shift register> of length $d$ has state
$$
(x_n,x_{n+1},\ldots,x_{n+d-1})\in\mathbb F_2^d
$$
and a feedback function $F:\mathbb F_2^d\to\mathbb F_2$. One update outputs the oldest bit, shifts the state, and inserts
$$
x_{n+d}=F(x_n,\ldots,x_{n+d-1}).
$$
The initial fill is $(x_0,\ldots,x_{d-1})$. It is a <linear-feedback shift register> when
$$
F(z_0,\ldots,z_{d-1})
=c_0z_0+\cdots+c_{d-1}z_{d-1}
$$
for fixed $c_i\in\mathbb F_2$.

The <Berlekamp-Massey algorithm> reads an intercepted sequence from left to right while maintaining the shortest connection polynomial that reproduces the prefix. At each new symbol it computes the discrepancy between the observed bit and that predicted by the current recurrence. A zero discrepancy leaves the polynomial unchanged; a nonzero discrepancy adds a suitably shifted copy of the connection polynomial saved at the previous increase in linear complexity. After at least twice the unknown register length, it recovers the shortest recurrence, after which the entire keystream can be predicted.

Applying those discrepancy updates to
$$
1,1,0,0,1,0,1,1
$$
returns the connection polynomial
$$
\boxed{C(D)=1+D^2+D^3}.
$$
Equivalently, the sequence obeys
$$
\boxed{x_n=x_{n-2}+x_{n-3}\pmod2\qquad(n\geq3)}.
$$
Indeed this predicts successively $0,1,0,1,1$. No recurrence of length one or two fits the prefix, so its linear complexity is three.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4I}

A <context-free grammar> is in <Chomsky normal form> when every production has one of the forms
$$
A\to BC,
\qquad
A\to a,
$$
where $A,B,C$ are nonterminals and $a$ is a terminal. One may additionally allow $S\to\epsilon$ when the empty word belongs to the language, usually with the restriction that the start symbol does not occur on a right-hand side.

An $\epsilon$-production has the form $A\to\epsilon$. A <unit production> has the form $A\to B$ for nonterminals $A,B$.

In $G_1$, the productions for $T$ give
$$
L_{G_1}(T)=\{c\}\{a,b\}^*.
$$
Consequently
$$
L(G_1)
=\{\epsilon\}
\cup
\{a,b\}c\{a,b\}^*a.
$$
In $G_2$, $X$ and $Y$ generate $a$ and $b$, while $T\to TX\mid TY\mid c$ again gives
$$
L_{G_2}(T)=\{c\}\{a,b\}^*.
$$
Since $Z\to TX$, the two start productions $S\to XZ\mid YZ$ generate exactly
$$
L(G_2)=\{a,b\}c\{a,b\}^*a.
$$
Thus
$$
\boxed{L(G_2)=L(G_1)\setminus\{\epsilon\}}.
$$
Every production of $G_2$ has the required binary-nonterminal or single-terminal form, so $G_2$ is the Chomsky-normal-form grammar for the nonempty part of $L(G_1)$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5J}

Writing $Y_i=\operatorname{medv}_i$, $L_i=\operatorname{lstat}_i$, and $A_i=\operatorname{age}_i$, the fitted <normal linear model> is
$$
Y_i
=\beta_0+\beta_1L_i+\beta_2A_i+\beta_3L_iA_i+\varepsilon_i,
\qquad
\varepsilon_i\stackrel{\rm iid}{\sim}N(0,\sigma^2).
$$
The R formula \c[lstat * age] includes both main effects and their <interaction term>.

Three notable features are:

* The fitted model is highly significant overall: its $F$ statistic is $209.3$ on $(3,502)$ degrees of freedom with $p<2.2\times10^{-16}$. Its $R^2=0.5557$, however, means that about $44\%$ of the observed response variation remains unexplained, and the residual standard deviation is about $6.15$ house-value units.

* At fixed age, the fitted slope with respect to lstat is
  $$
  -1.3921+0.004156\,A.
  $$
  The large negative main coefficient is strongly significant. The positive interaction is significant at the $5\%$ level ($p=0.0252$), indicating that this negative association becomes slightly weaker for older housing tracts.

* The age main-effect estimate is essentially zero with $p=0.9711$, but it is the age effect specifically at $L=0$ and should not be interpreted separately from the significant interaction. The diagnostic plots also show curved residual structure, increasing spread, a heavy upper tail with observations 215, 372, and 373, and substantial influence from observation 215. These features cast doubt on linearity, constant variance, and Gaussian residual assumptions.

Solved by gpt-5.6-sol high.

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

= a
{parent=6c}
{scope}

= Solution
{parent=a}

The first reaction is reversible, with reverse rate $k_{-1}c_1$. Applying the <law of mass action> to each reaction gives
$$
\boxed{
\begin{aligned}
\dot e&=-k_1es+k_{-1}c_1,\\
\dot s&=-k_1es+k_{-1}c_1-k_2c_1s,\\
\dot c_1&=k_1es-k_{-1}c_1-k_2c_1s+k_3c_2,\\
\dot c_2&=k_2c_1s-k_3c_2,\\
\dot p&=k_3c_2.
\end{aligned}}
$$
Each term has the sign dictated by whether the corresponding species is consumed or produced.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

Adding the equations for the free enzyme and its two complexes gives
$$
\frac d{dt}(e+c_1+c_2)=0.
$$
Thus
$$
\boxed{e(t)+c_1(t)+c_2(t)=e_0}.
$$
This <conservation law> says that the enzyme is neither created nor destroyed: each enzyme molecule is either free, bound in $C_1$, or bound in $C_2$.

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= Solution
{parent=c}

Use enzyme conservation to write
$$
e=e_0(1-v_1-v_2).
$$
Since $\tau=k_1e_0t$, $u=s/s_0$, $v_i=c_i/e_0$, and $\epsilon=e_0/s_0$, substitution into the mass-action equations gives
$$
\boxed{
f(u,v_1,v_2)
=-u(1-v_1-v_2)+\alpha v_1-\beta uv_1
},
$$
and
$$
\boxed{
\begin{aligned}
g_1(u,v_1,v_2)
&=u(1-v_1-v_2)-\alpha v_1-\beta uv_1+\gamma v_2,\\
g_2(u,v_1,v_2)
&=\beta uv_1-\gamma v_2,
\end{aligned}}
$$
where
$$
\alpha=\frac{k_{-1}}{k_1s_0},
\qquad
\beta=\frac{k_2}{k_1},
\qquad
\gamma=\frac{k_3}{k_1s_0}.
$$
The factors of $\epsilon$ multiplying $dv_i/d\tau$ identify the complex concentrations as fast variables in a <singular perturbation>.

Solved by gpt-5.6-sol high.

= d
{parent=6c}
{scope}

= Solution
{parent=d}

On times $\tau\gg\epsilon$, the <quasi-steady-state approximation> sets
$$
g_1=g_2=0.
$$
The second equation gives
$$
v_2=\frac{\beta}{\gamma}uv_1.
$$
In the first equation the terms $-\beta uv_1+\gamma v_2$ then cancel, leaving
$$
u(1-v_1-v_2)=\alpha v_1.
$$
Solving these two equations gives
$$
v_1=\frac{u}{\alpha+u+(\beta/\gamma)u^2},
\qquad
v_2=\frac{(\beta/\gamma)u^2}
{\alpha+u+(\beta/\gamma)u^2}.
$$
Since $\dot p=k_3c_2=k_3e_0v_2$,
$$
\boxed{
\frac{dp}{dt}
=A\frac{u^2}{\alpha+u+(\beta/\gamma)u^2},
\qquad
A=\frac{k_3e_0\beta}{\gamma}=k_2e_0s_0
}.
$$

At small substrate concentration this allosteric rate is quadratic, $dp/dt\sim(A/\alpha)u^2$, so its graph leaves the origin with zero slope. At large $u$ it saturates at $A\gamma/\beta=k_3e_0$. A <Michaelis-Menten equation> also saturates, but its rate $V_{\max}u/(K_M+u)$ is linear near the origin. The allosteric curve is therefore more sigmoidal:
$$
\begin{array}{c|cc}
&u\to0&u\to\infty\\ \hline
\text{allosteric}&O(u^2)&k_3e_0\\
\text{Michaelis--Menten}&O(u)&V_{\max}
\end{array}
$$
This is the <quasi-steady rate law for a two-substrate allosteric enzyme>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7E}

The <Papperitz symbol>
$$
P\left\{
\begin{array}{ccc}
z_1&z_2&z_3\\
\alpha_1&\alpha_2&\alpha_3\\
\beta_1&\beta_2&\beta_3
\end{array}
\right\}
$$
describes the two-dimensional solution space of a second-order homogeneous <Fuchsian differential equation> on the Riemann sphere with exactly three <regular singular point criterion for a second-order equation>[regular singular points] $z_1,z_2,z_3$. Near $z_i$, its two independent Frobenius behaviours are locally
$$
(z-z_i)^{\alpha_i}
\quad\text{and}\quad
(z-z_i)^{\beta_i},
$$
subject to logarithmic modifications in resonant cases. The six <characteristic exponent at a regular singular point>[characteristic exponents] obey the <Fuchs relation>
$$
\sum_{i=1}^3(\alpha_i+\beta_i)=1.
$$

For the <Gauss hypergeometric equation>, the exponents at infinity are $a$ and $b$. Put $w=z^{-1}$. Transforming the Papperitz symbol from $z$ to $w$ shows that
$$
y_a(z)
=z^{-a}F(a,1+a-c;1+a-b;z^{-1})
$$
and
$$
y_b(z)
=z^{-b}F(b,1+b-c;1+b-a;z^{-1})
$$
are local solutions near $z=\infty$, with respective leading behaviours $z^{-a}$ and $z^{-b}$. When $a-b\notin\mathbb Z$, these behaviours are distinct and the two solutions are linearly independent. They therefore form a basis of the solution space on any simply connected common domain with compatible branch choices.

The solution $F(a,b;c;z)$ normalized at zero can be analytically continued into that domain. Since it solves the same second-order equation, it must be a constant linear combination of the basis at infinity:
$$
\boxed{
F(a,b;c;z)
=Az^{-a}F(a,1+a-c;1+a-b;z^{-1})
+Bz^{-b}F(b,1+b-c;1+b-a;z^{-1})
}.
$$
The constants depend on $a,b,c$ and the branch convention, but not on $z$. This is the <hypergeometric connection formula at infinity>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=8B}

Taking downward distance as positive, the <Lagrangian> is
$$
\boxed{
L
=\frac12(3m)\dot z_1^2+\frac12(2m)\dot z_2^2
-\frac k2(z_1-l)^2
-\frac k2(z_2-z_1-l)^2
+3mgz_1+2mgz_2
}.
$$

At equilibrium the lower spring supports only the lower mass, while the upper spring supports both masses:
$$
k(z_2-z_1-l)=2mg,
\qquad
k(z_1-l)=5mg.
$$
Hence
$$
\boxed{
z_1^{(0)}=l+\frac{5mg}{k},
\qquad
z_2^{(0)}=2l+\frac{7mg}{k}
}.
$$

Let $q_i=z_i-z_i^{(0)}$. The linear terms cancel by the equilibrium equations, and the quadratic Lagrangian is
$$
L
=\frac12\dot{\mathbf q}^{\,T}T\dot{\mathbf q}
-\frac12\mathbf q^TV\mathbf q+\text{constant},
$$
with the <small-oscillation mass and stiffness matrices>
$$
\boxed{
T=m\begin{pmatrix}3&0\\0&2\end{pmatrix},
\qquad
V=k\begin{pmatrix}2&-1\\-1&1\end{pmatrix}
}.
$$

The <generalized eigenvalue problem for small oscillations> is
$$
(V-\omega^2T)\mathbf a=0.
$$
Its characteristic equation is
$$
\det(V-\omega^2T)
=k^2-7km\omega^2+6m^2\omega^4=0,
$$
so
$$
\boxed{
\omega_1=\sqrt{\frac{k}{6m}},
\qquad
\omega_2=\sqrt{\frac{k}{m}}
}.
$$
Corresponding displacement eigenvectors are
$$
\boxed{
\mathbf a_1=(2,3)^T,
\qquad
\mathbf a_2=(1,-1)^T
}.
$$

They are not orthogonal in the ordinary Euclidean inner product. They are orthogonal in the kinetic-energy, or mass-matrix, inner product:
$$
\boxed{
\mathbf a_1^TT\mathbf a_2
=(2,3)
m\begin{pmatrix}3&0\\0&2\end{pmatrix}
\binom{1}{-1}
=0
}.
$$
This is <mass-matrix orthogonality of normal modes>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=9A}

The constant $R$ is the curvature radius of the closed spatial slices when $a=1$; at general scale factor their curvature radius is $Ra$ and their scalar curvature is $6/(R^2a^2)$.

The <cosmological perfect-fluid continuity equation> is
$$
\dot\rho+3\frac{\dot a}{a}(\rho+P)=0.
$$
During expansion, $\dot a>0$, and therefore
$$
\frac d{dt}(\rho a^3)
=-3Pa^2\dot a\leq0.
$$
If $C=\rho(t_0)a(t_0)^3$, then throughout the later expanding phase
$$
\rho a^3\leq C.
$$
Multiplying the <Friedmann equation> by $a^2$ gives
$$
\dot a^2
=\frac{8\pi G}{3c^2}\rho a^2
-\frac{c^2}{R^2}
\leq
\frac{8\pi GC}{3c^2a}
-\frac{c^2}{R^2}.
$$
The right-hand side is negative once
$$
a>\frac{8\pi GCR^2}{3c^4},
$$
which is impossible because $\dot a^2\geq0$. Thus the scale factor cannot grow without bound.

In fact the expanding phase ends in finite time. Combining the Friedmann and continuity equations yields the <Friedmann acceleration equation>
$$
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)<0.
$$
The curvature term in the first Friedmann equation also implies
$$
\rho\geq\frac{3c^4}{8\pi GR^2a^2}.
$$
While $a$ remains below the finite bound just found, $\ddot a$ is bounded above by a strictly negative constant. Hence $\dot a$ reaches zero after finite time: the closed universe attains a maximum size and cannot expand forever. This is <recollapse of a closed Friedmann universe with nonnegative pressure>.

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
$$
H_n=H^{\otimes n},
\qquad
|s\rangle=H_n|0^n\rangle
=\frac1{\sqrt N}\sum_{x\in B_n}|x\rangle,
$$
where $H$ is the <Hadamard gate>. For a computational-basis string $z$, define the phase reflection
$$
I_z=I-2|z\rangle\langle z|.
$$
Thus $I_0$ changes the sign of $|0^n\rangle$, while $I_{x_0}$ changes the sign of the marked state.

Conjugating $I_0$ by $H_n$ reflects in the hyperplane perpendicular to $|s\rangle$, so
$$
-H_nI_0H_n=2|s\rangle\langle s|-I
$$
is the <Grover diffusion operator>, the reflection about $|s\rangle$. The <Grover search algorithm>[Grover iteration operator]
$$
Q=(2|s\rangle\langle s|-I)I_{x_0}
$$
is the product of two reflections. It acts as a rotation through $2\theta$ in the plane spanned by the marked state $|x_0\rangle$ and the normalized uniform superposition of unmarked states, where
$$
\sin\theta=\frac1{\sqrt N}.
$$
It is the identity up to sign on the orthogonal complement of that plane.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

Use $n$ <controlled-NOT gates> in parallel. For each $j=1,\ldots,n$, the $j$th input qubit is the control and the $j$th output-register qubit is the target. Their combined action is
$$
|x_1\cdots x_n\rangle|y_1\cdots y_n\rangle
\longmapsto
|x_1\cdots x_n\rangle
|y_1\mathbin\oplus x_1,\ldots,y_n\mathbin\oplus x_n\rangle,
$$
which is exactly
$$
\boxed{|x\rangle|y\rangle\mapsto|x\rangle|y\mathbin\oplus x\rangle=U_{\mathcal I}|x\rangle|y\rangle}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=10d}
{scope}

= Solution
{parent=c}

The known circuit from part (b) lets us apply $U_{\mathcal I}$ without querying the faulty oracle. Their composition satisfies
$$
U_{\mathcal I}U_f|x\rangle|y\rangle
=
\begin{cases}
|x\rangle|y\rangle,&x\ne x_0,\\
|x\rangle|y\mathbin\oplus a\rangle,&x=x_0.
\end{cases}
$$
Because $a=00\cdots01$, only the last answer qubit is flipped in the exceptional case.

Prepare the answer register as
$$
|0\rangle^{\otimes(n-1)}|-\rangle,
\qquad
|-\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2}.
$$
Since $X|-\rangle=-|-\rangle$, <quantum phase kickback> gives, for an arbitrary search-register state,
$$
(U_{\mathcal I}U_f)
|x\rangle|0^{n-1}\rangle|-\rangle
=(-1)^{[x=x_0]}
|x\rangle|0^{n-1}\rangle|-\rangle.
$$
The answer register is unchanged and factors out. The induced operation on the first register is therefore
$$
\boxed{I_{x_0}=I-2|x_0\rangle\langle x_0|}.
$$
This is the <marked-state phase oracle from a single faulty identity-oracle query>.

Solved by gpt-5.6-sol high.

= d
{parent=10d}
{scope}

= Solution
{parent=d}

Use the construction in part (c) for the marked-state phase reflection in each <Grover search algorithm> iteration. Every such reflection costs one query to $U_f$; the identity-oracle circuit, Hadamard gates, and $I_0$ are independent of $f$.

Starting from $|s\rangle$, after $r$ iterations the success probability is
$$
\sin^2((2r+1)\theta),
\qquad
\sin\theta=N^{-1/2},
$$
by the <Grover rotation angle>. Choose an integer
$$
r=\left\lfloor\frac{\pi}{4\theta}-\frac12\right\rceil.
$$
Then $(2r+1)\theta$ differs from $\pi/2$ by at most $\theta$, so
$$
\Pr(\text{measure }x_0)
\geq\cos^2\theta
=1-\frac1N.
$$
For every sufficiently large $N$ this is greater than $1/2$, while
$$
r=O(\theta^{-1})=O(\sqrt N).
$$
Thus measuring the search register determines the faulty input with the required constant success probability using $O(\sqrt N)$ oracle queries.

Solved by gpt-5.6-sol high.

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

= a
{parent=11i}
{scope}

= Solution
{parent=a}

For an odd prime $p$, the <Legendre symbol> is
$$
\left(\frac ap\right)
=
\begin{cases}
0,&p\mid a,\\
1,&a\not\equiv0\pmod p\text{ is a square},\\
-1,&a\text{ is a nonsquare}.
\end{cases}
$$
<Euler criterion> states that
$$
a^{(p-1)/2}\equiv\left(\frac ap\right)\pmod p.
$$

The <Gauss lemma (number theory)> says that, for $p\nmid a$, if $m$ of the least positive residues of
$$
a,2a,\ldots,\frac{p-1}{2}a
$$
lie in $(p/2,p)$, then
$$
\left(\frac ap\right)=(-1)^m.
$$
To prove it, replace every residue above $p/2$ by its negative. The resulting absolute residues are distinct up to sign and therefore form a permutation of
$$
1,2,\ldots,\frac{p-1}{2}.
$$
Multiplying the congruences gives
$$
a^{(p-1)/2}\left(\frac{p-1}{2}\right)!
\equiv
(-1)^m\left(\frac{p-1}{2}\right)!
\pmod p.
$$
Cancel the nonzero factorial and apply Euler's criterion.

For $a=2$, the residues are $2,4,\ldots,p-1$. The number above $p/2$ is
$$
m=\frac{p-1}{2}-\left\lfloor\frac p4\right\rfloor,
$$
whose parity gives the <second supplementary law for quadratic reciprocity>
$$
\left(\frac2p\right)
=(-1)^{(p^2-1)/8}.
$$
Consequently, for odd primes,
$$
\boxed{x^2\equiv2\pmod p
\quad\Longleftrightarrow\quad
p\equiv1\text{ or }7\pmod8}.
$$
The congruence is also soluble for $p=2$.

Solved by gpt-5.6-sol high.

= b
{parent=11i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Suppose only finitely many primes $q_1,\ldots,q_r$ are congruent to $7$ modulo $8$, and set
$$
n=8q_1\cdots q_r.
$$
Then
$$
M=\frac{n^2-2}{2}\equiv7\pmod8.
$$
If an odd prime $q$ divides $M$, then $n^2\equiv2\pmod q$, so part (a) gives
$$
q\equiv1\text{ or }7\pmod8.
$$
Since their product $M$ is $7$ modulo $8$, at least one prime divisor $q$ is $7$ modulo $8$ to an odd power. Yet no listed $q_i$ divides $M$, because $n\equiv0\pmod{q_i}$ gives
$$
M\equiv-1\pmod{q_i}.
$$
This produces a new prime congruent to $7$ modulo $8$, a contradiction. Hence
$$
\boxed{\pi_7(x)\longrightarrow\infty}.
$$
This is the <Euclid proof for infinitely many primes congruent to seven modulo eight>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $N=\lfloor\sqrt x\rfloor$ and form
$$
P=\prod_{\lceil N/2\rceil\leq n\leq N}(n^2-2).
$$
For sufficiently large $x$, every factor is positive and at most $x$. Every odd prime divisor $p$ of $P$ makes $2$ a <quadratic residue> modulo $p$, so part (a) gives $p\equiv1$ or $7\pmod8$.

For an odd prime power $p^a$, the congruence $n^2\equiv2\pmod{p^a}$ has at most two residue classes, because each root modulo $p$ is simple and lifts uniquely. Therefore
$$
v_p(P)
\leq
\sum_{a\geq1}2\left(\frac N{p^a}+1\right)
\leq
\frac{2N}{p-1}+\frac{2\log x}{\log p}.
$$
Since $\log p/(p-1)\leq\log3/2$ for odd $p$,
$$
v_p(P)\log p\leq N\log3+2\log x.
$$
The prime $2$ contributes at most $N\log2$, because $v_2(n^2-2)\leq1$. If
$$
K=\pi_1(x)+\pi_7(x)+1,
$$
then, for all sufficiently large $x$, $2\log x\leq N\log3/2$ and hence
$$
\log P\leq\frac32KN\log3.
$$

On the other hand, the product contains at least $N/3$ factors for large $N$, and every one satisfies
$$
n^2-2\geq\frac{N^2}{4}-2\geq x^{3/4}
$$
once $x$ is sufficiently large. Thus
$$
\log P\geq\frac N3\cdot\frac34\log x
=\frac N4\log x.
$$
Comparing the two estimates gives
$$
\boxed{
\pi_1(x)+\pi_7(x)+1
=K
\geq\frac{\log x}{6\log3}
}
$$
for every sufficiently large $x$. This is the <elementary logarithmic lower bound for primes in the quadratic-residue classes of two>.

Solved by gpt-5.6-sol high.

= 12G
{parent=Paper 4}
{scope}
{title2=Topics in Analysis}

= a
{parent=12g}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <Brouwer fixed-point theorem> says that every continuous self-map of the closed unit disc has a fixed point.

A union of two disjoint closed discs does not have the <fixed-point property>. Define a map that is constant from the first component to a point in the second and constant from the second component to a point in the first. It is continuous because the components are disjoint and has no fixed point.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The open interval does not have the fixed-point property. The continuous self-map
$$
(x,0)\longmapsto(x/2,0)
$$
has its only possible fixed point at the excluded endpoint $(0,0)$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The closed interval does have the fixed-point property. If $f(x,0)=(F(x),0)$, then $F(0)-0\geq0$ and $F(1)-1\leq0$. The <intermediate value theorem> supplies $x\in[0,1]$ with $F(x)-x=0$.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

The annulus does not have the fixed-point property. The antipodal map
$$
x\longmapsto-x
$$
is a continuous self-map of the annulus and has no fixed point there.

Solved by gpt-5.6-sol high.

= b
{parent=12g}
{scope}

= Solution
{parent=b}

Let $T:\mathbb R^2\to D$ be the nearest-point projection,
$$
T(y)=
\begin{cases}
y,&|y|\leq1,\\
y/|y|,&|y|>1.
\end{cases}
$$
It is continuous. The map $T\circ f:D\to D$ has a fixed point $x$ by the <Brouwer fixed-point theorem>.

If $|x|<1$, the identity $T(f(x))=x$ can hold only with $f(x)=x$, because the projection of a point outside $D$ lies on its boundary. If $|x|=1$, the hypothesis gives $|f(x)|\leq1$, so $T(f(x))=f(x)$ and again $f(x)=x$. Thus $f$ has a fixed point.

Solved by gpt-5.6-sol high.

= c
{parent=12g}
{scope}

= Solution
{parent=c}

The set $E=\Delta_2\times\Delta_2$ is a compact convex square. Define
$$
u_i(p,q)=\max\{0,A(e_i,q)-A(p,q)\},
\qquad
p'=\frac{p+u(p,q)}{1+u_1(p,q)+u_2(p,q)}.
$$
Similarly, set
$$
v_j(p,q)=\max\{0,B(p,e_j)-B(p,q)\},
\qquad
q'=\frac{q+v(p,q)}{1+v_1(p,q)+v_2(p,q)}.
$$
These formulae define a continuous self-map $(p,q)\mapsto(p',q')$ of $E$. The square is homeomorphic to a closed disc, so the <Brouwer fixed-point theorem> gives a fixed point $(p^*,q^*)$.

Let $U=u_1+u_2$ at that fixed point. From $p'=p$,
$$
u_i=p_iU.
$$
If $U>0$, every $i$ with $p_i>0$ has $A(e_i,q)-A(p,q)>0$. This contradicts bilinearity, because
$$
\sum_i p_i\bigl(A(e_i,q)-A(p,q)\bigr)=0.
$$
Hence $U=0$, all $u_i=0$, and
$$
A(p^*,q^*)\geq A(e_i,q^*)
$$
for both pure strategies. By linearity this holds for every mixed strategy $p$. The same argument with $v_j$ gives
$$
B(p^*,q^*)\geq B(p^*,q)
$$
for every $q$. Therefore
$$
\boxed{
\forall(p,q)\in E,\quad
A(p^*,q^*)\geq A(p,q^*),
\qquad
B(p^*,q^*)\geq B(p^*,q)
}.
$$

In game terms, $p$ and $q$ are the players' <mixed strategies>, and $A,B$ are their expected payoffs in a two-by-two <bimatrix game>. The inequalities say that neither player can improve unilaterally, so $(p^*,q^*)$ is a <Nash equilibrium>. This is the <Brouwer proof of Nash equilibrium for a two-by-two game>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=13J}

This is <attenuation bias from classical measurement error>, also called regression dilution. The latent predictor is $Z$, but $X_1$ and $X_2$ are noisy proxies. The noisier proxy produces a slope closer to zero.

For a regression with an intercept, the population slope is
$$
\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}.
$$
Here all variables are centred and the noises are independent. Therefore
$$
\operatorname{Cov}(Z,Y)=\beta\sigma_z^2,
\qquad
\operatorname{Var}(Z)=\sigma_z^2,
$$
so regressing $Y$ on $Z$ has slope $\beta=2$. For
$$
X_j=Z+\eta_j,
$$
we have
$$
\operatorname{Cov}(X_j,Y)=\beta\sigma_z^2,
\qquad
\operatorname{Var}(X_j)=\sigma_z^2+\sigma_{x_j}^2.
$$
Thus
$$
\boxed{
\beta_{\rm slope}(Y\sim X_j)
=\beta\frac{\sigma_z^2}
{\sigma_z^2+\sigma_{x_j}^2}
}.
$$
With the stated values, the slopes are
$$
2\frac1{1+0.5^2}=1.6,
\qquad
2\frac1{1+1^2}=1,
$$
matching the simulation.

Under this independent additive measurement-error model, the magnitude of the $Y$-on-$X_1$ slope is generally smaller than the magnitude of the $Y$-on-$Z$ slope whenever $\sigma_{x_1}^2>0$. Increasing $\sigma_y$ does not change the population slope, because response noise contributes neither to $\operatorname{Cov}(X_1,Y)$ nor to $\operatorname{Var}(X_1)$. It increases residual variance and the sampling variability of the estimate; with one million observations, doubling it should leave the displayed slope close to $1.6$ while increasing its standard error.

Solved by gpt-5.6-sol high.

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

= a
{parent=14c}
{scope}

= Solution
{parent=a}

Let
$$
M=\begin{pmatrix}f_u&f_v\\g_u&g_v\end{pmatrix}
$$
be the reaction Jacobian at the homogeneous equilibrium. Stability without diffusion gives
$$
f_u+g_v<0,
\qquad
J=\det M=f_ug_v-f_vg_u>0.
$$
For a spatial Fourier mode with $-\nabla^2$ eigenvalue $k^2$, the linearized matrix is
$$
M-k^2\begin{pmatrix}D_u&0\\0&D_v\end{pmatrix}.
$$
Its trace is even more negative than $\operatorname{tr}M$, so instability occurs exactly when its determinant becomes negative. Put $z=D_vk^2$ and $d=D_u/D_v$. Then
$$
\Delta(z)
=J-(f_u+dg_v)z+dz^2.
$$
This upward-opening quadratic is negative for some $z>0$ exactly when its minimum occurs at positive $z$ and lies below zero:
$$
f_u+dg_v>0,
\qquad
J-\frac{(f_u+dg_v)^2}{4d}<0.
$$
The second strict inequality implies the first when written with a positive square root, so the <two-species diffusion-driven instability criterion> becomes
$$
\boxed{f_u+dg_v>2\sqrt{dJ}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=14c}
{scope}

= Solution
{parent=b}

If $d=1$, the condition would require
$$
f_u+g_v>2\sqrt J>0.
$$
But homogeneous stability requires $f_u+g_v<0$. These inequalities are incompatible, so equal diffusivities cannot produce a <Turing instability>.

Solved by gpt-5.6-sol high.

= c
{parent=14c}
{scope}

= Solution
{parent=c}

When $D_u=D_v=D$, the linearized operator is
$$
\partial_t
\binom{\widetilde u}{\widetilde v}
=
\left(M+D\nabla^2I\right)
\binom{\widetilde u}{\widetilde v}.
$$
For a Fourier mode of wavenumber $k$, its two growth rates are
$$
\lambda_j(k)=\lambda_j(0)-Dk^2.
$$
Both homogeneous growth rates have negative real part, and subtracting the nonnegative scalar $Dk^2$ moves them farther into the left half-plane. Thus no spatial mode can become unstable. This is the <impossibility of a two-species Turing instability at equal diffusivities>.

Solved by gpt-5.6-sol high.

= d
{parent=14c}
{scope}

= Solution
{parent=d}

For
$$
M=
\begin{pmatrix}
-1&-1\\
1+\delta&1-\delta
\end{pmatrix},
$$
we have
$$
\operatorname{tr}M=-\delta<0,
\qquad
J=\det M=2\delta>0,
$$
so the homogeneous equilibrium is stable for $\delta>0$. The Turing condition is
$$
-1+d(1-\delta)>2\sqrt{2\delta d}.
$$
Writing $y=\sqrt d$, equality at threshold is
$$
(1-\delta)y^2-2\sqrt{2\delta}\,y-1=0.
$$
Its positive root is
$$
y_c
=\frac{\sqrt{2\delta}+\sqrt{1+\delta}}{1-\delta},
\qquad
\boxed{
d_c
=\left(
\frac{\sqrt{2\delta}+\sqrt{1+\delta}}{1-\delta}
\right)^2
}.
$$
For $d>d_c$ sufficiently close to the threshold the strict instability condition holds, while
$$
d_c=1+2\sqrt{2\delta}+O(\delta)\longrightarrow1
\qquad(\delta\to0^+).
$$
Thus unequal diffusivities are necessary, but their ratio can be arbitrarily close to one when the stable reaction matrix is itself arbitrarily close to marginal stability. This is the <near-unity diffusivity ratio for a Turing instability>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=15B}

The gravitational three-body <Lagrangian> is
$$
L
=\frac12\sum_{i=1}^3m_i|\dot{\mathbf r}_i|^2
+G\left(
\frac{m_1m_2}{|\mathbf r_1-\mathbf r_2|}
+\frac{m_1m_3}{|\mathbf r_1-\mathbf r_3|}
+\frac{m_2m_3}{|\mathbf r_2-\mathbf r_3|}
\right).
$$

Put $M_{12}=m_1+m_2$ and $M=M_{12}+m_3$. Solving the definitions of the <Jacobi coordinates for three particles> gives
$$
\begin{aligned}
\mathbf r_1&=\mathbf c+\frac{m_3}{M}\mathbf b
+\frac{m_2}{M_{12}}\mathbf a,\\
\mathbf r_2&=\mathbf c+\frac{m_3}{M}\mathbf b
-\frac{m_1}{M_{12}}\mathbf a,\\
\mathbf r_3&=\mathbf c-\frac{M_{12}}M\mathbf b.
\end{aligned}
$$
Substitution into the kinetic energy makes all cross terms cancel:
$$
\boxed{
T
=\frac12\alpha|\dot{\mathbf a}|^2
+\frac12\beta|\dot{\mathbf b}|^2
+\frac12\gamma|\dot{\mathbf c}|^2
},
$$
where
$$
\boxed{
\alpha=\frac{m_1m_2}{m_1+m_2},
\qquad
\beta=\frac{(m_1+m_2)m_3}{m_1+m_2+m_3},
\qquad
\gamma=m_1+m_2+m_3
}.
$$

The pair separations are
$$
\mathbf r_1-\mathbf r_2=\mathbf a,
\qquad
\mathbf r_1-\mathbf r_3
=\mathbf b+\frac{m_2}{M_{12}}\mathbf a,
\qquad
\mathbf r_2-\mathbf r_3
=\mathbf b-\frac{m_1}{M_{12}}\mathbf a.
$$
Hence the <potential energy>[gravitational potential energy] is
$$
V(\mathbf a,\mathbf b)
=-\frac{Gm_1m_2}{|\mathbf a|}
-\frac{Gm_1m_3}
{\left|\mathbf b+\frac{m_2}{M_{12}}\mathbf a\right|}
-\frac{Gm_2m_3}
{\left|\mathbf b-\frac{m_1}{M_{12}}\mathbf a\right|},
$$
which is independent of $\mathbf c$. Thus $\mathbf c$ is an <ignorable coordinate> and
$$
\frac d{dt}(\gamma\dot{\mathbf c})=0,
\qquad
\boxed{\ddot{\mathbf c}=0}.
$$
The center of mass moves uniformly because the isolated system has no external force.

Finally, substituting the inverse coordinate transformation into
$$
\mathbf L=\sum_i m_i\mathbf r_i\times\dot{\mathbf r}_i
$$
again cancels every cross term and yields
$$
\boxed{
\mathbf L
=\alpha\mathbf a\times\dot{\mathbf a}
+\beta\mathbf b\times\dot{\mathbf b}
+\gamma\mathbf c\times\dot{\mathbf c}
}.
$$
This is the <angular momentum decomposition in three-body Jacobi coordinates>.

Solved by gpt-5.6-sol high.

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

= a
{parent=16f}
{scope}

= Solution
{parent=a}

The <von Neumann hierarchy> is defined by <transfinite recursion>:
$$
V_0=\varnothing,
\qquad
V_{\alpha+1}=\mathcal P(V_\alpha),
\qquad
V_\lambda=\bigcup_{\beta<\lambda}V_\beta
$$
for a limit ordinal $\lambda$.

We prove by <transfinite induction> that every $V_\alpha$ is a <transitive set>. The claim is immediate for $V_0$. If $V_\alpha$ is transitive and $x\in y\in V_{\alpha+1}$, then $y\subseteq V_\alpha$, so $x\in V_\alpha$. Transitivity also gives $x\subseteq V_\alpha$, hence $x\in\mathcal P(V_\alpha)=V_{\alpha+1}$. At a limit stage, if $x\in y\in V_\lambda$, then $y\in V_\beta$ for some $\beta<\lambda$, so $x\in V_\beta\subseteq V_\lambda$.

Transitivity implies $V_\alpha\subseteq\mathcal P(V_\alpha)=V_{\alpha+1}$. Induction and unions at limit stages then give
$$
\boxed{\alpha\leq\beta\Longrightarrow V_\alpha\subseteq V_\beta}.
$$

By well-founded recursion on membership, define the <rank of a set>
$$
\operatorname{rank}(x)
=\sup_{y\in x}\bigl(\operatorname{rank}(y)+1\bigr).
$$
Induction on this rank gives $y\in V_{\operatorname{rank}(x)}$ for every $y\in x$. Thus
$$
x\subseteq V_{\operatorname{rank}(x)}
$$
and consequently
$$
\boxed{x\in V_{\operatorname{rank}(x)+1}}.
$$
Every set therefore occurs at some level of the hierarchy.

Solved by gpt-5.6-sol high.

= b
{parent=16f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

A relation $r$ on a set $x$ is <well-founded relation>[well-founded] when every nonempty subset of $x$ has an $r$-minimal element. It is <extensional relation>[extensional] when distinct elements have distinct predecessor sets:
$$
a\ne b
\Longrightarrow
\{z\in x:z\,r\,a\}
\ne
\{z\in x:z\,r\,b\}.
$$
The <Mostowski collapse theorem> states that every well-founded extensional relation on a set is uniquely isomorphic to membership on a transitive set; the collapse is recursively
$$
\pi(a)=\{\pi(b):b\,r\,a\}.
$$

For the requested example, let $A=V_{\omega+1}$ and define
$$
x_0=A,
\qquad
x_n=A\cup\{x_0,\ldots,x_{n-1}\}\quad(n\geq1),
\qquad
x=\{x_n:n\in\omega\}.
$$
No $x_n$ belongs to $A$, and therefore
$$
x_m\in x_n\quad\Longleftrightarrow\quad m<n.
$$
The map $x_n\mapsto n$ is an isomorphism from membership on $x$ to membership on the von Neumann ordinal $\omega$, so the Mostowski collapse is $\omega$. On the other hand, $x_0=V_{\omega+1}$ already has rank $\omega+1$, and hence the rank of $x$ is greater than $\omega$.

For statement (i), suppose $(x,r)$ is isomorphic to $(y,\in)$ with $y$ transitive. Membership is well-founded by the axiom of foundation. It is extensional on $y$ because transitivity ensures that the predecessors in $y$ of an element $a\in y$ are exactly the elements of $a$, and the axiom of extensionality distinguishes different sets. Both properties are preserved by isomorphism. Thus (i) is always true.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Statement (ii) is always true even when $y$ is not transitive. Given a nonempty subset $z\subseteq y$, the <axiom of foundation> supplies $a\in z$ with $a\cap z=\varnothing$. Thus no member of $z$ is a member of $a$, so $a$ is minimal for membership restricted to $y$. An isomorphic relation $r$ is therefore well-founded.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Statement (iii) can fail because a nontransitive ambient set can omit the elements that distinguish two of its members. Take
$$
y=\{\varnothing,\{\{\varnothing\}\}\}.
$$
Neither element of $y$ has an element that also lies in $y$: the second contains $\{\varnothing\}$, which is omitted. Their predecessor sets for membership restricted to $y$ are therefore both empty, although the two elements are distinct. Hence this restricted membership relation, and any isomorphic relation, is not extensional.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=17F}

In the <Erdős-Rényi model> $G(n,p)$, the vertex set is $[n]$ and every one of the $\binom n2$ possible edges is present independently with probability $p$.

Let $X$ count <isolated vertices>. If $p\geq(1+\varepsilon)\log n/n$, the <union bound> gives
$$
\mathbb P(\delta(G_n)=0)
\leq n(1-p)^{n-1}
\leq n e^{-p(n-1)}
\longrightarrow0.
$$
Thus $\mathbb P(E_n)\to1$.

<Chebyshev inequality> states that
$$
\mathbb P(|X-\mathbb EX|\geq t)
\leq\frac{\operatorname{Var}X}{t^2}.
$$
Now suppose $p\leq(1-\varepsilon)\log n/n$. Monotonicity lets us use the largest allowed $p$. With
$$
a=(1-p)^{n-1},
$$
the supplied lower exponential estimate shows
$$
\mathbb EX=na\longrightarrow\infty.
$$
For distinct vertices $i,j$,
$$
\mathbb P(i,j\text{ both isolated})
=(1-p)^{2n-3}
=\frac{a^2}{1-p}.
$$
Hence
$$
\operatorname{Var}X
=na(1-a)+n(n-1)a^2\frac p{1-p},
$$
and
$$
\frac{\operatorname{Var}X}{(\mathbb EX)^2}
\leq\frac1{na}+\frac p{1-p}
\longrightarrow0.
$$
Chebyshev with $t=\mathbb EX$ gives $\mathbb P(X=0)\to0$, so $\mathbb P(E_n)\to0$.

For connectivity, the lower-threshold result is immediate because a connected graph has no isolated vertex:
$$
F_n\subseteq E_n.
$$
For the upper threshold, if a graph is disconnected it has a component with vertex set $S$ of some size $1\leq k\leq n/2$. In particular, every one of the $k(n-k)$ edges from $S$ to its complement is absent. Therefore
$$
\mathbb P(F_n^c)
\leq
\sum_{k=1}^{\lfloor n/2\rfloor}
\binom nk(1-p)^{k(n-k)}.
$$
Choose
$$
\eta=\frac{\varepsilon}{2(1+\varepsilon)}.
$$
For $1\leq k\leq\eta n$, the standard bound $\binom nk\leq(en/k)^k$ gives
$$
\binom nk(1-p)^{k(n-k)}
\leq
\left[
\frac ek\,
n^{1-(1+\varepsilon)(1-\eta)}
\right]^k
=
\left(\frac ek n^{-\varepsilon/2}\right)^k.
$$
The sum of these terms tends to zero. For $\eta n\leq k\leq n/2$, use $\binom nk\leq2^n$ and $k(n-k)\geq\eta n^2/2$:
$$
\sum_{\eta n\leq k\leq n/2}
\binom nk(1-p)^{k(n-k)}
\leq
n\,2^n
\exp\left(-\frac{\eta}{2}pn^2\right)
\longrightarrow0.
$$
Thus $\mathbb P(F_n)\to1$ above the threshold. Together,
$$
\boxed{
p\geq(1+\varepsilon)\frac{\log n}{n}
\Longrightarrow\mathbb P(F_n)\to1,
\qquad
p\leq(1-\varepsilon)\frac{\log n}{n}
\Longrightarrow\mathbb P(F_n)\to0
}.
$$
This is the <connectivity threshold in the Erdős-Rényi model>.

Solved by gpt-5.6-sol high.

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

= a
{parent=18h}
{scope}

= Solution
{parent=a}

By <Dirichlet theorem on primes in arithmetic progressions>, choose a prime $\ell$ with
$$
\ell\equiv1\pmod r.
$$
The <cyclotomic field> $E=\mathbb Q(\zeta_\ell)$ is Galois over $\mathbb Q$, and
$$
\operatorname{Gal}(E/\mathbb Q)
\cong(\mathbb Z/\ell\mathbb Z)^\times
\cong C_{\ell-1},
$$
where the last group is cyclic because the multiplicative group of a finite field is cyclic.

The cyclic group $C_{\ell-1}$ has a unique subgroup $H$ of order $(\ell-1)/r$. By the <Galois correspondence>, its fixed field $L=E^H$ satisfies
$$
\operatorname{Gal}(L/\mathbb Q)
\cong \operatorname{Gal}(E/\mathbb Q)/H
\cong C_r.
$$
Normality follows because $H$ lies in an abelian group. Thus every $r>1$ occurs as the Galois group of a finite extension of $\mathbb Q$. This is the <cyclic Galois extension of the rational numbers of every finite degree>.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

The products $xy$ with $x\in L_1$ and $y\in L_2$ span the <field compositum> $L_1L_2$ over $K$, so
$$
[L_1L_2:K]\leq d_1d_2.
$$
The <tower law> gives
$$
[L_1L_2:K]
=d_1[L_1L_2:L_1]
=d_2[L_1L_2:L_2].
$$
Thus the degree is divisible by both $d_1$ and $d_2$. Since they are coprime, it is divisible by $d_1d_2$. Combining divisibility with the upper bound gives
$$
\boxed{[L_1L_2:K]=d_1d_2}.
$$
Equivalently, finite extensions of coprime degrees are <linearly disjoint field extensions>.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Take
$$
G=C_p\times C_{p-1}\cong C_{p(p-1)}
$$
and let $N=C_p\times\{1\}$. Take
$$
G'=\operatorname{AGL}_1(\mathbb F_p)
=\mathbb F_p\rtimes\mathbb F_p^\times,
$$
the <affine group over a finite field>, and let $N'$ be its translation subgroup. Both normal subgroups have order $p$, and
$$
G/N\cong C_{p-1}\cong G'/N'.
$$
The groups are not isomorphic: $G$ is abelian, while $G'$ is nonabelian.

For $p=3$, let
$$
L=\mathbb Q(\zeta_7).
$$
Then $\operatorname{Gal}(L/\mathbb Q)\cong C_6=G$. The fixed field of its subgroup $N$ of order three is its unique quadratic subfield,
$$
\boxed{L^N=\mathbb Q(\sqrt{-7})}.
$$

Let
$$
L'=\mathbb Q(\sqrt[3]{2},\zeta_3),
$$
the <splitting field> of $X^3-2$. Its Galois group is $S_3\cong\operatorname{AGL}_1(\mathbb F_3)=G'$. The normal subgroup $N'=A_3$ fixes the quadratic discriminant field, so
$$
\boxed{(L')^{N'}=\mathbb Q(\sqrt{-3})=\mathbb Q(\zeta_3)}.
$$

Now let $p>3$. Part (a), with $r=p(p-1)$, supplies a cyclic Galois extension with group $G$. For $G'$, take the splitting field
$$
L'=\mathbb Q(\zeta_p,2^{1/p})
$$
of $X^p-2$. The prime $2$ is unramified in $\mathbb Q(\zeta_p)$, so at every prime ideal above $2$ its valuation is one. It cannot therefore be a $p$th power in the cyclotomic field. Hence
$$
[L':\mathbb Q(\zeta_p)]=p,
\qquad
[L':\mathbb Q]=p(p-1).
$$
The automorphisms
$$
\sigma(2^{1/p})=\zeta_p2^{1/p},
\quad \sigma(\zeta_p)=\zeta_p,
$$
and, for $a\in\mathbb F_p^\times$,
$$
\tau_a(2^{1/p})=2^{1/p},
\quad \tau_a(\zeta_p)=\zeta_p^a
$$
satisfy
$$
\tau_a\sigma\tau_a^{-1}=\sigma^a.
$$
They generate all $p(p-1)$ automorphisms, proving
$$
\boxed{
\operatorname{Gal}(L'/\mathbb Q)
\cong C_p\rtimes\mathbb F_p^\times
\cong\operatorname{AGL}_1(\mathbb F_p)=G'
}.
$$
This is the <affine Galois group of the splitting field of x to the p minus two>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=19H}

The <Mackey restriction formula> says, for an $H$-representation $V$,
$$
\operatorname{Res}_H^G\operatorname{Ind}_H^GV
\cong
\bigoplus_{x\in H\backslash G/H}
\operatorname{Ind}_{H\cap xHx^{-1}}^H
\operatorname{Res}_{H\cap xHx^{-1}}^{xHx^{-1}}({}^xV).
$$
<Frobenius reciprocity> says
$$
\langle\operatorname{Ind}_H^G\alpha,\beta\rangle_G
=
\langle\alpha,\operatorname{Res}_H^G\beta\rangle_H.
$$
Applying both with an irreducible character $\chi$ gives
$$
\left\langle
\operatorname{Ind}_H^G\chi,
\operatorname{Ind}_H^G\chi
\right\rangle_G
=
\sum_{x\in H\backslash G/H}
\left\langle
\operatorname{Res}_{H\cap xHx^{-1}}\chi,
\operatorname{Res}_{H\cap xHx^{-1}}({}^x\chi)
\right\rangle.
$$
The identity double coset contributes one, and all terms are nonnegative integers. Therefore <Mackey irreducibility criterion> says that $\operatorname{Ind}_H^G\chi$ is irreducible exactly when every term belonging to a nonidentity double coset is zero.

Now take $G=SL_2(k)$. Write
$$
U=\left\{
\begin{pmatrix}1&b\\0&1\end{pmatrix}:b\in k
\right\},
\qquad
T=\left\{
\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}:a\in k^\times
\right\},
$$
so $B=T\ltimes U$. For $q\geq4$, commutators of $T$ with $U$ generate $U$, because conjugation sends $u(b)$ to $u(a^2b)$ and one can choose $a^2\ne1$. Every degree-one character of $B$ is therefore trivial on $U$ and has the form
$$
\boxed{
\chi_\theta
\begin{pmatrix}a&b\\0&a^{-1}\end{pmatrix}
=\theta(a)
},
$$
where $\theta:k^\times\to\mathbb C^\times$ is a multiplicative character.

The <Bruhat decomposition of SL2 over a finite field> has two double cosets,
$$
G=B\sqcup BwB,
\qquad
w=\begin{pmatrix}0&-1\\1&0\end{pmatrix},
$$
and $B\cap wBw^{-1}=T$. Conjugation by $w$ inverts $T$, so
$$
\chi_\theta^w|_T=\chi_{\theta^{-1}}|_T.
$$
The only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. Consequently
$$
\boxed{
\operatorname{Ind}_B^G\chi_\theta
\text{ is irreducible}
\quad\Longleftrightarrow\quad
\theta\ne\theta^{-1}
\quad\Longleftrightarrow\quad
\theta^2\ne1
}.
$$
Thus the trivial character is excluded, as is the unique quadratic character when $q$ is odd. All other degree-one characters of $B$ induce irreducibly; the induced representations have degree $[G:B]=q+1$. This is the <irreducible principal series of finite SL2>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=20H}

Two nonzero <fractional ideals> $I,J$ of $\mathcal O_K$ are in the same ideal class when
$$
I=(\alpha)J
$$
for some $\alpha\in K^\times$, equivalently when $IJ^{-1}$ is principal. Multiplication is well defined on classes:
$$
[I][J]=[IJ].
$$
The identity is $[\mathcal O_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because $\mathcal O_K$ is a Dedekind domain. Thus these classes form the abelian <ideal class group> $\operatorname{Cl}(K)$.

For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne\alpha\in I$ with
$$
|N_{K/\mathbb Q}(\alpha)|\leq c_KN(I).
$$
Since $(\alpha)\subseteq I$, the ideal
$$
J=(\alpha)I^{-1}
$$
is integral, represents $[I]^{-1}$, and satisfies
$$
N(J)
=\frac{|N_{K/\mathbb Q}(\alpha)|}{N(I)}
\leq c_K.
$$
Hence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\operatorname{Cl}(K)$ is finite. This is the <bounded-norm ideal representatives> argument.

Now take $K=\mathbb Q(\sqrt{-33})$. Since $-33\equiv3\pmod4$,
$$
\mathcal O_K=\mathbb Z[\sqrt{-33}],
\qquad
d_K=-132.
$$
The imaginary-quadratic <Minkowski bound for ideal classes> is
$$
\frac2\pi\sqrt{|d_K|}
=\frac2\pi\sqrt{132}<8.
$$
Every class is therefore represented by an integral ideal of norm at most seven.

The ramified prime ideals above two and three are
$$
\mathfrak p_2=(2,1+\sqrt{-33}),
\qquad
\mathfrak p_3=(3,\sqrt{-33}),
$$
with
$$
\mathfrak p_2^2=(2),
\qquad
\mathfrak p_3^2=(3).
$$
Neither is principal, since the norm equation
$$
a^2+33b^2=2\quad\text{or}\quad3
$$
has no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thus
$$
1,\quad[\mathfrak p_2],\quad[\mathfrak p_3],
\quad[\mathfrak p_2\mathfrak p_3]
$$
are four distinct classes, all of order at most two.

The prime five is inert, so it contributes no ideal of norm five. The prime seven splits as
$$
\mathfrak q_\pm=(7,\sqrt{-33}\mp3).
$$
Since
$$
(3+\sqrt{-33})
=\mathfrak p_2\mathfrak p_3\mathfrak q_-,
$$
comparison of norms, all equal to $42$, confirms the factorization and gives
$$
[\mathfrak q_-]=[\mathfrak p_2\mathfrak p_3].
$$
The conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[\mathfrak p_2\mathfrak p_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. Therefore
$$
\boxed{
\operatorname{Cl}(\mathbb Q(\sqrt{-33}))
\cong C_2\times C_2
}.
$$
This is the <ideal class group of Q of square root of minus thirty-three>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21I}

Cut the square model of the <Klein bottle> along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a <Möbius band>, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The <Seifert-van Kampen theorem> therefore gives
$$
\boxed{
\pi_1(K,k_0)
\cong\langle a,b\mid a^2b^{-2}=1\rangle
}.
$$

Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentation
$$
\langle x,y\mid yxy^{-1}=x^{-1}\rangle.
$$
The orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroup
$$
H=\langle a^2,a^{-1}b\rangle\cong\mathbb Z^2.
$$
By the classification of connected covering spaces, $H$ determines the <orientation double cover of the Klein bottle>
$$
p:(T^2,x_0)\longrightarrow(K,k_0).
$$
Choosing torus generators $\alpha,\beta$ along the two translation directions, we may take
$$
\boxed{
p_*(\alpha)=a^2,
\qquad
p_*(\beta)=a^{-1}b
}.
$$
These commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.

The space $Y$ is the <mapping cylinder> of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/{\sim}\cong K$. Hence
$$
\boxed{
\pi_1(Y,y_0)\cong\pi_1(K,k_0)
\cong\langle a,b\mid a^2b^{-2}\rangle
}.
$$

Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, to
$$
X=U\cup(X\setminus K).
$$
The intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to $\pi_1(U)\cong\pi_1(K)$ has image precisely
$$
p_*\pi_1(T^2)=H.
$$
The orientation homomorphism
$$
\pi_1(U)\longrightarrow\pi_1(U)/H\cong C_2
$$
is zero on this intersection, so it is compatible with the trivial homomorphism from $\pi_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjection
$$
\pi_1(X)\twoheadrightarrow C_2.
$$
Therefore
$$
\boxed{X\text{ cannot be simply connected}}.
$$
This is the <Klein-bottle mapping-cylinder obstruction to simple connectivity>.

Solved by gpt-5.6-sol high.

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

= a
{parent=22g}
{scope}

= Solution
{parent=a}

The sequence $(f_n)$ is <equicontinuous family of functions>[equicontinuous] on $[0,1]$ when, for every $\varepsilon>0$, there is a $\delta>0$ such that
$$
|s-t|<\delta
\Longrightarrow
|f_n(s)-f_n(t)|<\varepsilon
$$
for every $n$. The <Arzela-Ascoli theorem> says that a uniformly bounded equicontinuous sequence of continuous real-valued functions on a compact interval has a uniformly convergent subsequence.

Solved by gpt-5.6-sol high.

= b
{parent=22g}
{scope}

= Solution
{parent=b}

Choose $A>0$ and set
$$
M=\max_{|x|\leq A}|\phi(x)|.
$$
Take
$$
T_1\leq\frac A{\max\{M,1\}}.
$$
Inductively, if $|f_n(s)|\leq A$ on $[0,T_1]$, then
$$
|f_{n+1}(t)|
\leq\int_0^t|\phi(f_n(s))|\,ds
\leq MT_1\leq A.
$$
Since $f_0=0$, every iterate is bounded by $A$. Moreover,
$$
|f_{n+1}(t)-f_{n+1}(s)|
=\left|\int_s^t\phi(f_n(r))\,dr\right|
\leq M|t-s|.
$$
Thus $(f_n)_{n\geq1}$ is uniformly bounded and uniformly Lipschitz, hence equicontinuous, on $[0,T_1]$.

Solved by gpt-5.6-sol high.

= c
{parent=22g}
{scope}

= Solution
{parent=c}

The <Arzela-Ascoli theorem> makes the family of iterates relatively compact in $C([0,T_1])$. The standard successive-approximation proof for this scalar autonomous equation shows that, after decreasing to some $T_2\leq T_1$, the adjacent differences
$$
R_n=f_{n+1}-f_n
$$
tend uniformly to zero. One way to finish the existence argument without requiring a Lipschitz hypothesis on $\phi$ is the following direct scalar construction.

If $\phi(0)=0$, the function $f(t)=0$ is already a solution. If $\phi(0)\ne0$, continuity gives an interval $[-A_0,A_0]$ on which $\phi$ has constant sign and never vanishes. Define
$$
F(x)=\int_0^x\frac{du}{\phi(u)}.
$$
Then $F$ is continuously differentiable with $F'(0)\ne0$. By the <inverse function theorem>, it has a continuously differentiable local inverse on $[0,T_2]$ after choosing $T_2>0$ small enough and choosing the appropriate one-sided range. Put
$$
f(t)=F^{-1}(t).
$$
The chain rule gives
$$
F'(f(t))f'(t)=1,
$$
and hence
$$
\boxed{f(0)=0,\qquad f'(t)=\phi(f(t))}.
$$
Equivalently, $f(t)=\int_0^t\phi(f(s))\,ds$, the fixed-point equation approximated by the iterates. This is the <local existence for a scalar autonomous ordinary differential equation with continuous vector field>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=23G}

For $s\in\mathbb R$, the <Sobolev space> $H^s(\mathbb R^n)$ consists of tempered distributions $u$ such that
$$
\|u\|_{H^s}^2
=\int_{\mathbb R^n}
(1+|\xi|^2)^s|\widehat u(\xi)|^2\,d\xi
<\infty.
$$
For a multi-index $\alpha$,
$$
\widehat{D^\alpha u}(\xi)=(i\xi)^\alpha\widehat u(\xi).
$$
Since
$$
(1+|\xi|^2)^{s-|\alpha|}|\xi|^{2|\alpha|}
\leq(1+|\xi|^2)^s,
$$
we obtain the <Sobolev derivative estimate>
$$
\boxed{
\|D^\alpha u\|_{H^{s-|\alpha|}}
\leq\|u\|_{H^s}
}.
$$
Thus differentiation is bounded and linear between the stated spaces.

Taking the <Fourier transform> of
$$
-\Delta u+u=f
$$
gives
$$
(1+|\xi|^2)\widehat u(\xi)=\widehat f(\xi).
$$
The only possible solution is therefore
$$
\widehat u(\xi)=\frac{\widehat f(\xi)}{1+|\xi|^2}.
$$
Moreover,
$$
\|u\|_{H^{s+2}}^2
=\int(1+|\xi|^2)^{s+2}
\frac{|\widehat f(\xi)|^2}{(1+|\xi|^2)^2}\,d\xi
=\|f\|_{H^s}^2.
$$
This proves existence, uniqueness, and bounded dependence. Conversely, $u\mapsto(-\Delta+1)u$ maps $H^{s+2}$ boundedly to $H^s$, so
$$
\boxed{
-\Delta+1:H^{s+2}(\mathbb R^n)\longrightarrow H^s(\mathbb R^n)
}
$$
is a linear isomorphism with inverse multiplier $(1+|\xi|^2)^{-1}$. This is the <massive Laplacian isomorphism on Sobolev spaces>.

Finally, the assumed estimate and <Plancherel theorem> imply
$$
\int_{\mathbb R^n}
(1+|\xi|^2)^2|\widehat u_j(\xi)|^2\,d\xi
\leq
C\bigl(\|u_j\|_2^2+\|\Delta u_j\|_2^2\bigr)
\leq C K^2.
$$
Thus $(u_j)$ is bounded in $H^2(\mathbb R^n)$. The functions and all their first derivatives are supported in the fixed bounded set $\overline\Omega$. Their $H^1$ bounds give uniform translation estimates
$$
\|v(\,\cdot+h)-v\|_2
\leq|h|\,\|\nabla v\|_2
$$
for $v=u_j$ and $v=D_\ell u_j$. Tight support and the <Rellich-Kondrashov compactness theorem for H01>, equivalently the Fourier compactness criterion, therefore provide a common subsequence for which $u_j$ and every first derivative converge strongly in $L^2$. Hence
$$
\boxed{u_{j_k}\longrightarrow u\quad\text{strongly in }H^1(\mathbb R^n)}.
$$
This is <compactness from bounded support and an H2 bound>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24H}

For a <divisor on an algebraic curve> $D=\sum_pa_pp$ over an algebraically closed field, its <degree of a divisor>[degree] is
$$
\deg D=\sum_pa_p.
$$
For a nonzero <rational function> $f$, its <principal divisor on an algebraic curve>[principal divisor] is
$$
(f)=\operatorname{div}(f)=\sum_p\operatorname{ord}_p(f)p,
$$
where zeros have positive order and poles have negative order.

Now let $D=\sum_i a_ip_i$ have degree zero on the <projective line>. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since $\sum_i a_i=0$, the expression
$$
f=\prod_iL_i^{a_i}
$$
is homogeneous of degree zero and therefore defines a rational function on $\mathbb P^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, so
$$
(f)=\sum_i a_ip_i=D.
$$
Thus every degree-zero divisor on $\mathbb P^1$ is principal. If $E$ and $E'$ have the same degree, then $E-E'$ is principal, whence
$$
\boxed{E\sim E'}.
$$
This is the <divisor class on the projective line>.

Write $p_\infty=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; then
$$
dt=-s^{-2}ds.
$$
Hence the rational differential has one double pole at infinity and no other zero or pole:
$$
\boxed{(dt)=-2p_\infty},
$$
which represents the <canonical divisor of the projective line>.

If $D\sim mK_{\mathbb P^1}$, then $D\sim-2mp_\infty$, and <linear equivalence of divisors> gives $L(D)\cong L(-2mp_\infty)$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. Therefore
$$
\boxed{\ell(D)=\max\{1-2m,0\}}.
$$
This calculation uses only the rational functions on $\mathbb P^1$, rather than the <Riemann-Roch theorem>.

Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one <finite morphism> $f:C\to\mathbb P^1$. A degree-one finite morphism between smooth projective curves is an <isomorphism of algebraic varieties>[isomorphism]. It would follow that $C\cong\mathbb P^1$ and hence that $C$ has <geometric genus>[genus] zero, contrary to the hypothesis. Thus the <principal divisor with one simple zero and one simple pole> cannot occur on $C$:
$$
\boxed{p-q\text{ is not principal}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=25i}
{scope}

= Solution
{parent=a}

The <periodic Wirtinger inequality> says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, then
$$
\int_0^L f(s)^2\,ds
\leq \frac{L^2}{4\pi^2}\int_0^L f'(s)^2\,ds.
$$
Equality holds exactly for a linear combination of $\cos(2\pi s/L)$ and $\sin(2\pi s/L)$.

First suppose $\partial\Omega$ is one positively oriented simple closed curve, parametrized by <arc length> as $\gamma(s)=(x(s),y(s))$ for $0\leq s\leq L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. <Green theorem> and the <Cauchy-Schwarz inequality> give
$$
\begin{aligned}
2\operatorname{Area}(\Omega)
&=\int_0^L(xy'-yx')\,ds\\
&\leq
\left(\int_0^L(x^2+y^2)\,ds\right)^{1/2}
\left(\int_0^L((x')^2+(y')^2)\,ds\right)^{1/2}.
\end{aligned}
$$
Applying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x')^2+(y')^2=1$ yields
$$
2\operatorname{Area}(\Omega)
\leq\frac{L}{2\pi}\,L,
$$
and therefore the <planar isoperimetric inequality>
$$
\boxed{4\pi\operatorname{Area}(\Omega)\leq L^2}.
$$
For several boundary components, apply the simple-curve result to the relevant enclosed regions and use $\sum_jL_j^2\leq(\sum_jL_j)^2$; holes only decrease the area.

Equality in both inequalities forces
$$
(x,y)=A\cos(2\pi s/L)+B\sin(2\pi s/L)
$$
for constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.

Solved by gpt-5.6-sol high.

= b
{parent=25i}
{scope}

= Solution
{parent=b}

Let $A$ and $C$ be opposite interior angles of a convex quadrilateral with side lengths $a,b,c,d$, and let $s=(a+b+c+d)/2$. The <Bretschneider formula> gives its area $K$ as
$$
K^2=(s-a)(s-b)(s-c)(s-d)
-abcd\cos^2\frac{A+C}{2}.
$$
The fixed side lengths fix the first term, while the second term is nonpositive. Hence
$$
K^2\leq(s-a)(s-b)(s-c)(s-d),
$$
with equality exactly when $A+C=\pi$. Opposite angles are supplementary exactly when the quadrilateral is a <cyclic quadrilateral>. The given $Q$ is cyclic, so it attains this upper bound and
$$
\boxed{\operatorname{Area}(Q)\geq\operatorname{Area}(Q')}.
$$
A crossed or concave competitor can be uncrossed or reflected across a diagonal without changing its side lengths and without decreasing its unsigned area, so the same bound applies. Apart from degenerate coincidences, equality holds precisely when $Q'$ is also cyclic with the stated cyclic ordering of its side lengths.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=26G}

Because the centered <normal distribution> density $g_t$ is nonnegative and has integral one, <Fubini's theorem> and translation invariance give
$$
\begin{aligned}
\|f*g_t\|_1
&\leq\int_{\mathbb R}\int_{\mathbb R}|f(x-y)|g_t(y)\,dy\,dx\\
&=\int_{\mathbb R}g_t(y)\|f\|_1\,dy
=\|f\|_1.
\end{aligned}
$$
Thus $f*g_t\in L^1$.

Use the angular-frequency convention
$$
\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-ix\xi}\,dx.
$$
The <convolution theorem> and the Gaussian transform give
$$
\widehat{f*g_t}(\xi)=\widehat f(\xi)e^{-t\xi^2/2}.
$$
Since $|\widehat f(\xi)|\leq\|f\|_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:
$$
\begin{aligned}
\frac1{2\pi}\int_{\mathbb R}\widehat f(\xi)e^{-t\xi^2/2}e^{ix\xi}\,d\xi
&=\int_{\mathbb R}f(z)
\left[\frac1{2\pi}\int_{\mathbb R}e^{-t\xi^2/2}e^{i(x-z)\xi}\,d\xi\right]dz\\
&=\int_{\mathbb R}f(z)g_t(x-z)\,dz\\
&=(f*g_t)(x).
\end{aligned}
$$
This proves the <Fourier inversion theorem> for the convolution directly.

Now additionally suppose $\widehat f\in L^1$. For every $x$, the <dominated convergence theorem> gives
$$
\lim_{t\downarrow0}(f*g_t)(x)
=\frac1{2\pi}\int_{\mathbb R}\widehat f(\xi)e^{ix\xi}\,d\xi.
$$
Fourier inversion identifies the right-hand side with $f(x)$ for <Lebesgue almost everywhere> $x$. Therefore the <Gaussian approximate identity> satisfies
$$
\boxed{f*g_t(x)\longrightarrow f(x)\quad\text{for almost every }x.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=27j}
{scope}

= Solution
{parent=a}

The queue length is an <M-M-1 queue> and hence a <birth-death process> with birth rate $\lambda$ and death rate $\mu$ away from zero. The <detailed balance for a birth-death process>[detailed-balance] equations are
$$
\pi_n\lambda=\pi_{n+1}\mu,
\qquad n\geq0.
$$
Writing $\rho=\lambda/\mu$, they give $\pi_n=\pi_0\rho^n$. Since $\rho<1$, this measure is summable and normalization yields
$$
\boxed{\pi_n=(1-\rho)\rho^n,\qquad n\geq0}.
$$
The existence of this invariant probability distribution for the irreducible nonexplosive chain proves <positive recurrence of an M-M-1 queue>[positive recurrence].

Solved by gpt-5.6-sol high.

= b
{parent=27j}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

With state-dependent admission, the effective birth rate in state $n$ is
$$
b_n=\lambda p(n)=\frac{\lambda}{n+1},
$$
while the death rate is $\mu$ for $n\geq1$. The stationary ratios for an <M-M-1 queue with state-dependent admission> are therefore
$$
\frac{\pi_{n+1}}{\pi_n}=\frac{\rho}{n+1}.
$$
Induction gives $\pi_n=\pi_0\rho^n/n!$, and normalization identifies a <Poisson distribution> of mean $\rho$:
$$
\boxed{\pi_n=e^{-\rho}\frac{\rho^n}{n!}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Here $b_n=\lambda2^{-n}$, so detailed balance gives
$$
\frac{\pi_{n+1}}{\pi_n}=\rho2^{-n}.
$$
Consequently
$$
\pi_n=\pi_0\rho^n2^{-n(n-1)/2},
\qquad
\boxed{\pi_0=
\left(\sum_{n=0}^{\infty}\rho^n2^{-n(n-1)/2}\right)^{-1}}.
$$
The series converges for every $\rho>0$, so this defines the unique invariant distribution.

By the <Poisson arrivals see time averages>[PASTA property], an arrival joins with probability $\sum_{n\geq0}\pi_np(n)$. In equilibrium the accepted-arrival rate must also equal the departure rate. Departures occur at rate $\mu$ precisely when the queue is nonempty, hence
$$
\lambda\sum_{n\geq0}\pi_np(n)=\mu(1-\pi_0).
$$
The required joining probability is therefore
$$
\boxed{\sum_{n\geq0}\pi_np(n)=\frac{\mu(1-\pi_0)}{\lambda}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Generate $U\sim\operatorname{Uniform}[0,1]$ and set
$$
X=-\log U.
$$
For $x\geq0$,
$$
\mathbb P(X\leq x)
=\mathbb P(U\geq e^{-x})
=1-e^{-x},
$$
which is the distribution function of an <exponential distribution> with rate one. This is <inverse transform sampling>.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

In one proposal of <rejection sampling>, the probability of both proposing a value in $dx$ and accepting it is
$$
h(x)\,dx\,\frac{f(x)}{Mh(x)}
=\frac1M f(x)\,dx.
$$
Integrating shows that the acceptance probability is $1/M$. Conditional on acceptance, the density of the proposed value is therefore
$$
\frac{f(x)/M}{1/M}=f(x).
$$
Each unsuccessful proposal restarts independently, so the eventual output $Y$ has the same conditional density:
$$
\boxed{Y\sim f}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

Take the proposal density $h(x)=e^{-x}\mathbf1_{x\geq0}$. For the target <half-normal distribution>,
$$
\frac{f(x)}{h(x)}
=\sqrt{\frac2\pi}\exp\left(-\frac{x^2}{2}+x\right)
=\sqrt{\frac{2e}{\pi}}\exp\left(-\frac{(x-1)^2}{2}\right).
$$
The ratio is maximized at $x=1$, so the sharp envelope constant is
$$
\boxed{M=\sqrt{\frac{2e}{\pi}}}.
$$
Generate $X=-\log U_1$ and an independent uniform $U_2$, and accept $X$ when
$$
U_2\leq\frac{f(X)}{Mh(X)}
=e^{-(X-1)^2/2}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=28k}
{scope}

= Solution
{parent=d}

Each independent proposal is accepted with probability $1/M$. The number of proposals is therefore a <geometric distribution> with mean
$$
\boxed{M=\sqrt{\frac{2e}{\pi}}}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=28k}
{scope}

= Solution
{parent=e}

The accepted $Y$ has the <half-normal distribution>, so $Y=|G|$ in distribution for $G\sim N(0,1)$. Multiplying it by an independent random sign that is positive and negative with equal probabilities restores the two symmetric halves of the Gaussian density. Thus
$$
\boxed{Z\sim N(0,1)}.
$$

Solved by gpt-5.6-sol high.

= f
{parent=28k}
{scope}

= Solution
{parent=f}

For an exponential proposal of rate $\lambda>0$,
$$
h_\lambda(x)=\lambda e^{-\lambda x}\mathbf1_{x\geq0},
$$
and the smallest valid rejection constant is
$$
\begin{aligned}
M(\lambda)
&=\sup_{x\geq0}\frac{f(x)}{h_\lambda(x)}\\
&=\frac1\lambda\sqrt{\frac2\pi}
\sup_{x\geq0}\exp\left(-\frac{x^2}{2}+\lambda x\right)\\
&=\sqrt{\frac2\pi}\,\frac{e^{\lambda^2/2}}{\lambda},
\end{aligned}
$$
because the exponent is maximized at $x=\lambda$. Differentiating its logarithm gives
$$
\frac d{d\lambda}\log M(\lambda)
=\lambda-\frac1\lambda.
$$
The unique minimum occurs at $\lambda=1$. Since rejection efficiency is $1/M(\lambda)$, choosing any $\lambda\ne1$ makes the algorithm less efficient:
$$
\boxed{\text{the rate-one proposal is optimal within the exponential family}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=29k}
{scope}

= Solution
{parent=a}

Between times $n-1$ and $n$, the investor holds $\theta_n$ shares worth $\theta_nS_{n-1}$ and places the remaining wealth $X_{n-1}-\theta_nS_{n-1}$ in the risk-free asset. The <self-financing portfolio>[self-financing] condition therefore gives
$$
\begin{aligned}
X_n
&=\theta_nS_n+(1+r)(X_{n-1}-\theta_nS_{n-1})\\
&=(1+r)X_{n-1}
+\theta_n[S_n-(1+r)S_{n-1}].
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

Under a <risk-neutral probability in a binomial market>[risk-neutral measure], the discounted stock must be a <martingale>. If $q$ is the probability of the factor $1+b$, this condition is
$$
q(1+b)+(1-q)(1+a)=1+r,
$$
so
$$
\boxed{q=\frac{r-a}{b-a}}.
$$
The hypothesis $a<r<b$ ensures $0<q<1$, making this equivalent to the original probability measure. Since the $N$ returns are independent under $Q$, the number of up moves has a <binomial distribution>, and hence
$$
\boxed{
Q\!\left(S_N=S_0(1+b)^i(1+a)^{N-i}\right)
=\binom Niq^i(1-q)^{N-i}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

The <backward option pricing> recursion is
$$
V(n-1,s)
=\frac{qV(n,s(1+b))+(1-q)V(n,s(1+a))}{1+r},
$$
with terminal value $V(N,s)=g(s)$. At a node with $S_{n-1}=s$, define the <replicating portfolio in a binomial market>[replicating] stock holding by
$$
\boxed{
\theta_n
=\frac{V(n,s(1+b))-V(n,s(1+a))}
{s(b-a)}}.
$$
This is a <predictable process> because it depends only on information available at time $n-1$.

Indeed, if $X_{n-1}=V(n-1,s)$, substituting the risk-neutral formula for $q$ into the self-financing update shows separately in the up and down states that
$$
X_n=V(n,S_n).
$$
An induction over $n$ therefore gives $X_N=V(N,S_N)=g(S_N)$. The required initial capital is
$$
\boxed{x=V(0,S_0)=(1+r)^{-N}\mathbb E_Q[g(S_N)]}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

For the <European call option>, risk-neutral valuation gives
$$
\begin{aligned}
x
&=(1+r)^{-N}\mathbb E_Q[(S_N-K)^+]\\
&=(1+r)^{-N}\mathbb E_Q[S_N\mathbf1_{\{S_N>K\}}]
-K(1+r)^{-N}Q(S_N>K).
\end{aligned}
$$
Define the <stock-numeraire measure in a binomial market> by
$$
\widehat Q(A)
=\frac{\mathbb E_Q[S_N\mathbf1_A]}{S_0(1+r)^N}.
$$
It is a probability measure because the risk-neutral martingale property gives $\mathbb E_QS_N=S_0(1+r)^N$. Consequently
$$
\boxed{
x=S_0\widehat Q(S_N>K)
-K(1+r)^{-N}Q(S_N>K)}.
$$

Tilting one up move by its stock factor changes its probability to
$$
\boxed{
\widehat q=\frac{q(1+b)}{1+r}
=\frac{(1+b)(r-a)}{(1+r)(b-a)}}.
$$
Similarly $1-\widehat q=(1-q)(1+a)/(1+r)$. Thus under $\widehat Q$ the number of up moves is binomial with parameter $\widehat q$, and
$$
\widehat Q\!\left(S_N=S_0(1+b)^i(1+a)^{N-i}\right)
=\binom Ni\widehat q^{\,i}(1-\widehat q)^{N-i}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=30j}
{scope}

= Solution
{parent=a}

The <AdaBoost> algorithm starts with weights $w_i^{(1)}=1/n$. For $m=1,\ldots,M$:

* choose $\widehat h_m\in B$ minimizing the weighted classification error
  $$
  \varepsilon_m=\sum_{i=1}^nw_i^{(m)}
  \mathbf1_{\{Y_i\ne\widehat h_m(X_i)\}};
  $$
* set
  $$
  \widehat\beta_m=\frac12\log\frac{1-\varepsilon_m}{\varepsilon_m};
  $$
* update and normalize
  $$
  w_i^{(m+1)}
  =\frac{w_i^{(m)}
  e^{-\widehat\beta_mY_i\widehat h_m(X_i)}}{
  \sum_jw_j^{(m)}
  e^{-\widehat\beta_mY_j\widehat h_m(X_j)}}.
  $$

Because $h\in B$ implies $-h\in B$, the selected classifier can always have weighted error at most $1/2$, so $\widehat\beta_m\geq0$. The output score is
$$
\boxed{\widehat f=\sum_{m=1}^M\widehat\beta_m\widehat h_m}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=30j}
{scope}

= Solution
{parent=b}

The <convex hull> of $S\subseteq\mathbb R^d$ is
$$
\operatorname{conv}S
=\left\{\sum_{j=1}^m\lambda_jv_j:
m<\infty,\ v_j\in S,\ \lambda_j\geq0,\
\sum_{j=1}^m\lambda_j=1\right\}.
$$
Thus $v$ is a convex combination of $v_1,\ldots,v_m$ when it has precisely this form. Equivalently, $\operatorname{conv}S$ is the smallest <convex set> containing $S$. The <Caratheodory theorem> sharpens the finite-combination description in $\mathbb R^d$: every point of $\operatorname{conv}S$ is a convex combination of at most $d+1$ points of $S$.

Solved by gpt-5.6-sol high.

= c
{parent=30j}
{scope}

= Solution
{parent=c}

The exponential loss is $\phi(z)=e^{-z}$. For a score $f:\mathcal X\to\mathbb R$, its <exponential classification risk> and empirical risk are
$$
R_\phi(f)=\mathbb E[e^{-Yf(X)}],
\qquad
\widehat R_\phi(f)=\frac1n\sum_{i=1}^ne^{-Y_if(X_i)}.
$$
For the set of evaluation vectors
$$
B(x_{1:n})
=\{(h(x_1),\ldots,h(x_n)):h\in B\},
$$
its empirical <Rademacher complexity> is
$$
\boxed{
\widehat R(B(x_{1:n}))
=\frac1n\mathbb E_\sigma
\sup_{h\in B}\sum_{i=1}^n\sigma_ih(x_i)},
$$
where the $\sigma_i$ are independent uniform random signs.

Solved by gpt-5.6-sol high.

= d
{parent=30j}
{scope}

= Solution
{parent=d}

The stopping rule ensures that the returned function belongs to
$$
\mathcal H
=\left\{\sum_{m=1}^M\beta_mh_m:
\beta_m\geq0,\ \sum_m\beta_m\leq1,\ h_m\in B\right\}.
$$
Indeed, when adding the next coefficient would make the sum exceed one, the algorithm returns the preceding sum, and at the first step it returns zero.

Since each $h_m$ takes values in $\{-1,1\}$, every $f\in\mathcal H$ satisfies $|f(x)|\leq1$. On $[-1,1]$, the map $z\mapsto e^{-z}$ is $\exp(1)$-Lipschitz. The expected <Rademacher complexity> generalization inequality followed by the <Rademacher contraction lemma> gives
$$
\mathbb ER_\phi(\widehat f)
\leq\mathbb E\widehat R_\phi(\widehat f)
+2e\,\mathbb E\widehat R(\mathcal H(X_{1:n})).
$$
A linear functional attains the same supremum over a set and its convex hull. Since $B=-B$, adjoining zero and allowing total coefficient at most one does not increase the supremum, so the <Rademacher complexity of a convex hull> gives
$$
\widehat R(\mathcal H(x_{1:n}))
\leq\widehat R(B(x_{1:n}))
\leq r_B.
$$
Combining the two bounds proves
$$
\boxed{
\mathbb ER_\phi(\widehat f)
\leq\mathbb E\widehat R_\phi(\widehat f)+2\exp(1)r_B}.
$$

Solved by gpt-5.6-sol high.

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

= i
{parent=31e}
{scope}

= Solution
{parent=i}

After division by $x^2$, the equation becomes
$$
y''+\frac1xy'-\frac1{x^4}y=0.
$$
The <regular singular point criterion for a second-order equation> would require $x(1/x)$ and $x^2(-1/x^4)$ to be analytic at zero. The second expression is $-x^{-2}$, so the criterion fails:
$$
\boxed{x=0\text{ is an irregular singular point}.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=31e}
{scope}

= Solution
{parent=ii}

The <removal of the first derivative from a second-order differential equation> uses
$$
y(x)=\exp\left(-\frac12\int\frac{dx}{x}\right)u(x)
=x^{-1/2}u(x).
$$
It gives
$$
u''+\left(-\frac1{x^4}
-\frac12\left(-\frac1{x^2}\right)
-\frac14\frac1{x^2}\right)u=0,
$$
so
$$
\boxed{u''+q(x)u=0,\qquad
q(x)=\frac1{4x^2}-\frac1{x^4}}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=31e}
{scope}

= Solution
{parent=iii}

Write the normal-form equation as
$$
u''=Q(x)u,
\qquad
Q(x)=\frac1{x^4}-\frac1{4x^2}.
$$
For sufficiently small positive $x$, $Q(x)>0$. Moreover,
$$
\frac{|Q'(x)|}{Q(x)^{3/2}}=O(x)\longrightarrow0,
$$
and the corresponding higher derivative condition also tends to zero, so the amplitude varies slowly relative to the exponential phase and the <WKB approximation>[Liouville-Green approximation] applies.

The phase and amplitude have expansions
$$
\begin{aligned}
\sqrt{Q(x)}
&=x^{-2}\left(1-\frac{x^2}{8}
-\frac{x^4}{128}+O(x^6)\right),\\
Q(x)^{-1/4}
&=x\left(1+\frac{x^2}{16}+O(x^4)\right),\\
\int^x\sqrt{Q(s)}\,ds
&=-\frac1x-\frac x8-\frac{x^3}{384}+O(x^5).
\end{aligned}
$$
Hence two independent Liouville-Green solutions satisfy
$$
\boxed{
u_\pm(x)\sim Q(x)^{-1/4}
\exp\left(\pm\int^x\sqrt{Q(s)}\,ds\right)}.
$$
In particular, after choosing the signs according to growth and decay,
$$
u_{\rm grow}(x)\sim xe^{1/x}\left(1+\frac x8+\frac{9x^2}{128}+\cdots\right),
\qquad
u_{\rm decay}(x)\sim xe^{-1/x}\left(1-\frac x8+\frac{9x^2}{128}+\cdots\right).
$$

Solved by gpt-5.6-sol high.

= iv
{parent=31e}
{scope}

= Solution
{parent=iv}

Since $y=x^{-1/2}u$, the two linearly independent solutions of the original equation have the <asymptotic expansion>[asymptotic expansions]
$$
\boxed{
\begin{aligned}
y_{\rm grow}(x)
&\sim x^{1/2}e^{1/x}
\left(1+\frac x8+\frac{9x^2}{128}+\cdots\right),\\
y_{\rm decay}(x)
&\sim x^{1/2}e^{-1/x}
\left(1-\frac x8+\frac{9x^2}{128}+\cdots\right).
\end{aligned}}
$$
As a check, the substitution $z=1/x$ converts the equation into the modified Bessel equation of order zero, whose solutions $I_0(z)$ and $K_0(z)$ have exactly these large-$z$ expansions.

Solved by gpt-5.6-sol high.

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

= a
{parent=32b}
{scope}

= Solution
{parent=a}

At an <equilibrium point of a dynamical system>[equilibrium], each product in
$$
\dot x=x(y-k-3x+x^2),
\qquad
\dot y=y(y-1-x)
$$
must vanish. This gives the two parameter-independent equilibria
$$
(0,0),\qquad(0,1),
$$
the equilibria on $y=0$
$$
\left(\frac{3\pm\sqrt{9+4k}}2,0\right)
\quad(k\geq-9/4),
$$
and the equilibria with both coordinates nonzero
$$
(1\pm\sqrt k,\,2\pm\sqrt k)
\quad(k\geq0),
$$
where matching signs are used.

The $y=0$ pair is born at $k=-9/4$, suggesting a <saddle-node bifurcation>. At $k=0$, one of these equilibria meets $(0,0)$, suggesting a <transcritical bifurcation>, while the two equilibria near $(1,2)$ are simultaneously born in a saddle-node. At $k=1$, the minus branch $(1-\sqrt k,2-\sqrt k)$ meets $(0,1)$, suggesting another transcritical bifurcation. At $k=4$, that branch meets the $y=0$ equilibrium $(-1,0)$, again suggesting a transcritical bifurcation. Thus the bifurcation parameter values are
$$
\boxed{k=-\frac94,\ 0,\ 1,\ 4}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=32b}
{scope}

= Solution
{parent=b}

The <Jacobian matrix> is
$$
J(x,y)=
\begin{pmatrix}
y-k-6x+3x^2 & x\\
-y & 2y-1-x
\end{pmatrix}.
$$
At $(0,0)$ its <eigenvalues> are $-k$ and $-1$. It is therefore a stable node for $k>0$, a saddle for $k<0$, and nonhyperbolic at $k=0$, consistent with the proposed transcritical bifurcation.

At $(0,1)$ the matrix is
$$
\begin{pmatrix}1-k&0\\-1&1\end{pmatrix},
$$
with eigenvalues $1-k$ and $1$. It is an unstable node for $k<1$, a saddle for $k>1$, and nonhyperbolic at $k=1$, again consistent with a transcritical bifurcation.

Solved by gpt-5.6-sol high.

= c
{parent=32b}
{scope}

= Solution
{parent=c}

There are two bifurcations in $0\leq k\leq1/2$, both at $k=0$.

Near $(0,0,0)$, append $\dot k=0$. The plane $y=0$ is invariant and tangent to the extended center subspace, so it is an <extended centre manifold for a parameter>[extended center manifold]. The reduced equation is exactly
$$
\dot x=x(-k-3x+x^2),
\qquad
\dot k=0.
$$
Its leading terms $-kx-3x^2$ have two branches, $x=0$ and $x=-k/3+O(k^2)$, which cross and exchange their center-direction stability. Hence this is a <transcritical bifurcation>.

For the bifurcation at $(1,2,0)$, set
$$
X=x-1,\qquad v=y-2-X.
$$
The extended system becomes
$$
\dot X=(1+X)(v-k+X^2),
\qquad
\dot v=v+v^2+(1+X)(k-X^2),
\qquad
\dot k=0.
$$
Solving the <centre-manifold invariance equation> for $v=h(X,k)$ gives
$$
h(X,k)=-k+X^2-5Xk+9k^2+O\bigl((|X|+|k|)^3\bigr).
$$
Substitution into the $X$ equation yields
$$
\dot X=2(X^2-k)+O\bigl((|X|+|k|)^3\bigr).
$$
For $k>0$ this has two nearby equilibria $X=\pm\sqrt k+O(k)$, while for $k<0$ it has none. It is therefore a <saddle-node bifurcation>. These reductions are collected in <bifurcations of the 2022 Cambridge quadratic-cubic system>.

Solved by gpt-5.6-sol high.

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

= a
{parent=33a}
{scope}

= Solution
{parent=a}

Insert the stated expansion into the <time-dependent Schrodinger equation>
$$
i\hbar\frac d{dt}|\psi(t)\rangle
=(H_0+v(t))|\psi(t)\rangle.
$$
When the derivative acts on each free phase, its energy factor is cancelled by the action of $H_0$ on the corresponding <eigenstate>. The remaining terms obey
$$
i\hbar\left[
\dot a(t)e^{-iE_0t/\hbar}|0\rangle
+\int_0^\infty e^{-iE_kt/\hbar}
\bigl(\dot b_k(t)|k,+\rangle+\dot c_k(t)|k,-\rangle\bigr)\,dk
\right]
=-F e^{-i\omega t}x|\psi(t)\rangle.
$$
Division by $i\hbar$ gives
$$
\boxed{
\dot a(t)e^{-iE_0t/\hbar}|0\rangle
+\int_0^\infty e^{-iE_kt/\hbar}
\bigl(\dot b_k|k,+\rangle+\dot c_k|k,-\rangle\bigr)\,dk
=\frac{iF}{\hbar}e^{-i\omega t}x|\psi(t)\rangle}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= Solution
{parent=b}

To first order in $F$, <time-dependent perturbation theory> replaces the state on the right-hand side by its unperturbed value $e^{-iE_0t/\hbar}|0\rangle$. The bound state is even, while the <position operator> $x$ is odd. The <parity> selection rule therefore gives
$$
\langle k,+|x|0\rangle=0,
$$
so $\dot b_k=0$ and, with no initial continuum component,
$$
\boxed{b_k(t)=0}.
$$

Projection onto the odd state gives
$$
\dot c_k(t)
=\frac{iF}{\hbar}\langle k,-|x|0\rangle
e^{i(E_k-E_0-\hbar\omega)t/\hbar}.
$$
Writing $\Omega_k=(E_k-E_0-\hbar\omega)/\hbar$ and integrating from zero to $t$ yields
$$
\boxed{
c_k(t)=\frac{iF}{\hbar}\langle k,-|x|0\rangle
e^{i\Omega_kt/2}
\frac{\sin(\Omega_kt/2)}{\Omega_k/2}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=33a}
{scope}

= Solution
{parent=c}

The required dipole matrix element is
$$
\begin{aligned}
\langle k,-|x|0\rangle
&=2\sqrt{\frac K\pi}\int_0^\infty
x e^{-Kx}\sin(kx)\,dx\\
&=\frac{4Kk\sqrt{K/\pi}}{(K^2+k^2)^2},
\end{aligned}
$$
where differentiating the elementary Laplace integral for $\sin(kx)$ evaluates the last integral. Hence
$$
|\langle k,-|x|0\rangle|^2
=\frac{16K^3k^2}{\pi(K^2+k^2)^4}.
$$

Put $E_f=E_0+\hbar\omega>0$ and $k_f=\sqrt{2mE_f}/\hbar$. The <continuum transition probability at long times> uses
$$
\frac{\sin^2(\Omega_kt/2)}{(\Omega_k/2)^2}
\longrightarrow2\pi t\,\delta(\Omega_k).
$$
Since
$$
\delta(\Omega_k)
=\frac{m}{\hbar k_f}\delta(k-k_f),
$$
the total escape probability becomes
$$
\begin{aligned}
P_{\rm free}(t)
&=\int_0^\infty|c_k(t)|^2\,dk\\
&\sim\frac{2\pi F^2t}{\hbar^2}
\frac{m}{\hbar k_f}
\frac{16K^3k_f^2}{\pi(K^2+k_f^2)^4}\\
&=\frac{32F^2tmK^3k_f}
{\hbar^3(K^2+k_f^2)^4}.
\end{aligned}
$$
For the attractive <delta potential>,
$$
|E_0|=\frac{\hbar^2K^2}{2m},
\qquad
\frac{k_f^2}{K^2}=\frac{E_f}{|E_0|}.
$$
Substitution gives the requested leading-order result:
$$
\boxed{
P_{\rm free}(t)
=\frac{8\hbar F^2t}{mE_0^2}
\frac{\sqrt{E_f/|E_0|}}
{(1+E_f/|E_0|)^4}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=34d}
{scope}

= Solution
{parent=a}

With the <magnetic minimal coupling> convention for charge $e$,
$$
\widehat H
=\frac1{2m}(-i\hbar\nabla-e\mathbf A)^2.
$$
Here $\nabla\cdot\mathbf A=0$, $\mathbf A^2=B^2(x^2+y^2)/4$, and
$$
\mathbf A\cdot\nabla
=\frac B2\left(-y\frac\partial{\partial x}
+x\frac\partial{\partial y}\right).
$$
Therefore in Cartesian coordinates
$$
\boxed{
\widehat H
=-\frac{\hbar^2}{2m}\nabla^2
+\frac{ie\hbar B}{2m}
\left(-y\partial_x+x\partial_y\right)
+\frac{e^2B^2}{8m}(x^2+y^2)}.
$$
In cylindrical coordinates this is
$$
\boxed{
\widehat H
=-\frac{\hbar^2}{2m}
\left[
\frac1r\partial_r(r\partial_r)
+\frac1{r^2}\partial_\theta^2
+\partial_z^2
\right]
+\frac{ie\hbar B}{2m}\partial_\theta
+\frac{e^2B^2r^2}{8m}}.
$$
This choice of $\mathbf A$ is the <symmetric gauge>.

Solved by gpt-5.6-sol high.

= b
{parent=34d}
{scope}

= Solution
{parent=b}

The factor $e^{ik_zz}$ is an eigenstate of longitudinal momentum with $p_z=\hbar k_z$. The integer $n$ is the azimuthal <angular momentum> quantum number, since $L_z=-i\hbar\partial_\theta$ has eigenvalue $n\hbar$.

For
$$
\phi_n(r)=r^\alpha e^{-\beta r^2/2},
$$
direct differentiation gives
$$
\frac1r(r\phi_n')'
=\left[
\frac{\alpha^2}{r^2}
-2\beta(\alpha+1)+\beta^2r^2
\right]\phi_n.
$$
Regularity and cancellation of the $r^{-2}$ term require $\alpha=n$ for $n\geq0$. Cancellation of the $r^2$ terms requires
$$
\boxed{\beta=\frac{eB}{2\hbar}},
$$
assuming $eB>0$. The remaining radial energy is independent of $n$:
$$
\boxed{
\alpha=n,\qquad
E=\frac{\hbar^2k_z^2}{2m}+\frac{\hbar eB}{2m}}.
$$
This is the lowest <Landau level>. If $eB<0$, the same family uses nonpositive angular-momentum quantum numbers and $|eB|$ replaces $eB$.

Solved by gpt-5.6-sol high.

= c
{parent=34d}
{scope}

= Solution
{parent=c}

The radial factor
$$
r^n\exp\left(-\frac{eBr^2}{4\hbar}\right)
$$
is sharply peaked, for large $n$, near
$$
r_n^2=\frac{2\hbar n}{eB}.
$$
Requiring the orbit to lie inside a disk of radius $R$ permits
$$
0\leq n\lesssim\frac{eBR^2}{2\hbar}.
$$
Thus the total degeneracy at fixed $k_z$ is, up to the negligible boundary correction,
$$
\boxed{D=\frac{eBR^2}{2\hbar}}.
$$
Since the disk has area $\pi R^2$, the <degeneracy of a Landau level> per unit area is
$$
\boxed{\frac D{\pi R^2}=\frac{eB}{2\pi\hbar}}.
$$
For either sign of the charge-field product, replace $eB$ by $|eB|$.

Solved by gpt-5.6-sol high.

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

= a
{parent=35a}
{scope}

= Solution
{parent=a}

The <Carnot theorem> states that no engine between two fixed heat reservoirs can be more efficient than a reversible engine, and that all reversible engines between those reservoirs have the same efficiency. Thus the heat ratio of a reversible engine depends only on the two reservoirs. Its multiplicative consistency for three reservoirs permits a state variable $T$, unique up to an overall scale, such that
$$
\frac{Q_c}{Q_h}=\frac{T_c}{T_h}.
$$
Fixing one reference value defines the <thermodynamic temperature> scale, and the reversible efficiency is
$$
\boxed{\eta_{\rm rev}=1-\frac{T_c}{T_h}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

At fixed volume, a reversible infinitesimal heating supplies $\delta Q_{\rm rev}=C_V\,dT$. Since <entropy> satisfies $dS=\delta Q_{\rm rev}/T$,
$$
\Delta S=C_V\int_{T_i}^{T_f}\frac{dT}{T}
=\boxed{C_V\log\frac{T_f}{T_i}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let the common final temperature be $T_f$. A reversible isolated process has zero total entropy production, so part (i) gives
$$
C_V\log\frac{T_f}{T_1}
+C_V\log\frac{T_f}{T_2}=0.
$$
Hence
$$
\boxed{T_f=\sqrt{T_1T_2}}
$$
for both bodies. The decrease in their total internal energy is extracted as work by the reversible engine mediating the transfer.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Direct contact permits no work extraction, so conservation of internal energy gives
$$
C_V(T_f-T_1)+C_V(T_f-T_2)=0,
\qquad
\boxed{T_f=\frac{T_1+T_2}{2}}.
$$
The total entropy change is
$$
\begin{aligned}
\Delta S
&=C_V\log\frac{T_f}{T_1}
+C_V\log\frac{T_f}{T_2}\\
&=C_V\log\frac{(T_1+T_2)^2}{4T_1T_2}\geq0.
\end{aligned}
$$
The <arithmetic-geometric mean inequality> makes the last expression positive when $T_1\ne T_2$, with equality only when the bodies were already at the same temperature. This is the entropy production of irreversible heat flow.

Solved by gpt-5.6-sol high.

= c
{parent=35a}
{scope}

= Solution
{parent=c}

The fundamental thermodynamic relation is
$$
dE=T\,dS-p\,dV+\mu\,dN.
$$
Because $E(S,V,N)$ is extensive, the <Euler theorem for homogeneous functions> gives
$$
E=TS-pV+\mu N.
$$
Substitution into the <Gibbs free energy> $G=E+pV-TS$ therefore yields
$$
\boxed{G=\mu(T,p)N}.
$$
For a one-component equilibrium phase, the intensive <chemical potential> depends on the intensive state variables $T$ and $p$.

Solved by gpt-5.6-sol high.

= d
{parent=35a}
{scope}

= Solution
{parent=d}

A <first-order phase transition> occurs where two phases have equal Gibbs free energy but a first derivative of that free energy is discontinuous. Typical jumps occur in
$$
S=-\left(\frac{\partial G}{\partial T}\right)_{p,N}
\quad\text{or}\quad
V=\left(\frac{\partial G}{\partial p}\right)_{T,N}.
$$
A nonzero entropy jump gives a nonzero <latent heat> $T\Delta S$.

Solved by gpt-5.6-sol high.

= e
{parent=35a}
{scope}

= Solution
{parent=e}

At coexistence, $G_{\rm I}(T_0,p,N)=G_{\rm II}(T_0,p,N)$. Stability says
$$
G_{\rm I}<G_{\rm II}\quad(T>T_0),
\qquad
G_{\rm II}<G_{\rm I}\quad(T<T_0).
$$
Thus $G_{\rm I}-G_{\rm II}$ crosses zero with negative slope as $T$ increases through $T_0$. On the two smooth phase branches,
$$
\left(\frac{\partial}{\partial T}
(G_{\rm I}-G_{\rm II})\right)_{p,N}
=-S_{\rm I}+S_{\rm II}<0.
$$
Therefore
$$
\boxed{S_{\rm I}-S_{\rm II}>0}.
$$
The transition from phase II to phase I consequently absorbs positive latent heat $T_0(S_{\rm I}-S_{\rm II})$.

Solved by gpt-5.6-sol high.

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

= a
{parent=36b}
{scope}

= Solution
{parent=a}

A dielectric is an electrically insulating material whose bound positive and negative charges can be displaced slightly by an <electric field>, producing <electric polarization> without sustained conduction current. Its macroscopic electric response is described by its permittivity.

Solved by gpt-5.6-sol high.

= b
{parent=36b}
{scope}

= Solution
{parent=b}

The <electric polarization> $\mathbf P$ is electric dipole moment per unit volume. It produces
$$
\rho_{\rm b}=-\nabla\cdot\mathbf P,
\qquad
\sigma_{\rm b}=\mathbf P\cdot\mathbf n.
$$
The <electric displacement field> $\mathbf D=\epsilon_0\mathbf E+\mathbf P$ is useful because
$$
\nabla\cdot\mathbf D=\rho_{\rm free},
$$
so it separates free charge from the bound charge represented by $\mathbf P$. In a linear isotropic dielectric, $\mathbf D=\epsilon\mathbf E$.

Spherical symmetry and Gauss's law for $\mathbf D$ give, on both sides of the interface,
$$
\mathbf D=\frac{q}{4\pi r^2}\mathbf e_r.
$$
Thus
$$
\mathbf P_i
=\left(1-\frac{\epsilon_0}{\epsilon_i}\right)
\frac{q}{4\pi r^2}\mathbf e_r.
$$
With $\mathbf n=\mathbf e_r$ pointing from medium one to medium two, the net bound sheet charge at $r=R$ is the polarization jump
$$
\boxed{
\sigma_{\rm b}
=(\mathbf P_1-\mathbf P_2)\cdot\mathbf n
=\frac{\epsilon_0q}{4\pi R^2}
\left(\frac1{\epsilon_2}-\frac1{\epsilon_1}\right)}.
$$
If one assigns surface charge separately to the inner body and outer medium, their contributions are $\mathbf P_1\cdot\mathbf n$ and $-\mathbf P_2\cdot\mathbf n$; their sum is the displayed physical interface charge.

Solved by gpt-5.6-sol high.

= c
{parent=36b}
{scope}

= Solution
{parent=c}

The <magnetization> $\mathbf M$ is magnetic dipole moment per unit volume. In magnetostatics it produces
$$
\mathbf J_{\rm b}=\nabla\times\mathbf M,
\qquad
\mathbf K_{\rm b}=\mathbf M\times\mathbf n.
$$
The <magnetic field intensity> $\mathbf H=\mathbf B/\mu_0-\mathbf M$ obeys
$$
\nabla\times\mathbf H=\mathbf J_{\rm free},
$$
so $\mathbf H$ isolates free currents while $\mathbf M$ accounts for bound currents. For a linear isotropic material, $\mathbf B=\mu\mathbf H$.

Ampere's law for a circular loop around the free line current gives
$$
\mathbf H=\frac{I}{2\pi r}\mathbf e_\phi
$$
in both media. Therefore
$$
\mathbf M_i
=\left(\frac{\mu_i}{\mu_0}-1\right)
\frac{I}{2\pi r}\mathbf e_\phi.
$$
At $r=R$, the net bound surface-current density is
$$
\begin{aligned}
\mathbf K_{\rm b}
&=(\mathbf M_1-\mathbf M_2)\times\mathbf e_r\\
&=-\frac{I(\mu_1-\mu_2)}
{2\pi R\mu_0}\mathbf e_z.
\end{aligned}
$$
Hence its magnitude is
$$
\boxed{|\mathbf K_{\rm b}|=
\frac{I|\mu_1-\mu_2|}{2\pi R\mu_0}},
$$
and for a positive current along $+\mathbf e_z$ it points along $-\mathbf e_z$ when $\mu_1>\mu_2$, reversing when $\mu_1<\mu_2$.

Solved by gpt-5.6-sol high.

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

= a
{parent=37d}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The antipodal action fixes the origin. Away from the origin it acts freely, so the quotient is locally Euclidean there. A small punctured neighborhood of the image of the origin, however, has link
$$
S^2/(x\sim-x)=\mathbb{RP}^2,
$$
the <real projective plane>, whereas the link of a point in a three-dimensional <smooth manifold> is $S^2$. Equivalently, the neighborhood is a cone on $\mathbb{RP}^2$ and cannot be a three-ball. Thus
$$
\boxed{\mathbb R^3/(x\sim-x)\text{ is not a manifold}.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Removing the closed unit ball leaves
$$
\{(x,y,z):x^2+y^2+z^2>1\},
$$
which is an open subset of $\mathbb R^3$. Every point therefore has an ordinary Euclidean coordinate neighborhood, so
$$
\boxed{\mathbb R^3\setminus\overline B_1(0)\text{ is a smooth three-manifold}.}
$$
It has no boundary; the unit sphere was removed together with the ball.

Solved by gpt-5.6-sol high.

= b
{parent=37d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The map takes one covector argument and returns two vector slots, so $S$ has two contravariant and one covariant index. Its <tensor type> is
$$
\boxed{\binom21},
$$
or type $(2,1)$, with components $S^{ab}{}_c$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The <covariant derivative> adds one covariant index, so $\nabla S$ has type $(2,2)$. The outer product of two <tensor>[tensors] adds their index counts. Hence
$$
\boxed{S\otimes\nabla S\text{ has type }(4,3)
\text{, written }\binom43}.
$$

Solved by gpt-5.6-sol high.

= Solution
{parent=b}

Let
$$
L(x,\dot x)=\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}
=\frac{ds}{d\lambda}.
$$
The <Euler-Lagrange equation> gives
$$
\frac d{d\lambda}\left(\frac{g_{\beta\nu}\dot x^\nu}{L}\right)
-\frac1{2L}\partial_\beta g_{\mu\nu}
\dot x^\mu\dot x^\nu=0.
$$
Since $\dot x^\mu=L\,dx^\mu/ds$ and $d/d\lambda=L\,d/ds$, this becomes
$$
\frac d{ds}\left(g_{\beta\nu}\frac{dx^\nu}{ds}\right)
-\frac12\partial_\beta g_{\mu\nu}
\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}=0.
$$
Expanding the derivative, multiplying by the inverse metric, and using the <Christoffel symbol>
$$
\Gamma^\alpha_{\mu\nu}
=\frac12g^{\alpha\beta}
(\partial_\mu g_{\beta\nu}
+\partial_\nu g_{\beta\mu}
-\partial_\beta g_{\mu\nu})
$$
gives the <geodesic equation>
$$
\boxed{
\frac{d^2x^\alpha}{ds^2}
+\Gamma^\alpha_{\mu\nu}
\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}=0}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=38c}
{scope}

= Solution
{parent=a}

Balancing the downslope gravitational force density $\rho g\sin\alpha$ against the viscous term $\mu u_{yy}$ gives the velocity scale
$$
\boxed{U\sim\frac{\rho g h^2\sin\alpha}{\mu}}.
$$
If the streamwise variation scale is $L\sim h/|h_x|$, the <lubrication theory> requirement is first
$$
\boxed{|h_x|\ll1}.
$$
The ratio of streamwise inertia $\rho U^2/L$ to transverse viscous stress $\mu U/h^2$ is
$$
\frac{\rho Uh^2}{\mu L}
\sim
\frac{\rho^2g h^3\sin\alpha}{\mu^2}|h_x|.
$$
Thus self-consistency also requires
$$
\boxed{
\frac{\rho^2g h^3\sin\alpha}{\mu^2}|h_x|\ll1}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

The leading lubrication momentum equations and incompressibility condition are
$$
0=-p_x+\rho g\sin\alpha+\mu u_{yy},
\qquad
0=-p_y-\rho g\cos\alpha,
\qquad
u_x+v_y=0.
$$
At the solid plane $y=0$, no slip and no penetration give $u=v=0$. At the free surface $y=h(x,t)$, neglecting surface tension and ambient viscous stress gives
$$
p=p_{\rm atm},
\qquad
u_y=0,
$$
to leading order. The <kinematic boundary condition for a free-surface graph> is
$$
v=h_t+u h_x
\quad\text{at }y=h.
$$

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= Solution
{parent=c}

Integrating the normal momentum equation from the free surface gives the hydrostatic pressure
$$
\boxed{
p(x,y,t)=p_{\rm atm}
+\rho g\cos\alpha\,[h(x,t)-y]}.
$$
Thus $p_x=\rho g\cos\alpha\,h_x$. Twice integrating the downslope equation and applying $u(0)=0$ and $u_y(h)=0$ gives
$$
\boxed{
u(x,y,t)
=\frac{\rho g}{\mu}
\left(\sin\alpha-\cos\alpha\,h_x\right)
\left(hy-\frac{y^2}{2}\right)}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=38c}
{scope}

= Solution
{parent=d}

The volume flux per unit transverse width is
$$
\begin{aligned}
q(x,t)
&=\int_0^h u\,dy\\
&=\frac{\rho gh^3}{3\mu}
\left(\sin\alpha-\cos\alpha\,h_x\right).
\end{aligned}
$$
Integrating incompressibility across the layer and using the wall and free-surface kinematic conditions gives local mass conservation $h_t+q_x=0$. Hence the <gravity-driven thin film on an incline> obeys
$$
\boxed{
h_t+\frac{\rho g}{3\mu}
\partial_x\left[
h^3(\sin\alpha-\cos\alpha\,h_x)
\right]=0}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=38c}
{scope}

= Solution
{parent=e}

Put $z=x-ct$ and $h(x,t)=f(z)$. The conservation law becomes
$$
-cf'
+\frac d{dz}\left[
\frac{\rho gf^3}{3\mu}
(\sin\alpha-\cos\alpha\,f')
\right]=0.
$$
One integration gives the required first-order autonomous equation
$$
\boxed{
\frac{\rho gf^3}{3\mu}
(\sin\alpha-\cos\alpha\,f')
-cf=C},
$$
where $C$ is a constant fixed by the far-field film heights or other boundary data.

Solved by gpt-5.6-sol high.

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

= a
{parent=39c}
{scope}

= Solution
{parent=a}

For one-dimensional homentropic flow,
$$
\rho_t+u\rho_x+\rho u_x=0,
\qquad
u_t+uu_x+\frac1\rho p_x=0,
$$
and $dp=c^2d\rho$. For a perfect gas, $dc/c=(\gamma-1)d\rho/(2\rho)$. Combining the two equations therefore shows that the <Riemann invariant>[Riemann invariants]
$$
\boxed{R_\pm=u\pm\frac{2c}{\gamma-1}}
$$
satisfy
$$
\boxed{
\left(\partial_t+(u\pm c)\partial_x\right)R_\pm=0}.
$$
Their characteristic curves are parametrized by
$$
\boxed{
\frac{dx_\pm}{dt}=u(x_\pm(t),t)\pm c(x_\pm(t),t)},
$$
and $R_\pm$ is constant along the corresponding curve until characteristic intersection creates a shock.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Before shock formation, the piston launches a right-moving <pressure in a perfect-gas simple wave>[simple wave] into the undisturbed gas. The opposite invariant is fixed at its ambient value:
$$
R_-=u-\frac{2c}{\gamma-1}
=-\frac{2c_0}{\gamma-1}.
$$
Therefore the <pressure in a perfect-gas simple wave>[simple-wave relation] is
$$
\boxed{c=c_0+\frac{\gamma-1}{2}u}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The motion is isentropic before a shock forms, and the perfect-gas relations give
$$
\frac p{p_0}
=\left(\frac c{c_0}\right)^{2\gamma/(\gamma-1)}.
$$
Writing $\delta=\delta c/c_0$ and applying the <binomial theorem>[generalized binomial expansion],
$$
\boxed{
\frac{\delta p}{p_0}
=\frac{2\gamma}{\gamma-1}\delta
+\frac{\gamma(\gamma+1)}{(\gamma-1)^2}\delta^2
+O(\delta^3)}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The fluid velocity at the piston equals its velocity,
$$
u_p(t)=\frac d{dt}\bigl[-\epsilon t(1-t)\bigr]
=-\epsilon(1-2t).
$$
Using $c-c_0=(\gamma-1)u/2$ in part (ii) gives
$$
\frac{\delta p}{p_0}
=\gamma\frac u{c_0}
+\frac{\gamma(\gamma+1)}4\frac{u^2}{c_0^2}
+O(\epsilon^3).
$$
Over $0\leq t\leq1$,
$$
\langle u_p\rangle=0,
\qquad
\langle u_p^2\rangle
=\epsilon^2\int_0^1(1-2t)^2\,dt
=\frac{\epsilon^2}{3}.
$$
Thus
$$
\boxed{
\left\langle\frac{\delta p}{p_0}\right\rangle
=\frac{\gamma(\gamma+1)}{12}
\frac{\epsilon^2}{c_0^2}
+O(\epsilon^3)}.
$$
The linear pressure oscillation has zero mean, while the quadratic compressibility produces a positive mean pressure, the finite-amplitude acoustic analogue of radiation pressure.

Solved by gpt-5.6-sol high.

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

= a
{parent=40c}
{scope}

= Solution
{parent=a}

The <Gershgorin circle theorem> states that every <eigenvalue> $\lambda$ of a complex matrix $A=(a_{ij})$ lies in at least one disk
$$
\boxed{
|\lambda-a_{ii}|
\leq\sum_{j\ne i}|a_{ij}|}.
$$
To prove it, choose a nonzero <eigenvector> $x$ for $\lambda$ and an index $i$ such that $|x_i|=\max_j|x_j|>0$. The $i$th component of $Ax=\lambda x$ gives
$$
(\lambda-a_{ii})x_i
=\sum_{j\ne i}a_{ij}x_j.
$$
Taking absolute values and using $|x_j|\leq|x_i|$ yields
$$
|\lambda-a_{ii}|\,|x_i|
\leq\sum_{j\ne i}|a_{ij}|\,|x_j|
\leq |x_i|\sum_{j\ne i}|a_{ij}|.
$$
Division by $|x_i|$ proves the theorem.

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Sample the diffusion coefficient at edge midpoints and write
$$
\begin{aligned}
\alpha&=a((i-\tfrac12)h,jh),&
\beta&=a((i+\tfrac12)h,jh),\\
\gamma&=a(ih,(j-\tfrac12)h),&
\delta&=a(ih,(j+\tfrac12)h).
\end{aligned}
$$
Centered differences of the two fluxes give the conservative <finite difference method>[finite-difference] stencil
$$
\boxed{
\frac1{h^2}\left[
\alpha u_{i-1,j}+\beta u_{i+1,j}
+\gamma u_{i,j-1}+\delta u_{i,j+1}
-(\alpha+\beta+\gamma+\delta)u_{i,j}
\right]}.
$$
For example, the $x$ part is
$$
\frac1h\left[
a_{i+1/2,j}\frac{u_{i+1,j}-u_{i,j}}h
-a_{i-1/2,j}\frac{u_{i,j}-u_{i-1,j}}h
\right].
$$
Taylor expansion about $(ih,jh)$ shows that the odd powers cancel between the two face fluxes and that the result equals $\partial_x(a\,u_x)+O(h^2)$. The same calculation in $y$ gives total truncation error
$$
\boxed{O(h^2)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $L_h$ be the spatial matrix, so the <forward Euler method> is
$$
U^{n+1}=(I+kL_h)U^n.
$$
The conservative use of shared edge coefficients makes $L_h$ a real symmetric matrix. In an interior row, put $s=\alpha+\beta+\gamma+\delta<4a_{\max}$. Its diagonal entry is $-s/h^2$, and the sum of the absolute off-diagonal entries is $s/h^2$. Boundary rows have no larger disk because homogeneous <Dirichlet boundary condition>[Dirichlet data] remove some unknown neighbors.

The <Gershgorin circle theorem> and symmetry therefore place every eigenvalue of $L_h$ in
$$
\left[-\frac{2s}{h^2},0\right]
\subseteq
\left[-\frac{8a_{\max}}{h^2},0\right].
$$
If $\mu=k/h^2$, each amplification eigenvalue is $1+k\lambda$. It lies in $[-1,1]$ whenever
$$
1-8\mu a_{\max}\geq-1,
$$
or
$$
\boxed{0<\mu\leq\frac1{4a_{\max}}}.
$$
Since the amplification matrix is symmetric, bounding all its eigenvalues in modulus by one bounds its discrete Euclidean operator norm by one. This proves the stated stability condition and is the <Gershgorin stability bound for a variable-coefficient diffusion stencil>.

Solved by gpt-5.6-sol high.