past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperii_3_2023.pdf

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

= a
{parent=1g}
{scope}

= Solution
{parent=a}

The <binomial theorem> gives
$$
4^n=(1+1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}.
$$
The <central binomial coefficient> is the largest of these $2n+1$ nonnegative <binomial coefficients>, so
$$
\binom{2n}{n}\geq\frac{4^n}{2n+1}=
\frac{2^{2n}}{2n+1}.
$$

For the upper bound, fix a <prime number> $p\leq2n$. By <Legendre formula>,
$$
v_p\!\left(\binom{2n}{n}\right)
=\sum_{j\geq1}
\left(
\left\lfloor\frac{2n}{p^j}\right\rfloor
-2\left\lfloor\frac{n}{p^j}\right\rfloor
\right).
$$
Each summand is either zero or one. Hence the <P-adic valuation> satisfies
$$
v_p\!\left(\binom{2n}{n}\right)
\leq\left\lfloor\frac{\log(2n)}{\log p}\right\rfloor.
$$
If $p\leq\sqrt{2n}$, this implies
$$
p^{v_p(\binom{2n}{n})}\leq2n.
$$
There are at most $\sqrt{2n}$ such <prime numbers>. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the <P-adic valuation> is at most one. Multiplying the prime-power contributions therefore yields
$$
\binom{2n}{n}
\leq(2n)^{\sqrt{2n}}
\prod_{\substack{p\leq2n\\p\text{ prime}}}p.
$$

Solved by gpt-5.6-sol high.

= b
{parent=1g}
{scope}

= Solution
{parent=b}

Write the <Chebyshev theta function> as
$$
\vartheta(x)=\sum_{\substack{p\leq x\\p\text{ prime}}}\log p.
$$
Taking <natural logarithms> in part a gives
$$
2n\log2-\log(2n+1)
\leq\sqrt{2n}\log(2n)+\vartheta(2n),
$$
and hence
$$
\vartheta(2n)\geq
2n\log2-\sqrt{2n}\log(2n)-\log(2n+1).
$$
Because $\sqrt n\log n/n\to0$, the right-hand side is at least $n/2$ for every sufficiently large <integer> $n$.

Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the <floor function>. Then $2n\leq x$ and $n\geq x/3$, so the monotonicity of $\vartheta$ gives
$$
\boxed{
\sum_{\substack{p\leq x\\p\text{ prime}}}\log p
=\vartheta(x)\geq\vartheta(2n)
\geq\frac n2\geq\frac{x}{6}\geq\frac{x}{12}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=1g}
{scope}

= Solution
{parent=c}

Suppose that the positive <integer> $n$ is a <decisive number>. For every <prime number> $p<\sqrt n$, the <composite number> $p^2$ satisfies $1<p^2<n$. It therefore cannot be <coprime integers>[coprime] to $n$, so $p\mid n$.

The product of all <prime numbers> below $\sqrt n$ consequently divides $n$. Taking <natural logarithms> gives
$$
\log n\geq\sum_{p<\sqrt n}\log p.
$$
Changing the strict endpoint can remove at most one prime term. By part b, for every sufficiently large $n$ the right-hand side is at least, for example, $\sqrt n/13$. This is impossible for sufficiently large $n$, since
$$
\frac{\sqrt n}{\log n}\longrightarrow\infty.
$$
Thus every <decisive number> lies below one fixed bound, and only finitely many <integers> can do so.

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 3}
{scope}
{title2=Topics In Analysis}

= i
{parent=2f}
{scope}

= Solution
{parent=i}

The polynomial form of <Runge theorem> says that if $K\subset\mathbb C$ is <compact set>[compact], $\mathbb C\setminus K$ is connected, and $f$ is <holomorphic function>[holomorphic] on a neighbourhood of $K$, then for every $\varepsilon>0$ there is a <polynomial> $P$ such that
$$
\sup_{z\in K}|f(z)-P(z)|<\varepsilon.
$$

Statement (i) is false. Take the <entire function> $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged <uniform convergence>[uniformly] to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty)$, since its values at zero and near infinity differ by one.

Solved by gpt-5.6-sol high.

= ii
{parent=2f}
{scope}

= Solution
{parent=ii}

Statement (ii) is true. Let
$$
K_n=E\cap\{z:|z|\leq n\}.
$$
Each $K_n$ is <compact set>[compact] and has connected complement. By the <polynomial Runge theorem>, choose a <polynomial> $P_n$ with
$$
\sup_{K_n}|P_n-f|<\frac1n.
$$
Every fixed $z\in E$ belongs to $K_n$ for all sufficiently large $n$, so $P_n(z)\to f(z)$. Thus the convergence is <pointwise convergence>[pointwise] on $E$.

Solved by gpt-5.6-sol high.

= iii
{parent=2f}
{scope}

= Solution
{parent=iii}

Statement (iii) is false. Let $\Omega$ be a thin open annulus about the unit circle, let $f(z)=1/z$, and split the circle into two proper closed arcs $K_1,K_2$ whose union is the whole circle. Each arc has connected complement, so <polynomial approximation of the reciprocal on a proper circular arc> gives uniform polynomial approximations to $f$ on each $K_j$ separately.

If polynomials $P_n$ converged uniformly to $1/z$ on $K_1\cup K_2$, then uniform convergence on the unit circle and the <Cauchy integral theorem> would give
$$
0=\lim_{n\to\infty}\int_{|z|=1}P_n(z)\,dz
=\int_{|z|=1}\frac{dz}{z}=2\pi i,
$$
a contradiction.

Solved by gpt-5.6-sol high.

= iv
{parent=2f}
{scope}

= Solution
{parent=iv}

Statement (iv) is true. Apply the <Weierstrass approximation theorem> separately to the real and imaginary parts of the <continuous function> $f$ on the <compact>[compact interval] $[0,1]$. Combining the two real polynomial sequences gives complex polynomials converging <uniform convergence>[uniformly] to $f$ on $I$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3I}

Write $\mathbf1_n$ for the all-one column vector. The <parity-check extension of a linear code> is
$$
C^+=\{(c_1,\ldots,c_n,c_1+\cdots+c_n):c\in C\}.
$$
The defining map $C\to\mathbb F_2^{n+1}$ is <linear map>[linear], so $C^+$ is a binary <linear code>. If $G$ is a <generator matrix> and $H$ a <parity-check matrix> for $C$, then one may take
$$
G^+=\begin{pmatrix}G&G\mathbf1_n\end{pmatrix},
\qquad
H^+=
\begin{pmatrix}
H&0\\
\mathbf1_n^T&1
\end{pmatrix}.
$$

The <punctured code> in, say, the last coordinate is
$$
C^-=\{(c_1,\ldots,c_{n-1}):c\in C\}.
$$
Coordinate deletion is a <linear map>, so $C^-$ is <linear code>[linear]. Its <generator matrix> $G^-$ is obtained by deleting the last column of $G$. If the <minimum Hamming distance of a linear code>[minimum distance] is at least two, deletion is injective on $C$, and hence $C^-$ has the same dimension as $C$.

For its <parity-check matrix>, first use invertible row operations on $H$ to make its nonzero last column a coordinate vector. The last column is nonzero because otherwise the weight-one word supported there would belong to $C$, contrary to $d\geq2$. Delete the pivot row and the last column; the resulting matrix $H^-$ is a <parity-check matrix> for $C^-$. Equivalently, $(C^-)^\perp$ is obtained by shortening the <dual code> $C^\perp$ at the deleted coordinate.

Finally, the <shortened code> in the last coordinate is
$$
C'=\{(c_1,\ldots,c_{n-1}):(c_1,\ldots,c_{n-1},0)\in C\}.
$$
It is linear whenever $C$ is linear, because it is the puncture of the subspace $C\cap\{x_n=0\}$.

Solved by gpt-5.6-sol high.

= 4I
{parent=Paper 3}
{scope}
{title2=Automata & Formal Languages}

= a
{parent=4i}
{scope}

= Solution
{parent=a}

For the <formal grammar> $G=(\Sigma,V,P,S)$,
$$
\boxed{L(G)=\{w\in\Sigma^*:S\mathrel{\mathop{\Longrightarrow}\limits_G^*}w\}.}
$$
Thus $L(G)$ contains exactly the finite terminal words reachable from the start symbol by a finite <grammar derivation>.

Solved by gpt-5.6-sol high.

= b
{parent=4i}
{scope}

= Solution
{parent=b}

Two <formal grammars> over the same terminal alphabet are <equivalent grammar>[equivalent] when they generate the same language:
$$
L(G_0)=L(G_1).
$$

Solved by gpt-5.6-sol high.

= c
{parent=4i}
{scope}

= Solution
{parent=c}

Two <formal grammars> over the same terminal alphabet are <isomorphic grammar>[isomorphic] when there is a bijection $\phi:V_0\to V_1$ between their nonterminals such that $\phi(S_0)=S_1$ and
$$
\alpha\to\beta\in P_0
\quad\Longleftrightarrow\quad
\phi(\alpha)\to\phi(\beta)\in P_1,
$$
where $\phi$ is extended to words by fixing each terminal and acting symbol by symbol.

Solved by gpt-5.6-sol high.

= d
{parent=4i}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

The grammars are <isomorphic grammar>[isomorphic]. The bijection
$$
S\longmapsto S,
\qquad A\longmapsto D,
\qquad B\longmapsto C
$$
maps every production of $P_0$ to the corresponding production of $P_1$. They are therefore <equivalent grammar>[equivalent]. Notice that $B$ in the first grammar and $C$ in the second are <unreachable nonterminal>[unreachable]; the isomorphism also preserves this irrelevant component.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

The grammars are non-equivalent. In the first grammar, terminal derivations without $AB\to c$ give
$$
\{a^rb^s:r,s\geq1\}.
$$
After applying $B\to Bb$ any number of times while $A$ and $B$ remain adjacent, the production $AB\to c$ also gives
$$
\{cb^t:t\geq0\}.
$$
Hence, in particular, $ab\in L(G_0)$ and $c\in L(G_0)$.

In the second grammar, the fixed terminals $ab$ in $XabY$ prevent $X$ and $Y$ from ever becoming adjacent, so $XY\to c$ can never be used. The remaining productions give exactly
$$
L(G_1)=\{a^rb^s:r,s\geq2\}.
$$
Thus $ab,c\notin L(G_1)$, proving that the generated <formal language theory>[languages] differ.

Solved by gpt-5.6-sol high.

= iii
{parent=d}
{scope}

= Solution
{parent=iii}

The first grammar generates
$$
L(G_0)=\{ab^{2r+1}a:r\geq0\},
$$
because $A\to bAb$ adds two $b$'s and $A\to b$ terminates the derivation.

In the second grammar, $Z$ is a <nonproductive nonterminal>: every right-hand side of a $Z$-production still contains a $Z$. Consequently the branch $Y\to ZZ$ never yields a terminal word. Every terminal derivation from $Y$ instead uses $Y\to bYb$ repeatedly and ends with $Y\to b$, so
$$
L(G_1)=\{ab^{2r+1}a:r\geq0\}=L(G_0).
$$
The grammars are therefore <equivalent grammar>[equivalent], although they are not <isomorphic grammar>[isomorphic].

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5J}

A one-parameter <exponential family> with natural statistic $Y$ has density or probability mass function
$$
f_\theta(y)=h(y)\exp\{\theta y-K(\theta)\},
$$
where $\theta$ is the <natural parameter of an exponential family> and $K$ is the <cumulant function of an exponential family>. Its <mean parameter of an exponential family>[mean parameter] is
$$
\mu(\theta)=K'(\theta).
$$

The <exponential-family deviance> from $\theta_1$ to $\theta_2$ is twice the <Kullback-Leibler divergence>:
$$
D(\theta_1,\theta_2)
=2\mathbb E_{\theta_1}
\left[\log\frac{f_{\theta_1}(Y)}{f_{\theta_2}(Y)}\right].
$$
The carrier $h(Y)$ cancels from the likelihood ratio, so, with $\mu_1=\mathbb E_{\theta_1}Y$,
$$
\boxed{
D(\theta_1,\theta_2)
=2\{(\theta_1-\theta_2)\mu_1-K(\theta_1)+K(\theta_2)\}.}
$$

For the <Poisson distribution> with mean $\mu$,
$$
\theta=\log\mu,
\qquad
K(\theta)=e^\theta=\mu.
$$
Therefore
$$
\boxed{
D(\mu_1,\mu_2)
=2\left\{\mu_1\log\frac{\mu_1}{\mu_2}-\mu_1+\mu_2\right\}.}
$$
Writing $\mu_2=\mu_1+\delta$ and applying the <Taylor series> of $\log(1+\delta/\mu_1)$ gives
$$
D(\mu_1,\mu_1+\delta)
=\frac{\delta^2}{\mu_1}
+O(\delta^3),
$$
so the second-order approximation is
$$
\boxed{D(\mu_1,\mu_2)\simeq
\frac{(\mu_2-\mu_1)^2}{\mu_1}.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=6C}

With $s=0$, the <equilibrium of an autonomous differential equation>[equilibria] satisfy
$$
g\left(\frac{kg}{1+g^2}-1\right)=0.
$$
Besides $g=0$, the two positive equilibria are
$$
g_\pm=\frac{k\pm\sqrt{k^2-4}}2,
\qquad
g_-<1<g_+,
$$
where $g_-g_+=1$. Since
$$
\partial_gf(g,0)=\frac{2kg}{(1+g^2)^2}-1,
$$
the nonzero equilibrium relation $k=(1+g^2)/g$ gives
$$
\partial_gf(g,0)=\frac{1-g^2}{1+g^2}.
$$
The <fixed point stability for an autonomous differential equation> therefore shows that $g=0$ and $g_+$ are stable, while $g_-$ is unstable. The graph of $f(g,0)$ starts at zero with negative slope, crosses upward at $g_-$, crosses downward at $g_+$, and tends to $-\infty$ as $g\to\infty$.

For a constant input $s$, write the equilibrium equation as
$$
s=h(g),
\qquad
h(g)=g-\frac{kg^2}{1+g^2}.
$$
The low stable equilibrium and the intervening unstable equilibrium coalesce in a <saddle-node bifurcation>. More precisely, let $g_c$ be the first positive solution of
$$
h'(g_c)=1-\frac{2kg_c}{(1+g_c^2)^2}=0
$$
and define
$$
\boxed{s_c(k)=h(g_c)
=g_c-\frac{kg_c^2}{1+g_c^2}.}
$$
Equivalently, $f(g_c,s_c)=0$ and $\partial_gf(g_c,s_c)=0$, with $g_c$ on the low-concentration branch. If $s_1>s_c(k)$, that branch no longer exists. Holding the input long enough carries the trajectory into the <basin of attraction> of the high state. When the input returns to zero, the concentration converges to
$$
\boxed{g^*=g_+=\frac{k+\sqrt{k^2-4}}2.}
$$
This is the <saturating autocatalytic switch>.

For $k\gg1$, the threshold occurs at $g=O(k^{-1})$, so
$$
f(g,s)=s+kg^2-g+O(kg^4).
$$
The approximate double-root conditions are
$$
s+kg^2-g=0,
\qquad
2kg-1=0.
$$
Thus $g_c\simeq1/(2k)$ and
$$
\boxed{s_c(k)\simeq\frac1{4k}.}
$$
Hence the constant in $s_c(k)\simeq Ck^{-1}$ is $\boxed{C=1/4}$, as recorded by the <strong-autocatalysis switching threshold>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7E}

For
$$
w''+p(z)w'+q(z)w=0,
$$
a finite point $z_0$ is an <ordinary point criterion for a second-order equation>[ordinary point] exactly when $p$ and $q$ are <holomorphic function>[holomorphic] there. It is a <regular singular point criterion for a second-order equation>[regular singular point] exactly when
$$
(z-z_0)p(z)
\quad\hbox{and}\quad
(z-z_0)^2q(z)
$$
are holomorphic at $z_0$.

Put $\zeta=1/z$ and $W(\zeta)=w(1/\zeta)$. Direct differentiation gives
$$
w'=-\zeta^2W',
\qquad
w''=\zeta^4W''+2\zeta^3W',
$$
so the transformed equation is
$$
W''+\left(\frac2\zeta-
\frac{p(1/\zeta)}{\zeta^2}\right)W'
+\frac{q(1/\zeta)}{\zeta^4}W=0.
$$
Consequently $z=\infty$ is ordinary precisely when
$$
\frac2\zeta-\frac{p(1/\zeta)}{\zeta^2}
\quad\hbox{and}\quad
\frac{q(1/\zeta)}{\zeta^4}
$$
are holomorphic at $\zeta=0$. It is a <regular singular point at infinity> precisely when $zp(z)$ and $z^2q(z)$ are holomorphic functions of $1/z$ near infinity.

If zero and infinity are regular singular and every nonzero finite point is ordinary, the <Laurent series> of $p$ and $q$ can contain only the terms compatible with both endpoint bounds. Hence
$$
\boxed{p(z)=\frac az,
\qquad q(z)=\frac b{z^2}}
$$
for constants $a,b$. The equation is a <Cauchy-Euler differential equation>. Its indicial equation is
$$
\lambda(\lambda-1)+a\lambda+b=0.
$$
For distinct roots $\lambda_1,\lambda_2$, the general solution on a domain with a chosen logarithm branch is
$$
w(z)=C_1z^{\lambda_1}+C_2z^{\lambda_2}.
$$
For a repeated root $\lambda$, the <general solution of an Euler-Cauchy equation> is
$$
w(z)=z^\lambda(C_1+C_2\log z).
$$

Finally require infinity to be ordinary. In the transformed equation its coefficients become
$$
\frac{2-a}{\zeta},
\qquad
\frac b{\zeta^2}.
$$
Both are holomorphic at zero exactly when $a=2$ and $b=0$. Thus the further restriction is
$$
\boxed{p(z)=\frac2z,
\qquad q(z)=0.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8d}
{scope}

= Solution
{parent=a}

For smooth functions $f,g$ on the six-dimensional phase space with canonical coordinates $(q_i,p_i)$, the <Poisson bracket> is
$$
\boxed{
\{f,g\}=\sum_{i=1}^3
\left(
\frac{\partial f}{\partial q_i}
\frac{\partial g}{\partial p_i}
-\frac{\partial f}{\partial p_i}
\frac{\partial g}{\partial q_i}
\right).}
$$

Solved by gpt-5.6-sol high.

= b
{parent=8d}
{scope}

= Solution
{parent=b}

Hamilton's equations and the multivariable <chain rule> give
$$
\frac{df}{dt}
=\sum_{i=1}^3\left(
\frac{\partial f}{\partial q_i}\dot q_i
+\frac{\partial f}{\partial p_i}\dot p_i
\right)
=\sum_{i=1}^3\left(
\frac{\partial f}{\partial q_i}\frac{\partial H}{\partial p_i}
-\frac{\partial f}{\partial p_i}\frac{\partial H}{\partial q_i}
\right)
=\{f,H\}.
$$
Thus the <Hamiltonian conservation law> gives
$$
\boxed{f\text{ is conserved}\quad\Longleftrightarrow\quad\{f,H\}=0.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=8d}
{scope}

= Solution
{parent=c}

For the <phase-space dilation generator> $D=q\cdot p$,
$$
\frac{\partial D}{\partial p_i}=q_i,
\qquad
\frac{\partial D}{\partial q_i}=p_i.
$$
Therefore the condition $\{f,D\}=0$ is the first-order equation
$$
\boxed{q\cdot\nabla_qf-p\cdot\nabla_pf=0.}
$$

If $f(\lambda q,\lambda^{-1}p)=f(q,p)$ for every positive real $\lambda$, differentiate with respect to $\lambda$ at $\lambda=1$. The <chain rule> gives exactly
$$
q\cdot\nabla_qf-p\cdot\nabla_pf=0,
$$
so every smooth scale-invariant $f$ satisfies the constraint. The same conclusion for negative $\lambda$ follows wherever the stated invariance is defined; differentiation only needs the positive component containing $\lambda=1$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=9B}

In the nonrelativistic regime,
$$
E(p)=mc^2+\frac{p^2}{2m}+O\!\left(\frac{p^4}{m^3c^2}\right).
$$
The assumption $E-\mu\gg k_BT$ puts the <Fermi-Dirac distribution> in its <Maxwell-Boltzmann distribution>[Maxwell-Boltzmann limit], so
$$
\begin{aligned}
n&\simeq\frac{4\pi g_s}{h^3}
e^{(\mu-mc^2)/(k_BT)}
\int_0^\infty p^2e^{-p^2/(2mk_BT)}\,dp\\
&=\frac{4\pi g_s}{h^3}
e^{(\mu-mc^2)/(k_BT)}
\frac{\sqrt\pi}{4}(2mk_BT)^{3/2}.
\end{aligned}
$$
Therefore the <nonrelativistic Maxwell-Boltzmann number density> is
$$
\boxed{
n=g_s\left(\frac{2\pi mk_BT}{h^2}\right)^{3/2}
\exp\left(\frac{\mu-mc^2}{k_BT}\right).}
$$

For $p+e^-\leftrightarrow H+\gamma$, <chemical equilibrium> and the vanishing <photon chemical potential> imply
$$
\mu_p+\mu_e=\mu_H.
$$
Applying the number-density formula to each massive species and eliminating the chemical potentials gives
$$
\frac{n_en_p}{n_H}
=\frac{g_eg_p}{g_H}
\left(\frac{2\pi k_BT}{h^2}\right)^{3/2}
\left(\frac{m_em_p}{m_H}\right)^{3/2}
\exp\left(-\frac{(m_p+m_e-m_H)c^2}{k_BT}\right).
$$
Use charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_eg_p/g_H$. With the <hydrogen binding energy>
$$
E_{\rm bind}=(m_p+m_e-m_H)c^2,
$$
we obtain the <Saha ionization equation>
$$
\boxed{
\frac{n_e^2}{n_H}\simeq
\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
\exp\left(-\frac{E_{\rm bind}}{k_BT}\right).}
$$
The assumptions are <thermal equilibrium> and <chemical equilibrium> before <cosmological recombination>, a <nondegenerate gas>[nonrelativistic nondegenerate gas], zero <photon chemical potential>, <charge neutrality>, negligible proton-electron mass correction in the translational prefactor, and the stated <degeneracy factor>[degeneracy-factor approximation].

Solved by gpt-5.6-sol high.

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

= a
{parent=10d}
{scope}

= Solution
{parent=a}

For <coprime integers> $a,N$ with $N>1$, the <multiplicative order> of $a$ modulo $N$ is
$$
\boxed{\operatorname{ord}_N(a)
=\min\{r\geq1:a^r\equiv1\pmod N\}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

The <Fermat-Euler theorem> gives
$$
a^{\varphi(N)}\equiv1\pmod N,
$$
so the <multiplicative order> $r=\operatorname{ord}_N(a)$ exists. For every <integer> $k$,
$$
f(k+r)=a^{k+r}\equiv a^ka^r\equiv a^k=f(k)\pmod N.
$$
Thus the modular exponential is <periodic modular exponentiation>[periodic] with period $r$. If $t>0$ were a smaller period, then $f(t)=f(0)$ would give $a^t\equiv1\pmod N$, contradicting the minimality of $r$. Hence $r$ is its least positive period.

Solved by gpt-5.6-sol high.

= c
{parent=10d}
{scope}

= Solution
{parent=c}

Modulo $33$,
$$
10^2=100\equiv1,
\qquad
10\not\equiv1,
$$
so $\operatorname{ord}_{33}(10)=2$. This order is even and
$$
10^{2/2}=10\not\equiv-1\pmod{33}.
$$
The <factor extraction from an even modular order> therefore gives
$$
\gcd(10-1,33)=3,
\qquad
\gcd(10+1,33)=11.
$$
Thus
$$
\boxed{33=3\cdot11.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=10d}
{scope}

= Solution
{parent=d}

The <quantum Fourier transform> measurement supplies a rational approximation
$$
\frac{c}{2^m}\simeq\frac{s}{r},
$$
where $r$ is the desired <multiplicative order>. Here
$$
\frac{c}{2^m}=\frac{256}{512}=\frac12.
$$
Its reduced <continued-fraction convergent> has denominator two, so <continued-fraction recovery in quantum order finding> proposes $r=2$. Direct <order verification from divisors> confirms it:
$$
8^2=64\equiv1\pmod{21},
\qquad
8\not\equiv1\pmod{21}.
$$
Hence
$$
\boxed{\operatorname{ord}_{21}(8)=2.}
$$
As a check, the classical final step of the <Shor algorithm> gives
$$
\gcd(8-1,21)=7,
\qquad
\gcd(8+1,21)=3.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=11G}

Write an integral <binary quadratic form> as
$$
[a,b,c](x,y)=ax^2+bxy+cy^2,
\qquad
d=b^2-4ac<0.
$$
A positive definite form is a <reduced positive definite binary quadratic form> when
$$
|b|\leq a\leq c,
$$
with $b\geq0$ on the boundary $|b|=a$ or $a=c$. The <class number of a negative discriminant> $h(d)$ is the number of <proper equivalence of binary quadratic forms>[proper equivalence classes] of primitive positive definite integral forms of discriminant $d$.

Every such class has a reduced representative. For a reduced form,
$$
4a^2\leq4ac=b^2+|d|\leq a^2+|d|,
$$
so
$$
a\leq\sqrt{\frac{|d|}{3}}.
$$
There are only finitely many possible integers $a$ and $b$, and then
$$
c=\frac{b^2-d}{4a}
$$
is determined. Hence $h(d)<\infty$. There is at least one class: the principal form is
$$
[1,0,-d/4]\quad\hbox{if }d\equiv0\pmod4,
$$
and
$$
[1,1,(1-d)/4]\quad\hbox{if }d\equiv1\pmod4.
$$
Thus $1\leq h(d)<\infty$, in agreement with the <enumeration of reduced binary quadratic forms>.

Now let $q\equiv3\pmod8$ be prime. The two forms
$$
f_0=[1,0,2q],
\qquad
f_1=[2,0,q]
$$
are primitive, positive definite, reduced, and have discriminant $-8q$. They are not <proper equivalence of binary quadratic forms>[properly equivalent]: $f_0$ represents one, while the least positive value of $f_1$ is two. Therefore
$$
\boxed{h(-8q)\geq2.}
$$

Assume henceforth that $h(-8q)=2$, so these are the only two classes. If
$$
p=x^2+2qy^2
$$
for a prime $p>q$, then $x$ is odd and
$$
p\equiv x^2+6y^2\equiv
\begin{cases}
1\pmod8,&y\text{ even},\\
7\pmod8,&y\text{ odd}.
\end{cases}
$$
Moreover $p\equiv x^2\pmod q$, so $p$ is a <quadratic residue> modulo $q$. This proves the necessity.

Conversely, suppose $p\equiv\pm1\pmod8$ and $p$ is a quadratic residue modulo $q$. By <quadratic reciprocity>, these two conditions imply
$$
\left(\frac{-8q}{p}\right)=1.
$$
Indeed, for $p\equiv1\pmod8$ both relevant signs are positive, while for $p\equiv7\pmod8$ the signs from $(-8/p)$ and reciprocity with $q\equiv3\pmod4$ cancel. Choose an even integer $b$ such that
$$
b^2\equiv-8q\pmod{4p}.
$$
Then
$$
[p,b,c],
\qquad
c=\frac{b^2+8q}{4p},
$$
is a primitive positive definite integral form of discriminant $-8q$ and represents $p$. It belongs to one of the two classes $f_0,f_1$.

The second form cannot represent such a prime. If
$$
p=2x^2+qy^2,
$$
then $y$ is odd and, since $q\equiv3\pmod8$,
$$
p\equiv
\begin{cases}
3\pmod8,&x\text{ even},\\
5\pmod8,&x\text{ odd}.
\end{cases}
$$
This contradicts $p\equiv\pm1\pmod8$. Hence $[p,b,c]$ lies in the principal class $f_0$, and <proper equivalence of binary quadratic forms> preserves represented integers. Therefore
$$
\boxed{
p=x^2+2qy^2
\quad\Longleftrightarrow\quad
p\equiv\pm1\pmod8
\ \hbox{and}\
\left(\frac pq\right)=1.}
$$
This is the <prime representation when h of minus eight q equals two>.

Solved by gpt-5.6-sol high.

= 12I
{parent=Paper 3}
{scope}
{title2=Automata & Formal Languages}

= i
{parent=12i}
{scope}

= Solution
{parent=i}

Extend $\delta:Q\times\Sigma\to Q$ to
$$
\widehat\delta:Q\times\Sigma^*\to Q
$$
by
$$
\widehat\delta(q,\varepsilon)=q,
\qquad
\widehat\delta(q,wa)=\delta(\widehat\delta(q,w),a).
$$
The <extended transition function of a deterministic finite automaton> processes a whole word from left to right. The language accepted by $D$ is
$$
\boxed{L(D)=\{w\in\Sigma^*:\widehat\delta(q_0,w)\in F\}.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=12i}
{scope}

= Solution
{parent=ii}

Let $N=|Q|$. The <pumping lemma for regular languages> states that every $w\in L(D)$ with $|w|\geq N$ can be written
$$
w=xyz
$$
with
$$
|xy|\leq N,
\qquad
|y|\geq1,
\qquad
xy^iz\in L(D)\quad\text{for every }i\geq0.
$$
Thus one may take the pumping number to be exactly the number $N$ of states of the given automaton, whether or not all of them are accessible.

Solved by gpt-5.6-sol high.

= iii
{parent=12i}
{scope}

= Solution
{parent=iii}

For $0\leq j\leq9$, consider the ten prefixes
$$
u_j=b^j.
$$
If $0\leq i<j\leq9$, append
$$
z_j=b^{11-j}a.
$$
Then
$$
u_i z_j=b^{i+11-j}a\in L,
$$
because $i+11-j\leq10$, whereas
$$
u_jz_j=b^{11}a\notin L.
$$
Indeed, in a word ending in one $a$, the prefix before the final positive run of $a$'s must have length at most ten. Hence every pair $u_i,u_j$ is distinguished by some suffix. They occupy ten distinct <Myhill-Nerode equivalence> classes, so the <minimal deterministic finite automaton> for $L$ has at least ten states.

Solved by gpt-5.6-sol high.

= iv
{parent=12i}
{scope}

= Solution
{parent=iv}

Suppose first that $A$ is a <regular language>, accepted by a <deterministic finite automaton> $D$. If two words reach the same state, then every continuation is accepted from both or rejected from both. Thus they are equivalent under $\sim_A$, and the number of equivalence classes is at most the finite number of states of $D$.

Conversely, suppose $\sim_A$ has finitely many classes. Define a deterministic automaton by
$$
Q=\Sigma^*/{\sim_A},
\qquad
q_0=[\varepsilon],
\qquad
\delta([w],a)=[wa],
$$
and
$$
F=\{[w]:w\in A\}.
$$
The transition is well defined: if $v\sim_Aw$, then for every suffix $u$,
$$
vau\in A\Longleftrightarrow wau\in A,
$$
so $va\sim_Awa$. Likewise, membership in $F$ is independent of the representative by taking the empty suffix in the definition of $\sim_A$. Induction gives
$$
\widehat\delta([\varepsilon],w)=[w],
$$
so the automaton accepts exactly $A$. It has finitely many states, and therefore $A$ is regular. This proves the <Myhill-Nerode theorem> in the stated formulation. The assumption $\varepsilon\notin A$ merely says that the initial state is not accepting.

Solved by gpt-5.6-sol high.

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

= a
{parent=13c}
{scope}

= Solution
{parent=a}

Setting $\epsilon=0$ makes the inhibitor equation quasistatic:
$$
0=v_{xx}+u-v.
$$
Taking the spatial <Fourier transform> and using the <Fourier transform of a derivative> gives
$$
0=-k^2\widehat v+\widehat u-\widehat v.
$$
Therefore the <fast-inhibitor elimination in a reaction-diffusion system> is
$$
\boxed{\widehat v(k,t)=\frac{\widehat u(k,t)}{1+k^2}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=13c}
{scope}

= Solution
{parent=b}

Since $f(0)=0$ and $f'(0)=-r$, the <linearization> is
$$
f(u)=-ru+O(u^2).
$$
For a spatial <Fourier mode> of the perturbation, part a gives
$$
\frac{\partial\widehat u}{\partial t}
=\left[
-Dk^2-r+\rho\left(1-\frac1{1+k^2}\right)
\right]\widehat u.
$$
Thus $\widehat u(k,t)\propto e^{\sigma(k)t}$ with the <dispersion relation after fast-inhibitor elimination>
$$
\boxed{\sigma(k)=-Dk^2-r+\rho\frac{k^2}{1+k^2}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=13c}
{scope}

= Solution
{parent=c}

Put $y=k^2\geq0$. At the threshold of a <Turing instability>, the maximum of
$$
\sigma(y)=-Dy-r+\rho\frac{y}{1+y}
$$
touches zero. Hence
$$
\sigma(y_c)=0,
\qquad
\frac{d\sigma}{dy}(y_c)
=-D+\frac{\rho}{(1+y_c)^2}=0.
$$
The second equation gives $\rho=D(1+y_c)^2$. Substitution in the first gives
$$
0=-Dy_c-r+Dy_c(1+y_c)=Dy_c^2-r.
$$
Therefore
$$
y_c=\sqrt{\frac rD},
\qquad
\boxed{k_c=\left(\frac rD\right)^{1/4}},
$$
and
$$
\boxed{
\rho_c(r)=D\left(1+\sqrt{\frac rD}\right)^2
=(\sqrt r+\sqrt D)^2.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=14b}
{scope}

= Solution
{parent=a}

During matter domination,
$$
a(t)\propto t^{2/3},
\qquad
\frac{\dot a}{a}=\frac2{3t},
\qquad
\frac{4\pi G\bar\rho_c}{c^2}=\frac2{3t^2}.
$$
The perturbation equation becomes
$$
\ddot\delta_k+\frac4{3t}\dot\delta_k-\frac2{3t^2}\delta_k=0.
$$
Substitution of the power law $\delta_k\propto t^\alpha$ gives
$$
3\alpha^2+\alpha-2=0,
$$
whose roots are $\alpha=2/3$ and $\alpha=-1$. Hence the <cosmic-time matter density modes> give
$$
\boxed{
\delta_k(t)=A(k)\left(\frac{t}{t_{\rm eq}}\right)^{2/3}
+B(k)\left(\frac{t}{t_{\rm eq}}\right)^{-1}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=14b}
{scope}

= Solution
{parent=b}

Normalize $a(t_0)=1$ and put $k_0=2\pi/(ct_0)$. The <cosmological horizon crossing> condition is
$$
\frac{t_H}{t_0}=\frac{k_0}{k}a(t_H).
$$
For $t_H>t_{\rm eq}$, matter domination gives $a(t_H)=(t_H/t_0)^{2/3}$, and therefore
$$
\boxed{\frac{t_H}{t_0}\simeq\left(\frac{k_0}{k}\right)^3.}
$$

For $t_H<t_{\rm eq}$, radiation domination and continuity at equality give
$$
a(t)=a_{\rm eq}\left(\frac{t}{t_{\rm eq}}\right)^{1/2},
\qquad
a_{\rm eq}=\left(\frac{t_{\rm eq}}{t_0}\right)^{2/3}
=\frac1{1+z_{\rm eq}}.
$$
Substitution in the crossing condition yields
$$
\boxed{
\frac{t_H}{t_0}\simeq
\frac1{\sqrt{1+z_{\rm eq}}}
\left(\frac{k_0}{k}\right)^2.}
$$
These are the two branches of the <horizon-crossing time across matter-radiation equality>.

Solved by gpt-5.6-sol high.

= c
{parent=14b}
{scope}

= Solution
{parent=c}

Neglect the decaying mode from part a. A mode that enters during matter domination grows in amplitude by
$$
\frac{a(t_0)}{a(t_H)}
=\left(\frac{t_0}{t_H}\right)^{2/3},
$$
so its variance grows by $(t_0/t_H)^{4/3}$. For $k<k_{\rm eq}$, part b gives
$$
P(k)=\frac{C}{k^3}
\left(\frac{k}{k_0}\right)^4
=\boxed{\frac{Ck}{k_0^4}}.
$$

For $k>k_{\rm eq}$, the mode enters during radiation domination and is assumed not to grow significantly until equality. It then grows by $a(t_0)/a_{\rm eq}$. Since the equality mode satisfies
$$
\frac{t_{\rm eq}}{t_0}
=\left(\frac{k_0}{k_{\rm eq}}\right)^3,
$$
the post-equality variance growth is
$$
\left(\frac{a(t_0)}{a_{\rm eq}}\right)^2
=\left(\frac{t_0}{t_{\rm eq}}\right)^{4/3}
=\left(\frac{k_{\rm eq}}{k_0}\right)^4.
$$
Therefore the <broken matter power spectrum from horizon entry> is
$$
\boxed{
P(k)=V\langle|\delta_k(t_0)|^2\rangle
=
\begin{cases}
\dfrac{Ck}{k_0^4},&k<k_{\rm eq},\\[6pt]
\dfrac{Ck_{\rm eq}^4}{k^3k_0^4},&k>k_{\rm eq}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=15d}
{scope}

= Solution
{parent=a}

Apply the gates from left to right. The first <Hadamard gate> and the first <controlled-NOT gate> produce the <Bell state>
$$
\frac{|00\rangle+|11\rangle}{\sqrt2}.
$$
The parallel $Z\otimes H$ gates then produce
$$
\frac12\left(|00\rangle+|01\rangle-|10\rangle+|11\rangle\right).
$$
The final controlled-NOT has the lower qubit as control; it interchanges $|01\rangle$ and $|11\rangle$, whose amplitudes here are equal. Thus the output is
$$
\boxed{
|\psi_{\rm out}\rangle
=\frac12\left(|00\rangle+|01\rangle-|10\rangle+|11\rangle\right).}
$$
By <quantum measurement in the computational basis>, every basis outcome has probability equal to the squared modulus of its amplitude:
$$
\boxed{\Pr(00)=\Pr(01)=\Pr(10)=\Pr(11)=\frac14.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

Use the <inverse quantum circuit> rule. Every gate in $C$ is self-inverse, so reverse their order. From left to right, the circuit for $V^{-1}$ is:

1. a controlled-NOT with the lower qubit controlling the upper;
2. $Z$ on the upper qubit and $H$ on the lower qubit;
3. a controlled-NOT with the upper qubit controlling the lower;
4. $H$ on the upper qubit.

This is the requested circuit description, with the two wires and control directions unchanged under each individual gate's adjoint.

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

Use three <controlled-NOT gate>[controlled-NOT gates] with control-target directions
$$
1\longrightarrow2,
\qquad
2\longrightarrow1,
\qquad
1\longrightarrow2.
$$
On computational-basis bits they act as
$$
(x,y)\longmapsto(x,x\oplus y)
\longmapsto(y,x\oplus y)
\longmapsto(y,x).
$$
They therefore implement the <three-CNOT decomposition of the SWAP gate>, and linearity proves the identity on every two-qubit state.

Solved by gpt-5.6-sol high.

= d
{parent=15d}
{scope}

= Solution
{parent=d}

After the first <Hadamard gate> and the <controlled unitary gate>, the state is
$$
\frac1{\sqrt2}
\left(|0\rangle|\psi\rangle
+e^{i\theta}|1\rangle|\psi\rangle\right).
$$
The final Hadamard gate gives
$$
\boxed{
\frac12\left[
(1+e^{i\theta})|0\rangle
+(1-e^{i\theta})|1\rangle
\right]|\psi\rangle.}
$$
Hence the <Hadamard test> returns one with probability
$$
\boxed{
\Pr(1)=\frac{|1-e^{i\theta}|^2}{4}
=\frac{1-\cos\theta}{2}
=\sin^2\frac\theta2.}
$$
If $U|\psi\rangle=|\psi\rangle$, then $\theta=0$ and the outcome is zero with certainty. If $U|\psi\rangle=-|\psi\rangle$, then $\theta=\pi$ modulo $2\pi$ and the outcome is one with certainty. The single measurement therefore distinguishes the two promised cases exactly.

Solved by gpt-5.6-sol high.

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

= a
{parent=16h}
{scope}

= Solution
{parent=a}

Fix a set $x$ and define by recursion on $\omega$
$$
T_0=x,
\qquad
T_{n+1}=\bigcup T_n.
$$
The recursion theorem defines a function-class $F$ uniquely by $F(n)=T_n$: for each $n\in\omega$ there is exactly one finite sequence satisfying the displayed recursion through stage $n$. Thus $F$ is genuinely functional, and the <axiom schema of replacement> applied to the set $\omega$ gives the set $\{T_n:n\in\omega\}$.

Now define
$$
T=\bigcup_{n\in\omega}T_n.
$$
It contains every member of $x$. If $y\in T$, then $y\in T_n$ for some $n$; hence every $z\in y$ belongs to
$$
\bigcup T_n=T_{n+1}\subseteq T.
$$
Thus $T$ is transitive. If $U$ is any transitive set containing every member of $x$, induction gives $T_n\subseteq U$ for every $n$, so $T\subseteq U$. Therefore
$$
\boxed{T=\operatorname{TC}(x)}
$$
is the <transitive closure> of $x$.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

The <axiom of foundation> says that every nonempty set $A$ contains an $x$ with $x\cap A=\varnothing$. The principle of <epsilon induction> says that if a formula $\varphi$ is progressive,
$$
\forall x\left[
(\forall y\in x,\ \varphi(y))\Longrightarrow\varphi(x)
\right],
$$
then $\varphi$ holds for every set.

Assume foundation and let $\varphi$ be progressive. If $\varphi(a)$ failed for some $a$, use <axiom schema of specification>[separation] to form the nonempty set of failures inside $\operatorname{TC}(\{a\})$. Foundation supplies an $\in$-minimal failure $x$. Every member of $x$ lies in the transitive closure and is not a failure, so progressiveness gives $\varphi(x)$, a contradiction. Hence epsilon induction holds.

Conversely, assume epsilon induction and suppose a nonempty set $A$ has no $\in$-minimal member. Let
$$
\varphi(x)\quad\Longleftrightarrow\quad x\notin A.
$$
If every member of $x$ satisfies $\varphi$ but $x\in A$, then $x\cap A=\varnothing$, making $x$ an $\in$-minimal member of $A$, contrary to the assumption. Thus $\varphi$ is progressive. Epsilon induction says every set lies outside $A$, contradicting that $A$ is nonempty. Foundation follows.

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

True. If $\operatorname{TC}(\{x\})$ is <countable set>[countable], then each of its members is a subset of this transitive set and is therefore countable.

Conversely, if $x$ is <reasonable set>[reasonable], then $x$ itself and every set occurring at every finite membership depth below it are countable. Starting with the countable set $\{x\}$, each next level is a <countable union of countable sets>, hence countable. Their countable union is $\operatorname{TC}(\{x\})$, so the transitive closure is countable.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

True. By part (i), the reasonable sets are exactly the <hereditarily countable set>[hereditarily countable sets] $H_{\omega_1}$. The rank of a countable transitive closure is the supremum of countably many countable ordinals, plus at most a finite amount, and is therefore below $\omega_1$. Hence every reasonable set belongs to
$$
\boxed{V_{\omega_1}.}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

False. The reasonable sets form the transitive set $H_{\omega_1}$, but this structure fails the <axiom of power set>. The set $\omega$ is reasonable, and every subset of $\omega$ is itself countable and therefore reasonable. Its full power set $\mathcal P(\omega)$ is uncountable, so it is not reasonable. No member of $H_{\omega_1}$ can therefore contain internally every subset of $\omega$ that belongs to $H_{\omega_1}$. Thus the reasonable sets do not form a model of ZFC.

Solved by gpt-5.6-sol high.

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

= a
{parent=17h}
{scope}

= Solution
{parent=a}

We prove the <quadratic Turan edge bound> by induction. The case $r=1$ is immediate. For $r\geq2$, if $G$ contains no $K_r$, the induction hypothesis for $r-1$ gives the stronger bound. Otherwise choose a copy $S$ of $K_r$. Every vertex outside $S$ has at most $r-1$ neighbours in $S$, since one adjacent to all of $S$ would complete a $K_{r+1}$. Thus the number of edges having at least one endpoint in $S$ is at most
$$
\binom r2+(n-r)(r-1)
=(r-1)\left(n-\frac r2\right).
$$
The graph $G-S$ is still $K_{r+1}$-free, so induction on the number of vertices gives
$$
\begin{aligned}
e(G)
&\leq
\left(1-\frac1r\right)\frac{(n-r)^2}{2}
+(r-1)\left(n-\frac r2\right)\\
&=\boxed{\left(1-\frac1r\right)\frac{n^2}{2}}.
\end{aligned}
$$
This proves the stated form of <Turan theorem>.

Solved by gpt-5.6-sol high.

= b
{parent=17h}
{scope}

= Solution
{parent=b}

Let
$$
P=v_0v_1\cdots v_\ell
$$
be a longest <path in a graph>[path] in $G$. Every neighbour of either endpoint lies on $P$. Suppose $\ell<2\delta(G)$. Among the $\ell$ possible cut positions of the path, the sets
$$
\{i:v_0v_i\in E(G)\},
\qquad
\{i:v_{i-1}v_\ell\in E(G)\}
$$
have total size greater than $\ell$, so they intersect. The corresponding two edges close a cycle containing all vertices of $P$.

If $\ell<n-1$, connectivity supplies an edge from a vertex outside this cycle to a vertex on it. Breaking the cycle there and adjoining the outside vertex creates a path longer than $P$, a contradiction. Hence
$$
\ell\geq\min(2\delta(G),n-1).
$$
Since $\delta(G)\geq t/2$ and $t\leq n-1$, the <long path from minimum degree> gives $\ell\geq t$. Thus $G$ contains $P_t$.

Solved by gpt-5.6-sol high.

= c
{parent=17h}
{scope}

= Solution
{parent=c}

Colour the vertices of $K_{(s-1)t}$ in $s-1$ groups of size $t$. Colour edges within groups blue and edges between groups red. The red graph is complete $(s-1)$-partite, so it has no red $K_s$. Every blue component has only $t$ vertices, so it has no <path in a graph>[path] of length $t$. Therefore
$$
\boxed{r(K_s,P_t)\geq(s-1)t+1.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=17h}
{scope}

= Solution
{parent=d}

We first derive the <Erdos-Gallai path edge bound> from part b. If an $n$-vertex graph has no path of length $t$, then
$$
e(G)\leq\frac{(t-1)n}{2}.
$$
Induct on $n$, component by component. A component with at most $t$ vertices satisfies the bound trivially. A larger connected component cannot have minimum degree at least $t/2$ by part b, so delete a vertex of degree less than $t/2$ and apply induction; the integer degree removed is at most $(t-1)/2$, which preserves the bound.

Now set
$$
n=(s-1)t+1
$$
and consider a red-blue colouring of $K_n$. If there is no red $K_s$, part a with $r=s-1$ gives
$$
e_R\leq\left(1-\frac1{s-1}\right)\frac{n^2}{2}.
$$
Consequently
$$
\begin{aligned}
e_B
&=\binom n2-e_R\\
&\geq\frac12\left(\frac{n^2}{s-1}-n\right)\\
&=\frac n2\left(t-1+\frac1{s-1}\right)
>\frac{(t-1)n}{2}.
\end{aligned}
$$
The path edge bound forces a blue $P_t$. Together with part c this proves the <clique-path Ramsey number>
$$
\boxed{r(K_s,P_t)=(s-1)t+1.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=18i}
{scope}

= Solution
{parent=a}

Let $G$ be a finite subgroup of $K^\times$ and let $m$ be its <exponent of a finite group>. A finite abelian group contains an element whose order is its exponent: combine elements of maximal prime-power order across the primary components. Hence $m\leq|G|$.

Every $g\in G$ satisfies $g^m=1$, so all $|G|$ elements are roots in the field $K$ of the nonzero <polynomial> $X^m-1$. The <Lagrange root bound over a field> gives $|G|\leq m$. Therefore $|G|=m$, and the element of order $m$ generates $G$. This proves that a <finite subgroup of a field multiplicative group is cyclic>.

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= Solution
{parent=b}

A <primitive root of unity> is an $n$th root of unity of multiplicative order exactly $n$.

Let $\zeta_m,\zeta_n\in K$ be primitive and let $H$ be the finite subgroup of $K^\times$ that they generate. By part a, $H$ is cyclic. Its order is divisible by both $m$ and $n$, hence by $N=\operatorname{lcm}(m,n)$. On the other hand every element of $H$ has $N$th power one, so the order of the cyclic group $H$ divides $N$. Its order is therefore exactly $N$, and a generator of $H$ is a primitive $N$th root of unity in $K$.

Solved by gpt-5.6-sol high.

= c
{parent=18i}
{scope}

= Solution
{parent=c}

The <cyclotomic polynomial> is
$$
\Phi_n(X)=\prod_{\substack{1\leq a\leq n\\(a,n)=1}}
(X-\zeta_n^a),
$$
where $\zeta_n$ is primitive. Partitioning all $n$th roots by exact order gives the <cyclotomic factorization>
$$
X^n-1=\prod_{d\mid n}\Phi_d(X).
$$
Induct on $n$. The quotient of the monic integer polynomial $X^n-1$ by the product of the already constructed monic $\Phi_d$, $d<n$, lies in $\mathbb Q[X]$ and is monic. <Gauss lemma> then shows that this quotient $\Phi_n$ lies in $\mathbb Z[X]$.

If the prime $p$ does not divide $n$, then over $\mathbb F_p$,
$$
\gcd(X^n-1,nX^{n-1})=1.
$$
Thus $X^n-1$ has no repeated root. Every factor, including the reduction of $\Phi_n$, is square-free, proving the <separability of a cyclotomic polynomial modulo p>.

Solved by gpt-5.6-sol high.

= d
{parent=18i}
{scope}

= Solution
{parent=d}

Choose a primitive root $\zeta_n$. Since every root of $\Phi_n$ is a power of $\zeta_n$, its splitting field is $L=K(\zeta_n)$. Each $\sigma\in G$ sends $\zeta_n$ to another primitive root, uniquely of the form
$$
\sigma(\zeta_n)=\zeta_n^{a_\sigma},
\qquad
a_\sigma\in(\mathbb Z/n\mathbb Z)^\times.
$$
The assignment $\sigma\mapsto a_\sigma$ is a homomorphism. It is injective because an automorphism fixing $\zeta_n$ fixes $K(\zeta_n)=L$.

If $\Phi_n$ is irreducible over $K$, the orbit of $\zeta_n$ has all $\varphi(n)$ primitive roots, so the injection has image of size $\varphi(n)$ and is surjective. Conversely, surjectivity makes every primitive root a conjugate of $\zeta_n$. Its minimal polynomial over $K$ then has at least $\varphi(n)$ roots and divides the degree-$\varphi(n)$ polynomial $\Phi_n$, so it equals $\Phi_n$. This proves the <Galois embedding for a cyclotomic polynomial> and the claimed equivalence.

Solved by gpt-5.6-sol high.

= e
{parent=18i}
{scope}

= Solution
{parent=e}

Let $L=\mathbb Q(\zeta_n)$. We use the standard cyclotomic containment lemma: every root of unity in $\mathbb Q(\zeta_n)$ has order dividing $\operatorname{lcm}(2,n)$. One proof uses the conductor of a rational cyclotomic field. The conductor attached to a primitive $m$th root is $m$, except that it is $m/2$ when $m\equiv2\pmod4$; containment in $\mathbb Q(\zeta_n)$ forces this conductor to divide $n$, which gives $m\mid\operatorname{lcm}(2,n)$.

If $n$ is even, $\operatorname{lcm}(2,n)=n$, and the powers of $\zeta_n$ already give all $n$ possible roots of unity in $L$. If $n$ is odd, $-\zeta_n$ has order $2n$, so $L$ contains all $2n$th roots of unity, and the containment lemma shows there are no others. Thus the <roots of unity in a rational cyclotomic field> number
$$
\boxed{
\begin{cases}
n,&n\text{ even},\\
2n,&n\text{ odd}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=19h}
{scope}

= Solution
{parent=a}

The <Burnside lemma> states that for a finite group $G$ acting on a finite set $X$,
$$
|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|.
$$
To prove it, count
$$
\Omega=\{(g,x)\in G\times X:gx=x\}.
$$
Counting by $g$ gives $\sum_g|X^g|$. Counting by $x$ gives
$$
\sum_{x\in X}|G_x|
=\sum_{\mathcal O}|{\mathcal O}|\,|G_x|
=\sum_{\mathcal O}|G|
=|G|\,|X/G|,
$$
using the orbit-stabilizer theorem on each orbit $\mathcal O$. Dividing by $|G|$ proves the formula.

Let $\pi$ be the character of the <permutation representation> $\mathbb C[X]$. Then $\pi(g)=|X^g|$, and a second application of Burnside's lemma gives
$$
\langle\pi,\pi\rangle_G
=\frac1{|G|}\sum_g|X^g|^2
=\#(G\backslash(X\times X)).
$$
For a two-transitive action there are exactly two orbits on $X\times X$: the diagonal and the ordered pairs of distinct points. Hence $\langle\pi,\pi\rangle=2$. Transitivity says the trivial representation occurs once. Since the squared multiplicities of the irreducible constituents sum to two, there is exactly one further constituent, with multiplicity one, and it is irreducible and nontrivial. Thus
$$
\boxed{\mathbb C[X]\cong\mathbf1\oplus V}
$$
as in the <permutation representation of a two-transitive action>.

Solved by gpt-5.6-sol high.

= b
{parent=19h}
{scope}

= Solution
{parent=b}

For permutation characters, <Burnside lemma> on a product gives
$$
\langle\pi_k,\pi_l\rangle_G
=\#\bigl(G\backslash(X_k\times X_l)\bigr).
$$
Two pairs $(A,B)$ and $(A',B')$ lie in the same $S_n$-orbit exactly when
$$
|A\cap B|=|A'\cap B'|.
$$
When $0\leq l\leq k\leq n/2$, every value $0,1,\ldots,l$ is possible because $k+l\leq n$. Therefore the <symmetric-group subset permutation representation> satisfies
$$
\boxed{\langle\pi_k,\pi_l\rangle_G=l+1.}
$$

For $1\leq r\leq n/2$, the <inclusion map between subset permutation modules> embeds $\mathbb C[X_{r-1}]$ into $\mathbb C[X_r]$, so
$$
\chi_r=\pi_r-\pi_{r-1}
$$
is a character. Its norm is
$$
\begin{aligned}
\langle\chi_r,\chi_r\rangle
&=\langle\pi_r,\pi_r\rangle
-2\langle\pi_r,\pi_{r-1}\rangle
+\langle\pi_{r-1},\pi_{r-1}\rangle\\
&=(r+1)-2r+r=1.
\end{aligned}
$$
By <character orthogonality>, $\chi_r$ is the character of an irreducible representation.

For $r>n/2$, complementation is an $S_n$-equivariant bijection $X_r\cong X_{n-r}$, so
$$
\pi_r=\pi_{n-r}.
$$
Consequently
$$
\pi_r-\pi_{r-1}
=-\bigl(\pi_{n-r+1}-\pi_{n-r}\bigr),
$$
the negative of one of the irreducible characters just found, and is not itself the character of a representation.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=20G}

Write a point of the band $X$ as $(u,t)$ with $u\in S^2$, $|t|\leq1/2$, after rescaling the radius of the first three coordinates. The antipodal identification is
$$
(u,t)\sim(-u,-t).
$$
The homotopy $(u,t)\mapsto(u,(1-s)t)$ is equivariant under this identification, so it descends to a deformation retraction of $Y$ onto the central slice
$$
S^2/(u\sim-u)=\mathbb{RP}^2.
$$
Thus
$$
\boxed{Y\simeq\mathbb{RP}^2.}
$$
The two boundary spheres of $X$ are exchanged by the antipodal map, so $\partial Y\cong S^2$. This is the <mapping-cylinder model of punctured real projective three-space>.

The standard cellular chain complex for $\mathbb{RP}^3$, with one cell in each dimension zero through three, has boundary maps alternating between multiplication by two and zero:
$$
0\longrightarrow\mathbb Z
\xrightarrow{0}\mathbb Z
\xrightarrow{2}\mathbb Z
\xrightarrow{0}\mathbb Z
\longrightarrow0.
$$
Therefore the <integral homology of real projective three-space> is
$$
\boxed{
H_i(\mathbb{RP}^3;\mathbb Z)=
\begin{cases}
\mathbb Z,&i=0,3,\\
\mathbb Z/2,&i=1,\\
0,&\text{otherwise}.
\end{cases}}
$$
In particular,
$$
H_0(Y)=\mathbb Z,\qquad
H_1(Y)=\mathbb Z/2,\qquad
H_i(Y)=0\quad(i\geq2).
$$

Apply the <Mayer-Vietoris theorem> to
$$
Z=Y_1\cup_{\partial Y}Y_2,
\qquad
Y_1\cap Y_2\simeq S^2.
$$
The relevant pieces of the long exact sequence are
$$
0\to H_3(Z)\to H_2(S^2)=\mathbb Z\to0
$$
and
$$
0\to(\mathbb Z/2)^2\to H_1(Z)
\to H_0(S^2)\to H_0(Y_1)\oplus H_0(Y_2).
$$
The last map is injective, so
$$
\boxed{
H_i(Z;\mathbb Z)=
\begin{cases}
\mathbb Z,&i=0,3,\\
(\mathbb Z/2)^2,&i=1,\\
0,&\text{otherwise}.
\end{cases}}
$$

Since $\partial Y\cong S^2$ is simply connected and $\pi_1(Y)\cong\pi_1(\mathbb{RP}^2)\cong\mathbb Z/2$, the <Seifert-van Kampen theorem> gives
$$
\boxed{
\pi_1(Z,z_0)\cong
(\mathbb Z/2)*(\mathbb Z/2)
\cong D_\infty,}
$$
the <infinite dihedral group>. Topologically, $Z$ is the <double of punctured real projective three-space>, namely $\mathbb{RP}^3\#\mathbb{RP}^3$.

The universal cover of each copy of $Y$ is $S^3$ with two disjoint open balls removed, homeomorphic to $S^2\times[0,1]$. The universal cover of the double strings infinitely many such cylinders together according to the Cayley line of $D_\infty$. Hence the familiar covering space is
$$
\boxed{\widetilde Z\cong S^2\times\mathbb R.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=21f}
{scope}

= Solution
{parent=a}

The <Tietze extension theorem> states: if $X$ is a <normal topological space>, $A\subseteq X$ is closed, and $f:A\to[-M,M]$ is continuous, then there is a continuous $F:X\to[-M,M]$ with $F|_A=f$.

We first prove an approximation lemma. Given continuous $h:A\to[-r,r]$, the closed subsets
$$
A_-=\{x\in A:h(x)\leq-r/3\},
\qquad
A_+=\{x\in A:h(x)\geq r/3\}
$$
are disjoint and closed in $X$. By the <Urysohn lemma>, there is a continuous $g:X\to[-r/3,r/3]$ equal to $-r/3$ on $A_-$ and $r/3$ on $A_+$. Then
$$
|h-g|\leq\frac{2r}{3}\quad\hbox{on }A.
$$

Starting with $h_0=f$ and $r_0=M$, apply the lemma recursively to obtain continuous $g_j:X\to\mathbb R$ such that
$$
\|g_j\|_\infty\leq\frac{r_j}{3},
\qquad
h_{j+1}=h_j-g_j|_A,
\qquad
\|h_{j+1}\|_\infty\leq r_{j+1},
$$
where $r_j=M(2/3)^j$. The series
$$
F=\sum_{j=0}^\infty g_j
$$
converges uniformly by the Weierstrass test, so its sum is continuous. On $A$, its remainder after $j$ terms is $h_{j+1}$, which tends uniformly to zero; hence $F|_A=f$. The bounds also keep $F$ in $[-M,M]$ after the standard endpoint-preserving version of the construction, proving the theorem.

Solved by gpt-5.6-sol high.

= b
{parent=21f}
{scope}

= Solution
{parent=b}

Write
$$
A=\bigcap_{n=1}^\infty U_n
$$
with each $U_n$ open. The closed sets $A$ and $X\setminus U_n$ are disjoint. By the <Urysohn lemma>, choose a continuous $f_n:X\to[0,1]$ with
$$
f_n=0\text{ on }A,
\qquad
f_n=1\text{ on }X\setminus U_n.
$$
The uniformly convergent series
$$
f(x)=\sum_{n=1}^\infty2^{-n}f_n(x)
$$
defines a continuous map to $[0,1]$. It vanishes on $A$. If $x\notin A$, then $x\notin U_n$ for some $n$, so $f_n(x)=1$ and $f(x)>0$. Thus
$$
\boxed{f^{-1}(0)=A,}
$$
which is the <closed G-delta set as a zero set> construction.

Solved by gpt-5.6-sol high.

= c
{parent=21f}
{scope}

= Solution
{parent=c}

Apply part b separately to obtain continuous $g,h:X\to[0,1]$ with
$$
g^{-1}(0)=A,
\qquad
h^{-1}(0)=B.
$$
Because $A\cap B=\varnothing$, the two functions never vanish simultaneously. Hence
$$
\boxed{f=\frac{g}{g+h}}
$$
is continuous and takes values in $[0,1]$. It equals zero exactly where $g=0$, namely on $A$, and equals one exactly where $h=0$, namely on $B$.

Solved by gpt-5.6-sol high.

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

= a
{parent=22f}
{scope}

= Solution
{parent=a}

A weak solution is a function $u\in H_0^1(U)$ satisfying
$$
\boxed{
\int_U\left(\nabla u\cdot\nabla v+m^2uv\right)\,dx
=\int_Ufv\,dx
\quad\text{for every }v\in H_0^1(U).}
$$
The zero boundary condition is encoded by membership in $H_0^1(U)$.

On the Hilbert space $H_0^1(U)$ define
$$
a(u,v)=\int_U(\nabla u\cdot\nabla v+m^2uv),
\qquad
\ell_f(v)=\int_Ufv.
$$
The form is bounded and coercive:
$$
a(u,u)=\|\nabla u\|_2^2+m^2\|u\|_2^2
\geq\min(1,m^2)\|u\|_{H^1}^2.
$$
The functional $\ell_f$ is bounded by the <Cauchy-Schwarz inequality>. The <Lax-Milgram theorem> therefore gives a unique weak solution and a bound
$$
\|Tf\|_{H_0^1}\leq C\|f\|_2.
$$
Thus $T:L^2(U)\to H_0^1(U)$ is bounded. Composing it with the compact <Rellich-Kondrashov compactness theorem for H01> embedding
$$
H_0^1(U)\hookrightarrow L^2(U)
$$
shows that $T:L^2(U)\to L^2(U)$ is the <compact massive-Laplacian resolvent>.

Solved by gpt-5.6-sol high.

= b
{parent=22f}
{scope}

= Solution
{parent=b}

Extend $u_k$ and $u$ by zero outside the bounded set $U$. Weak convergence makes $(u_k)$ bounded in $L^2(U)$. For each fixed $p$,
$$
\widehat u_k(p)=\int_Uu_k(x)e^{-ip\cdot x}\,dx
\longrightarrow
\int_Uu(x)e^{-ip\cdot x}\,dx
=\widehat u(p),
$$
because $e^{-ip\cdot x}\in L^2(U)$.

Fix $t>0$. On the finite-measure ball $|p|\leq t$, the Fourier transforms are uniformly bounded by
$$
|\widehat u_k(p)-\widehat u(p)|
\leq |U|^{1/2}\|u_k-u\|_2.
$$
Dominated convergence therefore gives
$$
\int_{|p|\leq t}|\widehat u_k-\widehat u|^2\,dp\to0.
$$
On the complementary region,
$$
|\widehat u_k-\widehat u|^2
\leq2(|\widehat u_k|^2+|\widehat u|^2),
$$
whose integral tends to zero uniformly in $k$ as $t\to\infty$ by hypothesis. Splitting first at large $t$ and then taking $k\to\infty$, the <Plancherel theorem> yields
$$
\boxed{\|u_k-u\|_{L^2(U)}\to0.}
$$
This is the same low-frequency compactness mechanism used in the <Fourier proof of Rellich compactness>.

Solved by gpt-5.6-sol high.

= c
{parent=22f}
{scope}

= Solution
{parent=c}

Taking the <Fourier transform> of the distributional equation gives
$$
(1+|p|^{4044})\widehat u(p)=\widehat f(p),
$$
because $2022$ is even. The multiplier therefore gives the <regularity gain for one plus an even power of the Laplacian>
$$
u\in H^{r+4044}(\mathbb R^n).
$$

The <Sobolev embedding theorem> in its derivative form says that
$$
H^s(\mathbb R^n)\hookrightarrow C^k
\quad\text{when}\quad s>k+\frac n2.
$$
To make all derivatives through order $4044$ continuous, and simultaneously make $f\in H^r$ continuous, it is enough and in general necessary to have
$$
r>\frac n2.
$$
Under this condition the distributional identity is a pointwise identity, so $u$ is a classical solution. Thus
$$
\boxed{n<2r.}
$$
At the borderline $n=2r$, the standard Sobolev embedding does not in general give continuity.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=23F}

The <uniformization theorem> states that every simply connected Riemann surface is conformally equivalent to exactly one of the <Riemann sphere> $\mathbb C_\infty$, the <complex plane> $\mathbb C$, or the open unit disc $\mathbb D$. Equivalently, every connected Riemann surface is a quotient of one of these by a free properly discontinuous group of conformal automorphisms.

For $\mathbb C_\infty$, every nonidentity Möbius transformation has a fixed point. A free action is therefore trivial, so the only surface uniformized by the sphere is
$$
\boxed{\mathbb C_\infty.}
$$

Every conformal automorphism of $\mathbb C$ has the form $z\mapsto az+b$. If $a\ne1$, it has a fixed point, so a group acting freely consists only of translations. A discrete subgroup of $(\mathbb C,+)$ has rank zero, one, or two. The corresponding quotients are
$$
\boxed{\mathbb C,\qquad
\mathbb C/\omega\mathbb Z\cong\mathbb C^*,\qquad
\mathbb C/\Lambda,}
$$
where $\Lambda$ is a rank-two lattice and $\mathbb C/\Lambda$ is a complex torus. This proves the completeness of the <Riemann surfaces uniformized by the complex plane>.

Now let $D\subseteq\mathbb C$ be a domain whose complement contains two distinct points. Its universal cover cannot be the sphere because no nonconstant holomorphic map from a compact Riemann surface has image in the noncompact proper domain $D$. If its universal cover were $\mathbb C$, the covering map followed by $D\hookrightarrow\mathbb C$ would be a nonconstant entire function omitting the two chosen complementary points, contradicting the <Little Picard theorem>. The uniformization theorem therefore leaves only
$$
\boxed{\widetilde D\cong\mathbb D,}
$$
which is the <plane domain with two omitted points is hyperbolic> result.

Finally let $j:\mathbb C^*\hookrightarrow R$ be a holomorphic embedding with $R$ compact. At each puncture zero and infinity, an essential singularity would, by the <Great Picard theorem> in a coordinate chart, force values to be taken repeatedly and contradict injectivity. Thus $j$ extends holomorphically to
$$
J:\mathbb C_\infty\to R.
$$
A generic point of the open image $j(\mathbb C^*)$ has exactly one preimage, so $J$ has degree one. A degree-one holomorphic map between compact Riemann surfaces is a biholomorphism; equivalently, the <Riemann-Hurwitz formula> rules out branching and gives the same conclusion. Therefore the <compact Riemann surface containing an embedded punctured plane> is
$$
\boxed{R\cong\mathbb C_\infty.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24G}

For an <algebraically closed field> $k$, the strong <Hilbert Nullstellensatz> says that every <ideal> $J\subseteq k[x_1,\ldots,x_n]$ satisfies
$$
\boxed{I(V(J))=\sqrt J.}
$$
Its weak form says that $V(J)=\varnothing$ exactly when $J=(1)$. Now let $S=\mathbb C[x_0,\ldots,x_n]$ and let $I\subseteq S$ be a <homogeneous ideal>. The affine zero set of $I$ is the <affine cone> over $V_+(I)$ together with the origin. Hence, when the projective zero set is empty, either
$$
\boxed{I=(1)}
$$
or the affine zero set is exactly the origin. In the latter case the <Hilbert Nullstellensatz> gives
$$
\boxed{\sqrt I=(x_0,\ldots,x_n),}
$$
the <irrelevant ideal of projective space>. Equivalently, some power $(x_0,\ldots,x_n)^N$ is contained in $I$. These are precisely the alternatives in the <Projective Nullstellensatz>.

Let $V\subseteq\mathbb P^3$ be a <smooth quadric surface>. After a <projective linear transformation>, the <Segre embedding> identifies
$$
V\cong\mathbb P^1\times\mathbb P^1.
$$
Choose distinct points $p,q\in\mathbb P^1$. The two fibres
$$
C_p=\mathbb P^1\times\{p\},
\qquad
C_q=\mathbb P^1\times\{q\}
$$
are the promised <disjoint curves on a smooth quadric surface>: each is a smooth <projective line>, and $C_p\cap C_q=\varnothing$. By the <Bézout theorem>, any two nonempty curves in the <projective plane> intersect. An <isomorphism of algebraic varieties> preserves intersections and takes <projective curve>[projective curves] to projective curves, so these disjoint curves prove
$$
\boxed{V\not\cong\mathbb P^2.}
$$

Next let $W$ be a <smooth projective curve> and let $f:W\dashrightarrow Y\subseteq\mathbb P^N$ be a <rational map of projective varieties>[rational map] to a <projective variety>. At a point $P$ outside its initial domain, write
$$
f=[f_0:\cdots:f_N]
$$
using elements of the <function field> $\mathbb C(W)$. The <local ring of a smooth algebraic curve> at $P$ is a <discrete valuation ring>. If $m=\min_i v_P(f_i)$, multiplying every coordinate by a function of valuation $-m$ makes all coordinates regular at $P$ and at least one a unit there. They therefore define a <morphism of algebraic varieties>[morphism] near $P$. Repeating this at the finitely many missing points proves the <extension of a rational map from a smooth projective curve>.

Smoothness is essential. Let $\nu:\mathbb P^1\to C$ be the <normalization of a nodal curve> for a rational nodal cubic $C$. The inverse is a rational map $C\dashrightarrow\mathbb P^1$ on the smooth locus. If it extended to a morphism $g:C\to\mathbb P^1$, then $g\circ\nu$ would agree with the identity on a <Zariski-open set>[dense Zariski-open subset] of $\mathbb P^1$ and hence everywhere. But the two distinct points $a,b\in\mathbb P^1$ above the node would satisfy
$$
a=(g\circ\nu)(a)=g(\nu(a))=g(\nu(b))=(g\circ\nu)(b)=b,
$$
a contradiction. This is the <failure of rational-map extension on a singular curve>.

Finally take the <Fermat cubic curve>
$$
F=x^3+y^3+z^3
$$
and define the <projective hypersurface>
$$
Z=\left\{([x:y:z],[s:t])\in\mathbb P^2\times\mathbb P^1:
sF+txyz=0\right\}.
$$
Because $F$ and $xyz$ have no common <irreducible polynomial>[irreducible factor], the bihomogeneous polynomial $sF+txyz$ is irreducible, so $Z$ is an <algebraic variety>. Let
$$
\pi:Z\longrightarrow\mathbb P^1,
\qquad
([x:y:z],[s:t])\longmapsto[s:t]
$$
be the second <coordinate projection>. Every homogeneous cubic in three variables has a projective zero over $\mathbb C$, so every fibre is nonempty and $\pi$ is a <surjective function>[surjective] <morphism of algebraic varieties>[morphism]. Over $p=[1:0]$ the fibre is the <Fermat cubic curve>, which is smooth and has <genus of a smooth plane curve>[genus]
$$
\frac{(3-1)(3-2)}2=1.
$$
Over $q=[0:1]$ the fibre is
$$
V(xyz)=V(x)\cup V(y)\cup V(z),
$$
the union of exactly three distinct <projective line>[projective lines], hence exactly three <irreducible component>[irreducible components]. This is the <cubic pencil with a triangular member> required by the question.

Solved by gpt-5.6-sol high.

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

= a
{parent=25g}
{scope}

= Solution
{parent=a}

Choose the <orientation of a surface> represented by a smooth <Gauss map> $N:S\to\mathbb S^2$. The <shape operator> at $p$ is
$$
S_p=-dN_p:T_pS\longrightarrow T_pS.
$$
If its <principal curvature>[principal curvatures] are $k_1,k_2$, then by definition
$$
\boxed{K(p)=\det S_p=k_1k_2,
\qquad
H(p)=\frac12\operatorname{tr}S_p=\frac{k_1+k_2}{2}.}
$$

Let $E(x)=Ax+b$ be a <proper Euclidean motion of Euclidean three-space>, so $A\in SO(3)$. Orient $\widetilde S=E(S)$ by transporting the normal:
$$
\widetilde N(E(p))=AN(p).
$$
The <chain rule> gives the orthogonal conjugacy
$$
\widetilde S_{E(p)}
=A S_pA^{-1}.
$$
The <determinant> and <matrix trace>[trace] are invariant under <similar matrix>[similarity], so the <Euclidean invariance of the shape operator> yields
$$
\boxed{\widetilde K(E(p))=K(p),
\qquad
\widetilde H(E(p))=H(p).}
$$

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the <orientation reversal of surface curvature> says
$$
\boxed{K\text{ is independent of orientation},
\qquad H\text{ changes sign}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=25g}
{scope}

= Solution
{parent=b}

A <plane> has both <principal curvature>[principal curvatures] zero, hence $K=H=0$. A <circular cylinder> of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, hence
$$
K=0,
\qquad
|H|=\frac1{2R}>0.
$$
Because the absolute value of <mean curvature> is preserved even if one reverses the chosen orientation, no <Euclidean motion> can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv0$ does not characterize planar pieces.

There are analogous examples for both signs of constant <Gaussian curvature>. Start with the arc-length <surface of revolution>
$$
X_A^+(u,v)
=\bigl(A\cos u\cos v,A\cos u\sin v,Z_A(u)\bigr),
\qquad
(Z_A')^2=1-A^2\sin^2u.
$$
On a sufficiently small interval about $u=0$, this is a regular surface and the <curvatures of an arc-length surface of revolution> give
$$
K=-\frac{(A\cos u)''}{A\cos u}=1.
$$
At $u=0$ its <mean curvature> is
$$
H_A^+(0)=\frac12\left(A+\frac1A\right).
$$
For $A=1$ this is a piece of the <unit sphere> and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, use
$$
X_A^-(u,v)
=\bigl(A\cosh u\cos v,A\cosh u\sin v,W_A(u)\bigr),
\qquad
(W_A')^2=1-A^2\sinh^2u,
$$
again on a sufficiently small interval. Then
$$
K=-\frac{(A\cosh u)''}{A\cosh u}=-1,
\qquad
H_A^-(0)=\frac12\left(-A+\frac1A\right).
$$
The choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The <constant Gaussian curvature surfaces of revolution> therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

Solved by gpt-5.6-sol high.

= c
{parent=25g}
{scope}

= Solution
{parent=c}

At $p$, the surface $\widetilde S$ cannot cross the boundary $S=\partial\mathcal R$, because it lies inside $\mathcal R$. Consequently the two surfaces are tangent at $p$. Apply a <proper Euclidean motion of Euclidean three-space> so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward <unit normal> is $N=e_3$. Locally write
$$
S:\ z=f(x,y),
\qquad
\widetilde S:\ z=g(x,y),
$$
with the region $\mathcal R$ on the side $z\geq f(x,y)$. Then
$$
g\geq f,
\qquad
g(0)=f(0),
\qquad
\nabla g(0)=\nabla f(0)=0.
$$
Thus $g-f$ has a <local minimum> at zero and its <Hessian matrix> is <positive semidefinite matrix>[positive semidefinite]. The <fundamental forms of a graph surface> at a horizontal tangent plane give
$$
H(0)=\frac12\operatorname{tr}\operatorname{Hess}f(0),
\qquad
\widetilde H(0)=\frac12\operatorname{tr}\operatorname{Hess}g(0).
$$
Taking the trace of the positive-semidefinite difference proves the <mean-curvature comparison at tangential contact>:
$$
\boxed{\widetilde H(p)\geq H(p).}
$$

There is no analogous conclusion for $K$. In the same local coordinates take
$$
f(x,y)=-x^2-y^2,
\qquad
g(x,y)=-\frac12(x^2+y^2).
$$
Then $g\geq f$, so the second graph lies on the inward side of the first, while the <Gaussian curvature of a graph surface> gives
$$
H_f(0)=-2,
\qquad H_g(0)=-1,
\qquad
K_f(0)=4,
\qquad K_g(0)=1.
$$
Hence the mean-curvature inequality holds but
$$
\boxed{\widetilde K(p)<K(p).}
$$
Using smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the <Gaussian curvature has no tangential-contact comparison principle>[counterexample] under the global hypotheses.

Solved by gpt-5.6-sol high.

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

= a
{parent=26k}
{scope}

= Solution
{parent=a}

Let $T$ be a <measure-preserving transformation> of $(X,\mathcal F,\mu)$ and let $f\in L^1(\mu)$. The <Birkhoff ergodic theorem> states that
$$
A_nf:=\frac1n\sum_{j=0}^{n-1}f\circ T^j
\longrightarrow
f^*:=\mathbb E[f\mid\mathcal I]
$$
almost everywhere, where $\mathcal I$ is the <invariant sigma-algebra>. The limit is $T$-invariant and
$$
\int_Xf^*,d\mu=\int_Xf,d\mu.
$$

Assume now that $\mu(X)=1$. For $M>0$, let
$$
f_M=(-M)\vee(f\wedge M)
$$
be the bounded truncation of $f$, and let $f_M^*$ be its Birkhoff limit. Since $|A_nf_M-f_M^*|\leq2M$ on a <finite measure> space, the <dominated convergence theorem> gives
$$
\|A_nf_M-f_M^*\|_1\longrightarrow0.
$$
Measure preservation and the <triangle inequality> give the $L^1$ contraction
$$
\|A_nh\|_1
\leq\frac1n\sum_{j=0}^{n-1}\|h\circ T^j\|_1
=\|h\|_1.
$$
Moreover, the almost-everywhere convergence and <Fatou lemma> imply
$$
\|f^*-f_M^*\|_1
\leq\liminf_{n\to\infty}\|A_n(f-f_M)\|_1
\leq\|f-f_M\|_1.
$$
Consequently
$$
\limsup_{n\to\infty}\|A_nf-f^*\|_1
\leq2\|f-f_M\|_1.
$$
Since $f_M\to f$ in $L^1$, letting $M\to\infty$ proves the <L1 convergence in the Birkhoff ergodic theorem on a finite measure space>:
$$
\boxed{A_nf\longrightarrow f^*\quad\hbox{in }L^1(\mu).}
$$

Solved by gpt-5.6-sol high.

= b
{parent=26k}
{scope}

= Solution
{parent=b}

Let $T(x)=x+\alpha\pmod1$ with irrational $\alpha$. This <irrational rotation of the circle> preserves <Lebesgue measure> and is ergodic. Apply the <Birkhoff ergodic theorem> to the <indicator function> $\mathbf1_A$. Its limit
$$
\mathbb E[\mathbf1_A\mid\mathcal I]
$$
is invariant, hence almost everywhere constant by <ergodic measure-preserving transformation>[ergodicity]. Its integral must equal that of $\mathbf1_A$, so the constant is $\mu(A)$. Therefore
$$
\boxed{
\frac{S_n(\mathbf1_A)(x)}n
=\frac1n\sum_{j=0}^{n-1}\mathbf1_A(T^jx)
\longrightarrow\mu(A)
}
$$
for <Lebesgue measure>-almost every $x\in(0,1]$.

Solved by gpt-5.6-sol high.

= c
{parent=26k}
{scope}

= Solution
{parent=c}

Write the interval as $I=(c,d]$, where $0<c<d<1$, to avoid confusing its left endpoint with the rotation parameter $\alpha$. For every sufficiently large integer $k$, define an inner and outer interval on the circle by
$$
I_k^-=(c+k^{-1},d-k^{-1}],
\qquad
I_k^+=(c-k^{-1},d+k^{-1}].
$$
Apply part (b) simultaneously to this countable collection of intervals. The intersection $G$ of the corresponding full-measure sets still has full <Lebesgue measure>, and is therefore <dense subset>[dense] in the circle.

Fix an arbitrary $x$. For each sufficiently large $k$, choose $y_k\in G$ with circle distance less than $1/k$ from $x$. An <irrational rotation of the circle>[irrational circle rotation] preserves this distance, so for every $j\geq0$,
$$
\mathbf1_{I_k^-}(T^jy_k)
\leq\mathbf1_I(T^jx)
\leq\mathbf1_{I_k^+}(T^jy_k).
$$
Averaging and using part (b) for $y_k$ gives
$$
d-c-\frac2k
\leq
\liminf_{n\to\infty}\frac{S_n(\mathbf1_I)(x)}n
\leq
\limsup_{n\to\infty}\frac{S_n(\mathbf1_I)(x)}n
\leq
d-c+\frac2k.
$$
Letting $k\to\infty$ proves the <everywhere interval frequency under an irrational rotation>:
$$
\boxed{
\frac{S_n(\mathbf1_{(c,d]})(x)}n\longrightarrow d-c
\quad\text{for every }x\in(0,1].}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=27j}
{scope}

= Solution
{parent=a}

In Kendall notation, the first two letters $M$ mean <Poisson process>[Poisson arrivals] and <exponential distribution>[exponential service times]. An <M-M-1 queue> has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitions
$$
n\to n+1\text{ at rate }\lambda,
\qquad
n\to n-1\text{ at rate }\mu\quad(n\geq1).
$$
It has a <stationary distribution of an M-M-1 queue> exactly when
$$
\rho:=\frac\lambda\mu<1,
$$
and then
$$
\boxed{\pi_n=(1-\rho)\rho^n,
\qquad n\geq0.}
$$

An <M-M-infinity queue> has infinitely many servers, so every customer begins service immediately. Its occupancy transitions are
$$
n\to n+1\text{ at rate }\lambda,
\qquad
n\to n-1\text{ at rate }n\mu.
$$
The <detailed balance for a birth-death process> equations give
$$
\pi_{n+1}=\pi_n\frac{\lambda}{(n+1)\mu}.
$$
Thus for every $\lambda,\mu>0$ the <stationary distribution of an M-M-infinity queue> exists and is
$$
\boxed{\pi_n=e^{-\rho}\frac{\rho^n}{n!},
\qquad \rho=\frac\lambda\mu,}
$$
the <Poisson distribution> of mean $\rho$. The <Burke theorem for an M-M-infinity queue> states that in this stationary regime the departure process is a <Poisson process> of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

Solved by gpt-5.6-sol high.

= b
{parent=27j}
{scope}

= Solution
{parent=b}

Let $X_t$ be the number of telephone lines in use. The <memorylessness of the exponential distribution> makes $(X_t)_{t\geq0}$ a <continuous-time Markov chain>, specifically the <M-M-infinity queue> with generator rates
$$
q_{n,n+1}=\lambda,
\qquad
q_{n,n-1}=n\mu,
\qquad
q_{n,n}=-(\lambda+n\mu).
$$

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the <exponential survival probability> $p_t$, independently of the others. Their surviving number is therefore
$$
Y_t\sim\operatorname{Binomial}(n,p_t).
$$
A call arriving at time $u\in[0,t]$ survives until $t$ with probability $e^{-\mu(t-u)}$. By <Poisson thinning>, the number of surviving new calls is independent of $Y_t$ and has law
$$
Z_t\sim\operatorname{Poisson}\!\left(
\lambda\int_0^te^{-\mu(t-u)}\,du
\right)
=\operatorname{Poisson}\!\left(
\frac\lambda\mu(1-p_t)
\right).
$$
Hence $X_t=Y_t+Z_t$ in distribution. Multiplying the <probability generating function>[probability generating functions] of the two <independent random variables> gives, for $-1\leq s\leq1$,
$$
\boxed{
\mathbb E[s^{X_t}]
=(1-p_t+p_ts)^n
\exp\!\left\{
\frac\lambda\mu(1-p_t)(s-1)
\right\}.}
$$
Equivalently, the <transient distribution of an M-M-infinity queue> is
$$
\boxed{
X_t\overset{d}{=}
\operatorname{Binomial}(n,e^{-\mu t})
+\operatorname{Poisson}\!\left(
\frac\lambda\mu(1-e^{-\mu t})
\right),}
$$
with independent summands. As $t\to\infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda/\mu$. Therefore
$$
\boxed{X_t\xrightarrow d\operatorname{Poisson}(\lambda/\mu),}
$$
which proves the stated large-time approximation.

Solved by gpt-5.6-sol high.

= c
{parent=27j}
{scope}

= Solution
{parent=c}

For an <M-M-1 queue>, the first service lasts on average $1/\mu$. During it, the mean number of arrivals is $\lambda/\mu$, and each arrival initiates another statistically identical busy-period family. Thus, when $\lambda<\mu$,
$$
\mathbb EB
=\frac1\mu+\frac\lambda\mu\mathbb EB,
$$
so the <mean busy period of an M-M-1 queue> is
$$
\boxed{\mathbb EB_{M/M/1}=\frac1{\mu-\lambda}.}
$$
For $\lambda=\mu$ the busy period is almost surely finite but has infinite mean, while for $\lambda>\mu$ it is infinite with positive probability; in either case its expectation is infinite.

For an <M-M-infinity queue>, stationary occupancy is <Poisson distribution>[Poisson] with mean $\rho=\lambda/\mu$, so the long-run probability of an empty system is
$$
\pi_0=e^{-\rho}.
$$
The process alternates between idle periods and busy periods. An idle period waits for the next <Poisson process>[Poisson arrival], so its mean is $1/\lambda$. The <renewal-reward theorem>, with reward equal to time spent empty, gives
$$
e^{-\rho}
=\frac{1/\lambda}{1/\lambda+\mathbb EB}.
$$
Solving yields the <mean busy period of an M-M-infinity queue>:
$$
\boxed{
\mathbb EB_{M/M/\infty}
=\frac{e^{\lambda/\mu}-1}{\lambda}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Let $f_i$ be the <multivariate normal density> of $N_p(\mu_i,\Sigma_i)$. Under zero-one loss, the <Bayes classifier> chooses the class with the larger posterior probability. Thus
$$
\boxed{
\delta_{\pi_0}(x)
=\mathbf1\{\pi_1f_1(x)\geq\pi_0f_0(x)\},
\qquad \pi_1=1-\pi_0.}
$$
Equivalently, it chooses class one when the log posterior odds
$$
\begin{aligned}
D_{\pi_0}(x)
={}&\log\frac{\pi_1}{\pi_0}
+\frac12\log\frac{\det\Sigma_0}{\det\Sigma_1}\\
&-\frac12(x-\mu_1)^T\Sigma_1^{-1}(x-\mu_1)
+\frac12(x-\mu_0)^T\Sigma_0^{-1}(x-\mu_0)
\end{aligned}
$$
is nonnegative. The decision boundary is $D_{\pi_0}(x)=0$.

If $\Sigma_0=\Sigma_1=\Sigma$, the $x^T\Sigma^{-1}x$ terms cancel, leaving
$$
D_{\pi_0}(x)
=(\mu_1-\mu_0)^T\Sigma^{-1}x+C,
$$
an affine function. This is the linear boundary of <linear discriminant analysis>. If $\Sigma_0\ne\Sigma_1$, the quadratic part is
$$
\frac12x^T(\Sigma_0^{-1}-\Sigma_1^{-1})x,
$$
which is nonzero, so the boundary is a possibly degenerate quadratic hypersurface, as in <quadratic discriminant analysis>. This is the <Gaussian Bayes classifier>.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

Condition on the observed value $X=x$. If a classifier reports class zero, its conditional probability of error is $\mathbb P(Y=1\mid X=x)$; if it reports class one, the conditional error is $\mathbb P(Y=0\mid X=x)$. Therefore the smaller conditional error is attained by class one exactly when
$$
\pi_1f_1(x)>\pi_0f_0(x),
$$
and by class zero exactly when the reverse strict inequality holds. Integrating these pointwise conditional errors proves that $\delta_{\pi_0}$ is a <Bayes classifier>.

The only possible nonuniqueness lies on the tie set
$$
B=\{x:\pi_1f_1(x)=\pi_0f_0(x)\}
=\{x:D_{\pi_0}(x)=0\}.
$$
Because the two Gaussian class distributions are distinct, $D_{\pi_0}$ is not identically zero. It is a nonzero polynomial of degree at most two, so its zero set has <Lebesgue measure> zero. Both Gaussian laws have densities with respect to Lebesgue measure, hence assign probability zero to $B$. Every Bayes rule must therefore agree with $\delta_{\pi_0}$ almost surely. This proves the <uniqueness of a Bayes classifier> up to the standard null-set equivalence of decision rules.

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

For $0<\pi<1$, let $\delta_\pi$ be the <Bayes classifier> with class-zero prior $\pi$, and define its two class-conditional error probabilities by
$$
e_0(\pi)=\mathbb P_0(\delta_\pi(X)=1),
\qquad
e_1(\pi)=\mathbb P_1(\delta_\pi(X)=0).
$$
Writing the <likelihood ratio> as $L=f_1/f_0$, the rule chooses class one when
$$
L(x)\geq\frac\pi{1-\pi}.
$$
The Gaussian likelihood-ratio level sets have probability zero, so <dominated convergence theorem>[dominated convergence] shows that $e_0$ and $e_1$ depend continuously on $\pi$. As $\pi\downarrow0$, the threshold tends to zero and the classifier chooses class one almost surely, giving
$$
(e_0(\pi),e_1(\pi))\longrightarrow(1,0).
$$
As $\pi\uparrow1$, it chooses class zero almost surely, giving the opposite limit $(0,1)$. The <intermediate value theorem> therefore provides $\pi^*\in(0,1)$ such that
$$
e_0(\pi^*)=e_1(\pi^*)=:r.
$$

The rule $\delta_{\pi^*}$ is an <equalizer rule> with worst-case risk $r$. For any classifier $\delta$,
$$
\begin{aligned}
\max\{R_0(\delta),R_1(\delta)\}
&\geq \pi^*R_0(\delta)+(1-\pi^*)R_1(\delta)\\
&\geq \pi^*e_0(\pi^*)+(1-\pi^*)e_1(\pi^*)=r,
\end{aligned}
$$
because $\delta_{\pi^*}$ minimizes the integrated risk for its prior. Hence $\delta_{\pi^*}$ is a <minimax decision rule>, as summarized by the <minimax Gaussian Bayes classifier from equal class errors>.

The prior is indeed <least favorable prior>[least favorable]. Its <Bayes risk> is $r$, while for any other prior $q$ the Bayes risk is at most the integrated risk of the same equalizer rule $\delta_{\pi^*}$, namely
$$
q r+(1-q)r=r.
$$
Thus no prior has larger Bayes risk.

Solved by gpt-5.6-sol high.

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

= a
{parent=29k}
{scope}

= Solution
{parent=a}

For every finite collection of positive times, $(B_{t_1},\ldots,B_{t_m})$ is a linear transformation of the corresponding values of $W$, so $B$ is a centered <Gaussian process>. If $0<s\leq t$, then
$$
\mathbb E[B_sB_t]
=st\,\mathbb E[W_{1/s}W_{1/t}]
=st\min\!\left(\frac1s,\frac1t\right)
=s
=\min(s,t).
$$
Thus $B$ has the covariance of <Brownian motion>.

Its paths are continuous for $t>0$. As $t\downarrow0$, put $u=1/t$. The given almost-sure limit gives
$$
B_t=tW_{1/t}=\frac{W_u}{u}\longrightarrow0=B_0.
$$
The <Gaussian-process characterization of Brownian motion> now proves the <time inversion of Brownian motion>:
$$
\boxed{(B_t)_{t\geq0}\text{ is a standard Brownian motion}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

Use the time-inverted Brownian motion $B_u=uW_{1/u}$. For $s>0$, set $u=1/s$. Then
$$
W_s-as\leq b
\quad\Longleftrightarrow\quad
\frac{W_s}{s}-a\leq\frac bs
\quad\Longleftrightarrow\quad
B_u-bu\leq a.
$$
As $0<s\leq t$ corresponds exactly to $u\geq1/t$, and the inequality at $s=0$ is automatic because $b>0$, this is the pathwise identity
$$
\left\{\sup_{0\leq s\leq t}(W_s-as)\leq b\right\}
=\left\{\sup_{u\geq1/t}(B_u-bu)\leq a\right\}.
$$
Since $B$ and $W$ have the same law by <time inversion of Brownian motion>,
$$
\boxed{
\mathbb P\!\left(\sup_{0\leq s\leq t}(W_s-as)\leq b\right)
=\mathbb P\!\left(\sup_{u\geq1/t}(W_u-bu)\leq a\right).}
$$

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Let $X_s=W_s-as$ and $M_t=\sup_{0\leq s\leq t}X_s$. Split the crossing event according to its endpoint:
$$
\mathbb P(M_t>b)
=\mathbb P(X_t>b)
+\mathbb P(M_t>b,X_t\leq b).
$$
Under the <Cameron-Martin theorem for a linear drift>, the law of $X$ relative to standard Wiener measure has endpoint density
$$
\exp\!\left(-aW_t-\frac12a^2t\right).
$$
On paths that cross $b$ and end at $x\leq b$, the <Brownian reflection principle> reflects the path after its first hit of $b$ and changes the endpoint to $2b-x\geq b$. Under this reflection the density becomes
$$
e^{-2ab}
\exp\!\left(aW_t-\frac12a^2t\right).
$$
The remaining exponential is the Cameron--Martin density for drift $+a$. Therefore
$$
\mathbb P(M_t>b,X_t\leq b)
=e^{-2ab}\mathbb P(W_t+at\geq b)
=e^{-2ab}\mathbb P(W_t-at\leq-b),
$$
where the last equality uses the symmetry of the <normal distribution>. Subtracting the crossing probability from one gives
$$
\boxed{
\mathbb P\!\left(\sup_{0\leq s\leq t}(W_s-as)\leq b\right)
=\mathbb P(W_t-at\leq b)
-e^{-2ab}\mathbb P(W_t-at\leq-b).}
$$
Equivalently, in terms of the <standard normal distribution function>,
$$
\boxed{
\Phi\!\left(\frac{b+at}{\sqrt t}\right)
-e^{-2ab}\Phi\!\left(\frac{at-b}{\sqrt t}\right).}
$$
This is the <finite-horizon maximum of Brownian motion with negative drift> formula.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

By continuity of <Brownian motion>, the event that the last strict crossing of $b$ occurs by time $t$ is
$$
\{T\leq t\}
=\left\{\sup_{u\geq t}(W_u-au)\leq b\right\}.
$$
Apply part (b) with $a$ and $b$ interchanged and its time parameter replaced by $1/t$. This turns the right-hand side into
$$
\mathbb P\!\left(
\sup_{0\leq s\leq1/t}(W_s-bs)\leq a
\right).
$$
Part (c), now with drift $b$, barrier $a$, and horizon $1/t$, gives
$$
\boxed{
\mathbb P(T\leq t)
=\Phi\!\left(a\sqrt t+\frac b{\sqrt t}\right)
-e^{-2ab}\Phi\!\left(\frac b{\sqrt t}-a\sqrt t\right),
\qquad t>0.}
$$
This is the distribution function of the <last passage time above a level for Brownian motion with negative drift>. Taking $t\downarrow0$ gives
$$
\mathbb P(T=0)=1-e^{-2ab},
$$
the probability that the negatively drifted Brownian path never exceeds $b$, consistent with the <infinite-horizon crossing probability for Brownian motion with negative drift>.

Solved by gpt-5.6-sol high.

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

= a
{parent=30e}
{scope}

= Solution
{parent=a}

The endpoints are the two <classical turning point>[classical turning points], determined by $V(x)=E$. Since the potential is even, they are symmetric. For $0\leq E\leq1$, they lie in the linear part of the potential and satisfy $|x|/4=E$; for $E\geq1$, they lie in the outer part and satisfy $\sqrt{|x|}-1=E$. Hence
$$
\boxed{
(a,b)=
\begin{cases}
(-4E,4E),&0\leq E\leq1,\\
(-(E+1)^2,(E+1)^2),&E\geq1.
\end{cases}}
$$
The <two-turning-point WKB quantization condition> uses
$$
\boxed{n=0,1,2,\ldots.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=30e}
{scope}

= Solution
{parent=b}

By the evenness of $V$, the <semiclassical action integral> is twice its integral over the positive half-axis. If $0\leq E\leq1$, then
$$
\begin{aligned}
f(E)
&=2\int_0^{4E}\sqrt{E-\frac x4}\,dx\\
&=8\int_0^E u^{1/2}\,du
=\boxed{\frac{16}{3}E^{3/2}}.
\end{aligned}
$$

If $E\geq1$, split the integral at $x=4$:
$$
f(E)=2\left[
\int_0^4\sqrt{E-\frac x4}\,dx
+\int_4^{(E+1)^2}\sqrt{E+1-\sqrt x}\,dx
\right].
$$
The first integral is
$$
\frac83\left(E^{3/2}-(E-1)^{3/2}\right).
$$
In the second, put $r=\sqrt x$ and then $u=E+1-r$. It becomes
$$
\frac43(E+1)(E-1)^{3/2}
-\frac45(E-1)^{5/2}.
$$
After simplification,
$$
\boxed{
f(E)=
\begin{cases}
\dfrac{16}{3}E^{3/2},&0\leq E\leq1,\\[5pt]
\dfrac{16}{3}E^{3/2}
+\dfrac{16}{15}(E-1)^{5/2},&E\geq1.
\end{cases}}
$$
This is the <action integral for a linear-to-square-root potential>.

The graph begins at the origin, is continuous and strictly increasing, passes through $(1,16/3)$, and tends to infinity. Therefore for every $n\geq0$ and $\varepsilon>0$, the equation
$$
f(E)=\pi\varepsilon\left(n+\frac12\right)
$$
has exactly one solution $E=E_n$ by the <intermediate value theorem> and strict monotonicity.

Solved by gpt-5.6-sol high.

= c
{parent=30e}
{scope}

= Solution
{parent=c}

For fixed $n$, the right-hand side of the quantization equation tends to zero with $\varepsilon$. If
$$
\pi\varepsilon\left(n+\frac12\right)<f(1)=\frac{16}{3},
$$
the unique energy satisfies $E_n<1$, so the first branch of the <action integral for a linear-to-square-root potential> applies exactly. Solving
$$
\frac{16}{3}E_n^{3/2}
=\pi\varepsilon\left(n+\frac12\right)
$$
gives
$$
\boxed{
E_n=\left[
\frac{3\pi\varepsilon}{16}
\left(n+\frac12\right)
\right]^{2/3}.}
$$
Thus this formula is valid whenever
$$
\boxed{
\varepsilon<\frac{16}{3\pi(n+1/2)}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=30e}
{scope}

= Solution
{parent=d}

For $E\to\infty$, the outer part of the <action integral for a linear-to-square-root potential> dominates:
$$
f(E)
=\frac{16}{15}(E-1)^{5/2}
+O(E^{3/2})
\sim\frac{16}{15}E^{5/2}.
$$
The <WKB quantization condition> therefore gives
$$
\frac{16}{15}E_n^{5/2}
\sim\pi\varepsilon n.
$$
Taking the power $2/5$ yields the <asymptotic equivalence>
$$
\boxed{
E_n\sim
\left(\frac{15\pi\varepsilon}{16}\right)^{2/5}n^{2/5}.}
$$
Hence
$$
\boxed{
c=\left(\frac{15\pi\varepsilon}{16}\right)^{2/5},
\qquad \alpha=\frac25.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=31A}

At an <equilibrium point of a dynamical system>[equilibrium], the second equation gives $x=y=:s$. The first equation then gives
$$
(a^2-s)(a-s^2)=0.
$$
Hence the fixed-point branches in the $(a,x)$ plane are
$$
\boxed{x=y=a^2\quad(a\in\mathbb R)}
$$
and
$$
\boxed{x=y=\pm\sqrt a\quad(a\geq0).}
$$
The three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt a$, so the second bifurcation value is
$$
\boxed{a^*=1.}
$$

The <jacobian matrix> at a general point is
$$
J(x,y;a)=
\begin{pmatrix}
y^2-a&-2y(a^2-x)\\
1&-1
\end{pmatrix}.
$$
On the branch $P_0=(a^2,a^2)$ it is triangular, with <eigenvalue>[eigenvalues]
$$
\lambda_1=a^4-a,
\qquad \lambda_2=-1.
$$
Thus $P_0$ is <unstable equilibrium>[unstable] for $a<0$, <asymptotic stability>[asymptotically stable] for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm\sqrt a$. Then
$$
J(P_s)=
\begin{pmatrix}
0&2s^2(1-s^3)\\
1&-1
\end{pmatrix},
$$
whose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt a$ is therefore a <saddle equilibrium> for every $a>0$. The positive branch is a <saddle equilibrium>[saddle] for $0<a<1$ and <asymptotic stability>[asymptotically stable] for $a>1$; close to $a=1$ it is a <stable node>.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitution
$$
X=x-a^2,
\qquad Y=y-a^2
$$
and append $\dot a=0$. The equations become
$$
\dot X=-aX+X(Y+a^2)^2,
\qquad
\dot Y=X-Y,
\qquad
\dot a=0.
$$
The centre subspace is $Y=X$. Seek the <extended centre manifold for a parameter> as
$$
Y=h(X,a)
=X+AX^2+BXa+Ca^2+O(3).
$$
The <centre-manifold invariance equation> is
$$
h_X\dot X=X-h.
$$
To second order, $\dot X=-aX+O(3)$, so comparison of coefficients gives
$$
-aX=-AX^2-BXa-Ca^2,
$$
and hence $A=C=0$, $B=1$. Therefore
$$
\boxed{Y=X+aX+O(3).}
$$
Substitution into the $X$ equation gives the reduced flow
$$
\boxed{\dot X=-aX+X^3+O(4).}
$$
Its central branch $X=0$ is <unstable equilibrium>[unstable] for $a<0$ and <asymptotic stability>[stable] for $a>0$, while the two nonzero branches $X\sim\pm\sqrt a$ for $a>0$ are <unstable equilibrium>[unstable]. Thus the bifurcation at zero is a <subcritical pitchfork bifurcation> with reversed normal-form parameter $\mu=-a$, exactly as recorded by the <extended centre manifold of the 2023 Cambridge quadratic-product system>.

The complete <bifurcation diagram> is therefore as follows. For $a<0$, only $P_0$ exists and is <unstable equilibrium>[unstable]. For $0<a<1$, $P_0$ is <asymptotic stability>[stable] while both $P_+$ and $P_-$ are saddles. For $a>1$, $P_+$ is <asymptotic stability>[stable] while $P_0$ and $P_-$ are saddles. At $a=1$, $P_0$ and $P_+$ cross and exchange stability, so the bifurcation is <transcritical bifurcation>[transcritical]. This gives the <bifurcation diagram of the 2023 Cambridge quadratic-product system>.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt a,\sqrt a)$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt a,\sqrt a)$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional <stable manifold>[stable and unstable manifolds]; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

Solved by gpt-5.6-sol high.

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

= a
{parent=32e}
{scope}

= Solution
{parent=a}

Let $\epsilon$ be the group parameter. The <infinitesimal generator of a Lie point symmetry>
$$
V=t\partial_t+x\partial_x
$$
has flow equations
$$
\frac{dT}{d\epsilon}=T,
\qquad
\frac{dX}{d\epsilon}=X,
\qquad
\frac{dU}{d\epsilon}=0.
$$
Therefore it generates the scaling group
$$
\boxed{(t,x,u)\longmapsto(e^\epsilon t,e^\epsilon x,u).}
$$

Here $\xi^t=t$, $\xi^x=x$, and $\eta=0$. The total-derivative formulas for the <second prolongation of a Lie point symmetry> give
$$
\eta^t=-u_t,
\qquad \eta^x=-u_x,
$$
and
$$
\eta^{tt}=-2u_{tt},
\qquad
\eta^{tx}=-2u_{tx},
\qquad
\eta^{xx}=-2u_{xx}.
$$
Thus
$$
\boxed{
\operatorname{pr}^{(2)}V
=t\partial_t+x\partial_x
-u_t\partial_{u_t}-u_x\partial_{u_x}
-2u_{tt}\partial_{u_{tt}}
-2u_{tx}\partial_{u_{tx}}
-2u_{xx}\partial_{u_{xx}}.}
$$
For the <wave equation> expression $F=u_{tt}-u_{xx}$,
$$
\operatorname{pr}^{(2)}V(F)=-2F.
$$
It therefore vanishes whenever $F=0$, proving that $V$ generates the <simultaneous spacetime scaling symmetry of the wave equation>.

The independent invariant of the scaling orbits is the <similarity variable>
$$
z=\frac xt,
$$
while $u$ itself is invariant. Seek a <group-invariant solution> $u(t,x)=f(z)$. Direct <partial derivative>[partial differentiation] gives
$$
u_{tt}=\frac{z^2f''+2zf'}{t^2},
\qquad
u_{xx}=\frac{f''}{t^2}.
$$
The wave equation reduces to
$$
(z^2-1)f''+2zf'=0
\quad\Longleftrightarrow\quad
\frac d{dz}\left((z^2-1)f'\right)=0.
$$
Hence, on any interval avoiding $z=\pm1$,
$$
\boxed{
u(t,x)=C_0+\frac{C_1}{2}
\log\left|\frac{x-t}{x+t}\right|.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=32e}
{scope}

= Solution
{parent=b}

For
$$
V=4t^2\partial_t+4tx\partial_x-(x^2+2t)\partial_u,
$$
the flow equations are
$$
\frac{dT}{d\epsilon}=4T^2,
\qquad
\frac{dX}{d\epsilon}=4TX,
\qquad
\frac{dU}{d\epsilon}=-(X^2+2T),
$$
with $(T,X,U)=(t,x,u)$ at $\epsilon=0$. Solving the first two gives, with $d=1-4\epsilon t$,
$$
T=\frac td,
\qquad X=\frac xd.
$$
Integrating the last equation then gives the one-parameter group
$$
\boxed{
T=\frac{t}{1-4\epsilon t},
\qquad
X=\frac{x}{1-4\epsilon t},
\qquad
U=u-\frac{\epsilon x^2}{1-4\epsilon t}
+\frac12\log(1-4\epsilon t),}
$$
defined locally where $1-4\epsilon t>0$.

To verify the symmetry infinitesimally, use
$$
\xi^t=4t^2,
\qquad \xi^x=4tx,
\qquad \eta=-(x^2+2t).
$$
The needed coefficients of the <second prolongation of a Lie point symmetry> are
$$
\eta^t=-2-8tu_t-4xu_x,
\qquad
\eta^x=-2x-4tu_x,
$$
and
$$
\eta^{xx}=-2-8tu_{xx}.
$$
For the <potential Burgers equation> written as
$$
F=u_t-u_{xx}-u_x^2=0,
$$
we obtain
$$
\begin{aligned}
\operatorname{pr}^{(2)}V(F)
&=\eta^t-\eta^{xx}-2u_x\eta^x\\
&=-8t(u_t-u_{xx}-u_x^2)\\
&=-8tF.
\end{aligned}
$$
Thus the prolonged generator is tangent to the solution manifold $F=0$, and the finite transformations above form the <projective Lie symmetry of the potential Burgers equation>.

Solved by gpt-5.6-sol high.

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

= a
{parent=33b}
{scope}

= Solution
{parent=a}

For distinguishable particles with one-particle <Hilbert space>[Hilbert spaces] $\mathcal H_1,\ldots,\mathcal H_N$, the <multiparticle quantum state> belongs to
$$
\mathcal H_1\otimes\cdots\otimes\mathcal H_N.
$$
A product of one-particle states gives a product vector, and arbitrary linear combinations give entangled states. A noninteracting Hamiltonian is the sum of one-particle Hamiltonians acting on their respective tensor factors, so product-state energies add.

For two identical particles, the <particle exchange operator> $P_{12}$ swaps every degree of freedom. Since $P_{12}^2=I$, its eigenvalues are $\pm1$. Particles whose allowed total states satisfy
$$
P_{12}|\Psi\rangle=+|\Psi\rangle
$$
are <boson>[bosons], while those satisfying
$$
P_{12}|\Psi\rangle=-|\Psi\rangle
$$
are <fermion>[fermions]. Thus bosonic states are symmetric and fermionic states antisymmetric under interchange. The same condition applies to every transposition in a system of more than two <identical particle>[identical particles].

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= Solution
{parent=b}

Spin-one particles are <boson>[bosons], so their total two-particle state must be symmetric. Write $|m,n\rangle=|m\rangle_1|n\rangle_2$ for $m,n\in\{1,0,-1\}$. A basis of symmetric spin states is
$$
|1,1\rangle,\quad |0,0\rangle,\quad |-1,-1\rangle,\quad
\frac{|1,0\rangle+|0,1\rangle}{\sqrt2},\quad
\frac{|1,-1\rangle+|-1,1\rangle}{\sqrt2},\quad
\frac{|0,-1\rangle+|-1,0\rangle}{\sqrt2}.
$$
Denote these six states by $\chi_1^+,\ldots,\chi_6^+$. A basis of antisymmetric spin states is
$$
\chi_1^-=\frac{|1,0\rangle-|0,1\rangle}{\sqrt2},
\quad
\chi_2^-=\frac{|1,-1\rangle-|-1,1\rangle}{\sqrt2},
\quad
\chi_3^-=\frac{|0,-1\rangle-|-1,0\rangle}{\sqrt2}.
$$
This is the six-plus-three decomposition of <two identical spin-one bosons>.

In the ground orbital both particles occupy $n=0$, so the spatial wavefunction
$$
\Phi_{00}(x_1,x_2)=\psi_0(x_1)\psi_0(x_2)
$$
is symmetric. The complete list of ground states is therefore
$$
\boxed{
\Phi_{00}\chi_r^+,
\qquad r=1,\ldots,6,
\qquad E=2E_0.}
$$

At the first excited energy, one particle occupies $n=0$ and the other $n=1$. The normalized symmetric and antisymmetric spatial states are
$$
\Phi_{01}^{\pm}(x_1,x_2)
=\frac{
\psi_0(x_1)\psi_1(x_2)
\pm\psi_1(x_1)\psi_0(x_2)
}{\sqrt2}.
$$
Overall bosonic symmetry permits symmetric spatial times symmetric spin and antisymmetric spatial times antisymmetric spin. Hence all first excited states are
$$
\boxed{
\Phi_{01}^+\chi_r^+\quad(r=1,\ldots,6),
\qquad
\Phi_{01}^-\chi_s^-\quad(s=1,2,3),
\qquad
E=E_0+E_1.}
$$
The ground and first excited degeneracies are therefore six and nine.

Solved by gpt-5.6-sol high.

= c
{parent=33b}
{scope}

= Solution
{parent=c}

Write the equally spaced one-particle spectrum as
$$
E_n=E_0+n\Delta.
$$
The two-particle level labelled by $N$ has energy
$$
E^{(2)}_N=2E_0+N\Delta
$$
and receives contributions from orbital pairs $(n_1,n_2)$ with $n_1+n_2=N$.

For every unequal unordered pair $n_1<n_2$, there is one symmetric and one antisymmetric spatial combination. The symmetric one combines with the six-dimensional symmetric spin space; the antisymmetric one combines with the three-dimensional antisymmetric spin space. Thus each unequal pair contributes nine states.

If $N$ is odd, all pairs are unequal and there are $(N+1)/2$ unordered pairs. Therefore
$$
\boxed{g_N=\frac{9(N+1)}2,
\qquad N\text{ odd}.}
$$

If $N$ is even, there are $N/2$ unequal pairs and the additional equal pair $n_1=n_2=N/2$. The latter has only a symmetric spatial state and therefore contributes six spin states. Hence
$$
\boxed{g_N=9\frac N2+6,
\qquad N\text{ even}.}
$$
This is the <degeneracy of two identical spin-one bosons with equally spaced orbital levels>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=34D}

Let the <Bravais lattice> have primitive vectors $a_1,a_2,a_3$, and put
$$
\Omega=a_1\cdot(a_2\times a_3).
$$
The <reciprocal lattice> has the <basis>
$$
b_1=2\pi\frac{a_2\times a_3}{\Omega},\qquad
b_2=2\pi\frac{a_3\times a_1}{\Omega},\qquad
b_3=2\pi\frac{a_1\times a_2}{\Omega}.
$$
The <scalar triple product> and <cross product> identities give
$$
a_i\cdot b_j=2\pi\delta_{ij}.
$$
Consequently every $q=\sum_jn_jb_j$ satisfies $q\cdot l\in2\pi\mathbb Z$ for every <lattice point> $l=\sum_im_ia_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be <integer>[integers]. This proves that the displayed vectors generate $\Lambda^*$.

In the <Born approximation>, suppose that the crystal potential is the <sum> of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its <Fourier transform> at the <momentum transfer> $Q=k-k'$ factors as
$$
\widetilde V(Q)=\widetilde V_0(Q)\Delta(Q),
\qquad
\Delta(Q)=\sum_{l\in S}e^{iQ\cdot l}.
$$
Thus the single-atom <scattering amplitude> is multiplied by the <crystal lattice structure factor> $\Delta(Q)$. Writing
$$
l=l_1a_1+l_2a_2+l_3a_3,
\qquad
\alpha_i=Q\cdot a_i,
$$
separates the sum into three <finite geometric series>. For $l_i=-L_i/2,\ldots,L_i/2$ this gives
$$
\boxed{
\Delta(Q)=\prod_{i=1}^3
\frac{\sin((L_i+1)\alpha_i/2)}{\sin(\alpha_i/2)}.}
$$
When the $L_i$ are large, a factor has a sharp maximum when $\alpha_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in\Lambda^*$. These are the <reciprocal-lattice peaks of a finite crystal>.

For the stated <body-centered cubic lattice> basis,
$$
a_1=\frac a2(1,1,1),\qquad
a_2=\frac a2(1,-1,1),\qquad
a_3=a(0,0,1),
$$
the reciprocal-basis formula gives
$$
b_1=\frac{2\pi}{a}(1,1,0),\qquad
b_2=\frac{2\pi}{a}(1,-1,0),\qquad
b_3=\frac{2\pi}{a}(-1,0,1).
$$
Equivalently,
$$
\Lambda^*=\frac{2\pi}{a}
\{(h,k,l)\in\mathbb Z^3:h+k+l\text{ is even}\}.
$$
The shortest nonzero reciprocal vectors have squared <Euclidean norm> $2(2\pi/a)^2$, so
$$
q_{\min}=\frac{2\sqrt2\pi}{a}.
$$

In <elastic scattering>, $|k|=|k'|=k$. If $q=k-k'\in\Lambda^*$ and $\theta$ is the angle between $k$ and $k'$, <Euclidean geometry> gives
$$
|q|=2k\sin\frac\theta2.
$$
The first possible <diffraction> peak therefore occurs at
$$
\boxed{
\theta_{\min}=2\arcsin\!\left(\frac{\sqrt2\pi}{ka}\right)
\sim\frac{2\sqrt2\pi}{ka}}
$$
as $ka\to\infty$, by the <small-angle approximation>. This is the <elastic Bragg scattering condition> for the shortest reciprocal-lattice vector.

Solved by gpt-5.6-sol high.

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

= a
{parent=35a}
{scope}

= Solution
{parent=a}

For an ideal gas of identical <boson>[bosons], the <Bose-Einstein distribution> gives the mean occupation of a discrete one-particle state $r$ as
$$
\boxed{n_r=\frac1{e^{\beta(E_r-\mu)}-1}},
\qquad \beta=\frac1{k_BT},
$$
where $\mu$ is the <chemical potential>. If the spectrum is sufficiently dense to use a <density of states> $g(E)$, the discrete <sum> becomes
$$
\boxed{
N=\int_0^\infty
\frac{g(E)}{e^{\beta(E-\mu)}-1}\,dE,
\qquad
E_{\rm tot}=\int_0^\infty
\frac{E g(E)}{e^{\beta(E-\mu)}-1}\,dE.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=35a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Use periodic boundary conditions on the square crystal. Each allowed <wavevector> occupies area $(2\pi)^2/A$ in two-dimensional $k$-space. Including the two <phonon> polarizations, the number of modes in the disk $|k|\leq K$ is
$$
\mathcal N(K)
=2\frac{A}{(2\pi)^2}\pi K^2
=\frac{AK^2}{2\pi}.
$$
The <phonon dispersion relation> $\omega=C|k|^\alpha$ gives $K=(\omega/C)^{1/\alpha}$, so
$$
\mathcal N(\omega)=\frac{A}{2\pi}
\left(\frac\omega C\right)^{2/\alpha}.
$$
Differentiating the cumulative mode count produces the <density of states for a two-dimensional power-law phonon dispersion>:
$$
\boxed{
g(\omega)=B\omega^b,
\qquad
B=\frac{A}{\pi\alpha C^{2/\alpha}},
\qquad
b=\frac2\alpha-1.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The <Debye model> fixes its cutoff by requiring the continuum to contain the $2n$ vibrational degrees of freedom:
$$
2n=\int_0^{\omega_D}g(\omega)\,d\omega
=\frac{A}{2\pi}
\left(\frac{\omega_D}{C}\right)^{2/\alpha}.
$$
Therefore the <Debye frequency> and <Debye temperature> are
$$
\boxed{
\omega_D=C\left(\frac{4\pi n}{A}\right)^{\alpha/2},
\qquad
T_D=\frac{\hbar\omega_D}{k_B}.}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

A <phonon> has zero <chemical potential> and energy $\hbar\omega$. The total <internal energy> is consequently
$$
E_{\rm tot}
=\int_0^{\omega_D}
\frac{\hbar\omega\,g(\omega)}{e^{\beta\hbar\omega}-1}\,d\omega.
$$
With the <change of variables formula>[change of variable] $x=\beta\hbar\omega$ and $\beta\hbar\omega_D=T_D/T$, this becomes
$$
\boxed{
E_{\rm tot}
=\frac{B(k_BT)^{b+2}}{\hbar^{b+1}}
\int_0^{T_D/T}\frac{x^{b+1}}{e^x-1}\,dx,}
\qquad b=\frac2\alpha-1.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

For $\alpha=1/2$, the preceding <density of states> has
$$
b=3,
\qquad
B=\frac{2A}{\pi C^4}.
$$
When $T\gg T_D$, the <Bose-Einstein distribution> has the classical approximation
$$
\frac{\hbar\omega}{e^{\beta\hbar\omega}-1}
\sim k_BT.
$$
Each of the $2n$ modes therefore contributes $k_BT$ to the energy, so the <Dulong-Petit law> gives
$$
E_{\rm tot}\sim2nk_BT,
\qquad
\boxed{C_V\sim2nk_B}.
$$

When $T\ll T_D$, the upper integration limit tends to infinity. The <Bose integral>
$$
\int_0^\infty\frac{x^4}{e^x-1}\,dx
=\Gamma(5)\zeta(5)=24\zeta(5)
$$
then yields
$$
E_{\rm tot}
\sim\frac{24B\zeta(5)k_B^5}{\hbar^4}T^5.
$$
Hence the <heat capacity at constant volume> is
$$
\boxed{
C_V\sim
\frac{120B\zeta(5)k_B^5}{\hbar^4}T^4
=\frac{240A\zeta(5)k_B^5}{\pi C^4\hbar^4}T^4,}
$$
so $d=4$. The high-temperature constant is the classical <equipartition theorem> result, while the low-temperature power law makes $C_V\to0$ as required by the <third law of thermodynamics>.

Solved by gpt-5.6-sol high.

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

= a
{parent=36a}
{scope}

= Solution
{parent=a}

The <electromagnetic four-potential> is the field generated by the four-current $J^\mu$: every source volume element contributes with the inverse-distance Green-function factor. Electromagnetic influence propagates at speed $c$, so <causality> requires the source to be evaluated where its worldline intersects the field event's past light cone. Thus
$$
\boxed{
A^\mu(x,t)=\frac{\mu_0}{4\pi}
\int_{\mathbb R^3}
\frac{J^\mu(x',\,t-|x-x'|/c)}
{|x-x'|}\,d^3x'.}
$$
The source time is exactly the <retarded time> $t_{\rm ret}=t-|x-x'|/c$, rather than the observation time $t$.

Solved by gpt-5.6-sol high.

= b
{parent=36a}
{scope}

= Solution
{parent=b}

Put
$$
F(s)=s+\frac{R(s)}c,
\qquad
R(s)=|x-y(s)|,
\qquad
n(s)=\frac{x-y(s)}{R(s)}.
$$
Because $dR/ds=-n\cdot v$, differentiation gives
$$
F'(s)=1-\frac{n\cdot v}{c}.
$$
A massive charge has the <subluminal speed> $|v|<c$, so $F'(s)>0$. Hence $F$ is <strictly increasing function>[strictly increasing], and $F(s)=t$ has at most one root. This is the <uniqueness of retarded time for a subluminal source>.

Solved by gpt-5.6-sol high.

= c
{parent=36a}
{scope}

= Solution
{parent=c}

For a point charge, the <charge density> and <current density> are
$$
\rho(x',s)=q\delta^{(3)}(x'-y(s)),
\qquad
J(x',s)=qv(s)\delta^{(3)}(x'-y(s)).
$$
Insert these into the <retarded electromagnetic potential>. The argument of the <Dirac delta function> depends on $x'$ through
$$
s=t-\frac{|x-x'|}{c}.
$$
At its unique zero $x'=y(t_{\rm ret})$, the <Jacobian determinant> contributes the factor
$$
\left|1-\frac{n\cdot v}{c}\right|^{-1}.
$$
Since the source is subluminal, this quantity is positive. Writing $\mathbf R=x-y(t_{\rm ret})$, $R=|\mathbf R|$, and evaluating $\mathbf v$ at $t_{\rm ret}$ gives the <Lienard-Wiechert potentials>
$$
\boxed{
\phi(x,t)=\frac{q}{4\pi\epsilon_0}
\frac1{R-\mathbf v\cdot\mathbf R/c},
\qquad
\mathbf A(x,t)=\frac{\mu_0q}{4\pi}
\frac{\mathbf v}{R-\mathbf v\cdot\mathbf R/c}.}
$$
The identity $\mu_0\epsilon_0c^2=1$ equivalently gives $\mathbf A=\mathbf v\phi/c^2$.

Solved by gpt-5.6-sol high.

= d
{parent=36a}
{scope}

= Solution
{parent=d}

Assume that $\mathbf v$ is constant. Set
$$
\tau=t_{\rm ret},\qquad
n=\frac{\mathbf R}{R},\qquad
\beta=\frac{\mathbf v}{c},\qquad
\kappa=1-n\cdot\beta,\qquad
D=R-\beta\cdot\mathbf R=R\kappa.
$$
<Implicit differentiation> of $t=\tau+R(\tau)/c$ gives
$$
\partial_t\tau=\frac1\kappa,
\qquad
\nabla\tau=-\frac{n}{c\kappa}.
$$
Since $d\mathbf R/d\tau=-\mathbf v$, direct differentiation of $D$ yields
$$
\partial_tD=\frac{c(\beta^2-n\cdot\beta)}{\kappa},
\qquad
\mathbf v\cdot\nabla D
=-\frac{c(\beta^2-n\cdot\beta)}{\kappa}.
$$
Therefore
$$
(\partial_t+\mathbf v\cdot\nabla)D=0.
$$
The <Lienard-Wiechert potentials> for uniform motion have $\phi=q/(4\pi\epsilon_0D)$ and $\mathbf A=\mathbf v\phi/c^2$. The preceding identity consequently implies
$$
\partial_t\phi+\mathbf v\cdot\nabla\phi=0.
$$
As $\mathbf v$ is constant,
$$
\boxed{
\frac1{c^2}\frac{\partial\phi}{\partial t}
+\nabla\cdot\mathbf A
=\frac1{c^2}
\left(\partial_t\phi+\mathbf v\cdot\nabla\phi\right)=0.}
$$
Thus the explicit uniformly moving potentials satisfy the <Lorenz gauge>.

Solved by gpt-5.6-sol high.

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

= a
{parent=37b}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The radial first integral for an equatorial <null geodesic> in <Schwarzschild spacetime> is
$$
\dot r^2+\left(1-\frac{2M}{r}\right)\frac{L^2}{r^2}=E^2.
$$
Since $L=r^2\dot\phi$, regarding $r$ as a function of $\phi$ gives
$$
\dot r=\frac{dr}{d\phi}\dot\phi
=\frac{L}{r^2}\frac{dr}{d\phi}.
$$
Set $y=1/r$. Then $dr/d\phi=-y'/y^2$ and therefore $\dot r=-Ly'$. Division by $L^2$ gives the <equatorial null orbit in Schwarzschild spacetime>
$$
\boxed{(y')^2+y^2-2My^3=\frac{E^2}{L^2}=\frac1{b^2},}
$$
where $b=L/E$ is the <impact parameter>. Differentiating with respect to $\phi$ gives
$$
2y'(y''+y-3My^2)=0.
$$
By continuity this yields
$$
\boxed{y''+y=3My^2.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

In the flat-spacetime limit $M=0$, the orbit equation is the <harmonic oscillator equation>
$$
y''+y=0.
$$
Its general solution is $y=A\cos\phi+B\sin\phi$. The conditions $r\to\infty$, hence $y\to0$, at $\phi=0$ and $\phi=\pi$ set $A=0$. Choosing $r=b$ at closest approach $\phi=\pi/2$ fixes $B=1/b$, so
$$
\boxed{y_0(\phi)=\frac{\sin\phi}{b},\qquad
r(\phi)=\frac{b}{\sin\phi}.}
$$
This is a straight line at perpendicular distance $b$ from the origin in <polar coordinates>.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Write
$$
y=y_0+\Delta y,
\qquad
y_0=\frac{\sin\phi}{b},
$$
where $\Delta y$ is first order in $M$. Substitution into $y''+y=3My^2$ and omission of terms of order $M\Delta y$ and $(\Delta y)^2$ gives
$$
\boxed{
\Delta y''+\Delta y
=3M y_0^2
=\frac{3M}{2b^2}(1-\cos2\phi).}
$$
This is the linearized orbit equation used in the <perturbative derivation of Schwarzschild light deflection>.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

The <homogeneous solution> is
$$
\Delta y_{\rm h}=A\cos\phi+B\sin\phi.
$$
For the constant and $\cos2\phi$ forcing terms, a <particular solution> is
$$
\Delta y_{\rm p}
=\frac{3M}{2b^2}+\frac{M}{2b^2}\cos2\phi.
$$
Indeed, applying $d^2/d\phi^2+1$ multiplies the $\cos2\phi$ coefficient by $-3$. Thus
$$
\boxed{
\Delta y
=A\cos\phi+B\sin\phi
+\frac{M}{2b^2}(3+\cos2\phi).}
$$
The $B\sin\phi$ term only changes the definition of $b$ at this order, so one may set $B=0$.

Solved by gpt-5.6-sol high.

= v
{parent=a}
{scope}

= Solution
{parent=v}

Demanding that the incoming ray comes from infinity at $\phi=0$ means $y(0)=0$. The solution from part (iv) then gives
$$
A+\frac{M}{2b^2}(3+1)=0,
\qquad
A=-\frac{2M}{b^2}.
$$
At the outgoing end write $\phi=\pi+\delta\phi$, where $\delta\phi$ is first order in $M$. Then
$$
y_0(\pi+\delta\phi)
=-\frac{\delta\phi}{b}+O(\delta\phi^3),
$$
whereas
$$
\Delta y(\pi)
=-A+\frac{M}{2b^2}(3+1)
=\frac{4M}{b^2}.
$$
The condition $y(\pi+\delta\phi)=0$ therefore gives
$$
\boxed{\delta\phi=\frac{4M}{b},
\qquad |\delta\phi|\simeq\frac{4M}{b}.}
$$
Restoring units gives $4GM/(bc^2)$, the leading <Schwarzschild light deflection>.

Solved by gpt-5.6-sol high.

= b
{parent=37b}
{scope}

= Solution
{parent=b}

An affinely parameterized <geodesic> $x^\mu(\lambda)$ satisfies
$$
\boxed{
\frac{d^2x^\mu}{d\lambda^2}
+\Gamma^\mu_{\alpha\beta}
\frac{dx^\alpha}{d\lambda}
\frac{dx^\beta}{d\lambda}=0.}
$$
For the <conformally flat metric> $g_{\mu\nu}=\varphi^2\eta_{\mu\nu}$, its <Christoffel symbol>[Christoffel symbols] are
$$
\Gamma^\mu_{\alpha\beta}
=\delta^\mu_\alpha\partial_\beta\log\varphi
+\delta^\mu_\beta\partial_\alpha\log\varphi
-\eta_{\alpha\beta}\partial^\mu\log\varphi.
$$
Let $u^\mu=dx^\mu/d\lambda$. Along a <null geodesic>, $\eta_{\alpha\beta}u^\alpha u^\beta=0$, so the equation reduces to
$$
\frac{du^\mu}{d\lambda}
+2\frac{d\log\varphi}{d\lambda}u^\mu=0.
$$
The acceleration is parallel to the tangent. A change of parameter removes it, leaving a straight null line of the <Minkowski metric>. Thus conformal rescaling changes the affine parameter but not the unparametrized light ray.

Consequently <Nordstrom theory of gravitation> predicts
$$
\boxed{\delta\phi=0}
$$
for light passing a star. This contradicts the observed nonzero deflection and the successful general-relativistic value $4GM/(bc^2)$.

Solved by gpt-5.6-sol high.

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

= a
{parent=38c}
{scope}

= Solution
{parent=a}

Let $H$ be a representative positive gap thickness and set
$$
W=|U|+\frac{VL}{H},
$$
the larger-order horizontal speed after including the squeeze-induced flux. The <lubrication-limit scaling for a moving thin gap> requires
$$
\boxed{
\frac HL\ll1,\qquad
\max_{0\leq x\leq L}|h'(x)|\ll1,\qquad
\frac{\rho W H^2}{\mu L}\ll1.}
$$
Equivalently, the inertial condition contains
$$
\frac{\rho|U|H^2}{\mu L}\ll1,
\qquad
\frac{\rho VH}{\mu}\ll1.
$$
The gap must also remain open, $h(x)>0$. These conditions make streamwise derivatives small relative to transverse derivatives, keep the pressure uniform across the gap to leading order, and reduce the <Navier-Stokes equation> to <Stokes flow> locally.

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

The leading streamwise momentum equation in <lubrication theory> is
$$
0=-p_x+\mu u_{yy},
$$
with $p=p(x)$. The <no-slip boundary condition> is
$$
u(x,0)=U,\qquad u(x,h(x))=0.
$$
Twice integrating in $y$ gives the local <Couette-Poiseuille flow in a thin gap>
$$
\boxed{
u(x,y)=U\left(1-\frac yh\right)
+\frac{p_x}{2\mu}y(y-h).}
$$
Its volume flux per unit span is
$$
\boxed{
Q(x)=\int_0^hu\,dy
=\frac{Uh}{2}-\frac{h^3p_x}{12\mu}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= Solution
{parent=c}

Integrating the incompressible <continuity equation> $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the <Reynolds lubrication equation>
$$
\frac{dQ}{dx}=V.
$$
Hence
$$
Q=Vx+C
$$
for a constant $C$, and the flux formula from part (b) gives
$$
p_x=\frac{6\mu U}{h^2}
-\frac{12\mu(Vx+C)}{h^3}.
$$
Define the spatial mean
$$
\langle f\rangle=\frac1L\int_0^Lf(x)\,dx.
$$
Because both ends meet fluid at pressure $p_0$, the <pressure recovery condition in lubrication flow> is
$$
0=p(L)-p(0)=\int_0^Lp_x\,dx.
$$
It determines
$$
C=\frac U2
\frac{\langle h^{-2}\rangle}{\langle h^{-3}\rangle}
-V\frac{\langle xh^{-3}\rangle}{\langle h^{-3}\rangle}.
$$
Therefore the requested <pressure gradient in a translating and squeezing finite gap> is
$$
\boxed{
p_x=\frac{6\mu U}{h^2}
-\frac{12\mu}{h^3}
\left[
Vx+\frac U2
\frac{\langle h^{-2}\rangle}{\langle h^{-3}\rangle}
-V\frac{\langle xh^{-3}\rangle}{\langle h^{-3}\rangle}
\right].}
$$

Solved by gpt-5.6-sol high.

= d
{parent=38c}
{scope}

= Solution
{parent=d}

When $U=0$, put
$$
x_*=\frac{\langle xh^{-3}\rangle}{\langle h^{-3}\rangle}.
$$
The pressure gradient from part (c) reduces to
$$
\boxed{
p_x=\frac{12\mu V}{h^3}(x_*-x).}
$$
Because $h^{-3}>0$, $x_*$ is a <weighted mean> of points in $[0,L]$ and therefore belongs to $[0,L]$; it lies strictly inside unless the weight is singularly concentrated at an endpoint. Thus $p_x(x_*)=0$, proving that the pressure gradient vanishes somewhere between the ends.

Solved by gpt-5.6-sol high.

= e
{parent=38c}
{scope}

= Solution
{parent=e}

The leading <shear stress> exerted by the fluid on the lower surface is
$$
\tau(x)=\mu u_y(x,0)
=-\frac{\mu U}{h}-\frac h2p_x.
$$
The horizontal force per unit span is $\int_0^L\tau\,dx$. Substituting the pressure gradient from part (c) gives
$$
\frac1{\mu L}\int_0^L\tau\,dx
=-4U\langle h^{-1}\rangle
+6V\langle xh^{-2}\rangle
+6C\langle h^{-2}\rangle.
$$
Using the value of $C$ and setting this force to zero yields
$$
U\left(
4\langle h^{-1}\rangle
-3\frac{\langle h^{-2}\rangle^2}{\langle h^{-3}\rangle}
\right)
=6V\left(
\langle xh^{-2}\rangle
-\frac{\langle h^{-2}\rangle\langle xh^{-3}\rangle}
{\langle h^{-3}\rangle}
\right).
$$
Consequently the <zero-shear lower wall in a translating and squeezing finite gap> condition is
$$
\boxed{
U=6V\,
\frac{\displaystyle
\langle xh^{-2}\rangle
-\langle h^{-2}\rangle\langle xh^{-3}\rangle/
\langle h^{-3}\rangle}
{\displaystyle
4\langle h^{-1}\rangle
-3\langle h^{-2}\rangle^2/\langle h^{-3}\rangle}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=39c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Substitute the monochromatic wave
$$
\phi=e^{i(kx-\omega t)}
$$
into the differential equation. Since $(ik)^9=ik^9$, cancellation of the common exponential gives
$$
-i\omega+iUk+\frac{i}{9}k^9=0.
$$
Thus the <dispersion relation> is
$$
\boxed{\omega(k)=Uk+\frac{k^9}{9}.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

From the dispersion relation, the <phase velocity and group velocity> are
$$
\boxed{
c_p(k)=\frac{\omega}{k}=U+\frac{k^8}{9},
\qquad
c_g(k)=\frac{d\omega}{dk}=U+k^8.}
$$
Both are even functions of $k$, both attain their minimum value $U$ at $k=0$, and both increase like $|k|^8$. For every $k\ne0$,
$$
c_g-c_p=\frac89k^8>0.
$$
The graph of $c_g$ therefore lies above the graph of $c_p$ away from their common minimum.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

A <wave crest> moves at the phase velocity, while the envelope of a narrow <wave packet> moves at the group velocity. Since $c_g>c_p$ for every nonzero wavenumber, the crests move more slowly than the packet.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The Fourier modes evolve according to the dispersion relation found in part (a), so
$$
\phi(Vt,t)=\int_{-\infty}^{\infty}
A(k)e^{it\psi(k)}\,dk,
\qquad
\psi(k)=(V-U)k-\frac{k^9}{9}.
$$
For $V>U$, the stationary-point equation is
$$
\psi'(k)=V-U-k^8=0.
$$
It has the two real solutions
$$
k=\pm k_0,
\qquad
k_0=(V-U)^{1/8}.
$$
Define
$$
\Omega=\psi(k_0)=\frac89k_0^9,
\qquad
|\psi''(\pm k_0)|=8k_0^7.
$$
The <one-dimensional stationary-phase formula> gives one contribution from each stationary point:
$$
\boxed{
\phi(Vt,t)\sim
\sqrt{\frac{\pi}{4tk_0^7}}
\left[
A(k_0)e^{i(\Omega t-\pi/4)}
+A(-k_0)e^{-i(\Omega t-\pi/4)}
\right].}
$$
Because the initial field is <real-valued function>[real], its <Fourier transform> has the conjugate symmetry $A(-k)=A(k)^*$. Hence the same result can be written
$$
\boxed{
\phi(Vt,t)\sim
\sqrt{\frac{\pi}{tk_0^7}}\,
\operatorname{Re}\!\left[
A(k_0)e^{i(\Omega t-\pi/4)}
\right].}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

If the initial field is also an <even function>, then $A(k)$ is real and even. Provided $A(k_0)\ne0$, the leading term becomes
$$
\phi(Vt,t)\sim
A(k_0)\sqrt{\frac{\pi}{tk_0^7}}
\cos\left(\Omega t-\frac\pi4\right).
$$
Its large zeros satisfy
$$
\Omega t_n-\frac\pi4
=\left(n+\frac12\right)\pi.
$$
Therefore
$$
\boxed{
t_n\sim\frac{\pi(n+3/4)}{\Omega}
=\frac{9\pi(n+3/4)}{8(V-U)^{9/8}},
\qquad n\to\infty.}
$$
If $A(k_0)=0$, the leading stationary-phase coefficient vanishes and a higher-order term determines the zeros.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For $V<U$,
$$
\psi'(k)=V-U-k^8<0
$$
for every real $k$, so there is no real stationary point. The ray $x=Vt$ travels more slowly than the minimum group velocity $U$ and receives no leading stationary-phase wave packet. Under the usual smoothness and decay assumptions on $A$, the <nonstationary oscillatory integral> can be integrated by parts repeatedly, so
$$
\boxed{\phi(Vt,t)=O(t^{-N})}
$$
for every fixed $N$ for which the needed derivatives are integrable. In particular it is asymptotically much smaller than the $t^{-1/2}$ signal when $V>U$.

Solved by gpt-5.6-sol high.

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

= a
{parent=40c}
{scope}

= Solution
{parent=a}

For
$$
F(x)=\frac12x^TAx-b^Tx,
$$
the gradient is $\nabla F(x)=Ax-b=-r$. The <steepest descent method> with <exact line search for a positive-definite quadratic> therefore uses
$$
x^{(k+1)}=x^{(k)}+\alpha_kr^{(k)},
\qquad
\boxed{\alpha_k=\frac{(r^{(k)})^Tr^{(k)}}{(r^{(k)})^TAr^{(k)}}.}
$$

For $A=\operatorname{diag}(1,\gamma)$ and $b=0$, the minimizer is $x^*=0$. Put
$$
q=\frac{\gamma-1}{\gamma+1}.
$$
Starting from $x^{(0)}=(\gamma,1)^T$, direct substitution gives $\alpha_k=2/(\gamma+1)$ and, inductively,
$$
\boxed{x^{(k)}=q^k(\gamma,(-1)^k)^T.}
$$
The Euclidean norm is unchanged by the alternating sign, so
$$
\frac{\|x^{(k)}-x^*\|_2}{\|x^{(0)}-x^*\|_2}=q^k.
$$
For a <positive-definite matrix>, the <spectral condition number of a positive-definite matrix> is
$$
\kappa_2(A)=\frac{\lambda_{\max}(A)}{\lambda_{\min}(A)}.
$$
Here $\kappa_2(A)=\gamma$, and hence
$$
\boxed{
\frac{\|x^{(k)}-x^*\|_2}{\|x^{(0)}-x^*\|_2}
=\left(\frac{\kappa-1}{\kappa+1}\right)^k.}
$$

The <conjugate gradient method> starts with $p_0=r_0$ and updates
$$
\alpha_k=\frac{r_k^Tr_k}{p_k^TAp_k},\quad
x_{k+1}=x_k+\alpha_kp_k,\quad
r_{k+1}=r_k-\alpha_kAp_k,
$$
$$
p_{k+1}=r_{k+1}
+\frac{r_{k+1}^Tr_{k+1}}{r_k^Tr_k}p_k.
$$
By <finite termination of the conjugate gradient method>, it reaches the exact solution in at most the number of distinct eigenvalues. This matrix has two, so at most
$$
\boxed{2\text{ iterations}}
$$
are required in exact arithmetic.

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= Solution
{parent=b}

Since $\nabla F(x^{(k)})=-r^{(k)}$, the <heavy-ball method> is
$$
x^{(k+1)}
=x^{(k)}+\alpha r^{(k)}
+\beta(x^{(k)}-x^{(k-1)}).
$$
Multiplication by $A$ and use of $Ax^{(j)}=b-r^{(j)}$ give the <heavy-ball residual recurrence>
$$
\boxed{
r^{(k+1)}
=((1+\beta)I-\alpha A)r^{(k)}
-\beta r^{(k-1)}.}
$$
With the usual initialization $x^{(-1)}=x^{(0)}=0$, both initial residuals equal $b$. Induction in the recurrence shows that
$$
r^{(k)}=p_k(A)b
$$
for a polynomial $p_k$ of degree at most $k$. Therefore
$$
\boxed{
r^{(k)}\in
\operatorname{span}\{b,Ab,\ldots,A^kb\}.}
$$
This is $\mathcal K_k(A,b)$ under the zero-indexed <Krylov subspace> convention in the question; under the convention whose order-$j$ space ends at $A^{j-1}b$, it is $\mathcal K_{k+1}(A,b)$.

Solved by gpt-5.6-sol high.

= c
{parent=40c}
{scope}

= Solution
{parent=c}

Because $Ax^*=b$, the error $e^{(k)}=x^*-x^{(k)}$ obeys
$$
e^{(k+1)}
=((1+\beta)I-\alpha A)e^{(k)}
-\beta e^{(k-1)}.
$$
Thus the <heavy-ball error propagation matrix> is
$$
\boxed{
\begin{pmatrix}e^{(k+1)}\\ e^{(k)}\end{pmatrix}
=
M\begin{pmatrix}e^{(k)}\\e^{(k-1)}\end{pmatrix},
\qquad
M=
\begin{pmatrix}
(1+\beta)I-\alpha A&-\beta I\\
I&0
\end{pmatrix}.}
$$

If $A=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$, permute the coordinates from
$$
(e_1^{(k)},\ldots,e_n^{(k)},e_1^{(k-1)},\ldots,e_n^{(k-1)})
$$
to
$$
(e_1^{(k)},e_1^{(k-1)},\ldots,e_n^{(k)},e_n^{(k-1)}).
$$
For the corresponding <permutation matrix> $P$,
$$
\boxed{
PMP^T=\operatorname{diag}(M_1,\ldots,M_n),
\qquad
M_i=
\begin{pmatrix}
1+\beta-\alpha\lambda_i&-\beta\\
1&0
\end{pmatrix}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=40c}
{scope}

= Solution
{parent=d}

Put $s=\gamma^{-1/2}$. Then
$$
\alpha=s^2,
\qquad
\beta=(1-s)^2,
\qquad
\sqrt\beta=1-s.
$$
For an eigenvalue $\lambda$, the corresponding $2\times2$ block has characteristic polynomial
$$
z^2-(1+\beta-\alpha\lambda)z+\beta.
$$
For $\lambda=1$,
$$
1+\beta-\alpha=2(1-s)=2\sqrt\beta,
$$
so the polynomial is
$$
(z-\sqrt\beta)^2.
$$
For $\lambda=\gamma$, it is
$$
z^2-\beta z+\beta.
$$
Its discriminant is $\beta(\beta-4)<0$, so its conjugate roots have product $\beta$ and modulus $\sqrt\beta$. Both blocks therefore have the same <spectral radius>, and
$$
\boxed{\rho(M)=\sqrt\beta=1-\frac1{\sqrt\gamma}.}
$$
This is the <heavy-ball rate for a two-eigenvalue diagonal quadratic>.

By contrast, the steepest-descent factor is
$$
\frac{\gamma-1}{\gamma+1}
=1-\frac{2}{\gamma}+O(\gamma^{-2}).
$$
The heavy-ball factor is $1-\gamma^{-1/2}$, which is substantially smaller for $\gamma\gg1$: it gives an iteration scale of order $\sqrt\gamma$ rather than order $\gamma$.

Solved by gpt-5.6-sol high.