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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3.bigb
= Paper 3
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperii_3_2021.pdf
= 1I
{parent=Paper 3}
{scope}
{title2=Number Theory}
= Solution
{parent=1I}
Starting with $\theta_0=\theta$, define
$$
a_n=\lfloor\theta_n\rfloor,
\qquad
\theta_{n+1}=\frac1{\theta_n-a_n}
$$
while the denominator is nonzero. This gives the <continued fraction>[simple continued fraction]
$$
\theta=[a_0;a_1,a_2,\ldots].
$$
A finite expansion is rational by evaluating it from the bottom. Conversely, for rational $\theta$, these steps are the <Euclidean algorithm> applied to numerator and denominator, so the remainders eventually vanish.
Define convergents by
$$
p_{-2}=0, p_{-1}=1,quad q_{-2}=1, q_{-1}=0,
$$
$$
p_n=a_np_{n-1}+p_{n-2},\qquad
q_n=a_nq_{n-1}+q_{n-2}.
$$
The recurrence gives
$$
p_nq_{n-1}-p_{n-1}q_n
=-(p_{n-1}q_{n-2}-p_{n-2}q_{n-1}),
$$
so induction yields
$$
\boxed{p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}}.
$$
In particular,
$$
\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}\right|
=\frac1{q_nq_{n+1}}.
$$
Since an irrational $\theta$ lies strictly between these convergents, the two approximation errors sum to this distance. If both displayed bounds in the question failed, their sum would be at least
$$
\frac1{2q_n^2}+\frac1{2q_{n+1}^2}
\geq\frac1{q_nq_{n+1}},
$$
by the arithmetic-geometric mean inequality, contradicting strict betweenness. Thus at least one bound holds.
Solved by gpt-5.6-sol high.
= 2H
{parent=Paper 3}
{scope}
{title2=Topics in Analysis}
<Runge theorem>[Runge approximation theorem] in its polynomial form says that if $K\subset\mathbb C$ is compact with connected complement and $f$ is holomorphic on a neighborhood of $K$, then polynomials approximate $f$ uniformly on $K$.
= i
{parent=2h}
{scope}
= Solution
{parent=i}
The assertion is true when “uniform” means on all of the unbounded quadrant $\Omega$. A uniformly convergent sequence $(P_n)$ is uniformly Cauchy. For sufficiently large $m,n$, the polynomial $P_n-P_m$ is bounded on $\Omega$. Every nonconstant polynomial is unbounded along some ray contained in $\Omega$, so $P_n-P_m$ is constant. Fixing one large $m$, the limit is therefore $P_m$ plus the limit of constants, hence is a polynomial.
Solved by gpt-5.6-sol high.
= ii
{parent=2h}
{scope}
= Solution
{parent=ii}
The assertion is true. Exhaust the quadrant by compact sets $K_n$ with connected complements and with every compact subset of $\Omega$ eventually contained in the interior of $K_n$. Runge's theorem gives a polynomial $P_n$ satisfying
$$
\sup_{K_n}|P_n-f|<2^{-n}.
$$
Thus $P_n\to f$ locally uniformly. For a fixed $z\in\Omega$, choose a small circle about $z$ eventually contained in every $K_n$. The <Cauchy derivative formula>[Cauchy integral formula for derivatives] applied to $P_n-f$ gives
$$
P_n^{(r)}(z)\longrightarrow f^{(r)}(z)
$$
for every $r\geq0$.
Solved by gpt-5.6-sol high.
= 3K
{parent=Paper 3}
{scope}
{title2=Coding and Cryptography}
= Solution
{parent=3K}
Let $H$ be the $d\times(2^d-1)$ matrix whose columns are the distinct nonzero vectors of $\mathbb F_2^d$. The <Hamming code>[binary Hamming code] is
$$
C=\ker H.
$$
No column is zero and no two columns agree, so its minimum distance is three. A radius-one Hamming ball contains
$$
1+(2^d-1)=2^d
$$
words, while $|C|=2^{2^d-1-d}$. The radius-one balls about codewords are disjoint and their total size is $2^{2^d-1}$, so they partition the ambient space. Hence $C$ is a <perfectness of a Hamming code>[perfect code].
Let the received word be all ones except in the last coordinate. The sum of all nonzero vectors of $\mathbb F_2^d$ is zero for $d\geq2$, so its syndrome is the last column of $H$. Minimum-distance decoding therefore flips the last bit and returns the all-one word. A Hamming code corrects every single error because each nonzero syndrome identifies its unique erroneous coordinate.
Solved by gpt-5.6-sol high.
= 4F
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}
= Solution
{parent=4F}
A <regular expression> is built from $\emptyset$, $\epsilon$, and alphabet symbols using union, concatenation, and Kleene star; its language is defined by applying the corresponding set operations recursively. A <deterministic finite automaton> is a tuple $(Q,\Sigma,\delta,q_0,F)$ and accepts
$$
L(D)=\{w:\widehat\delta(q_0,w)\in F\}.
$$
<Kleene theorem> says that the languages denoted by regular expressions are exactly those accepted by finite automata.
Regular languages are closed under finite union, using a product automaton or nondeterministic choice, and under finite intersection, using the product automaton with accepting set $F_L\times F_M$.
They are not closed under countable unions or intersections. Every language over a finite alphabet is a countable union of singleton languages, each regular, so a nonregular language such as $\{a^nb^n:n\geq0\}$ is a counterexample. It is also
$$
\bigcap_{w\notin L}(\Sigma^*\setminus\{w\}),
$$
a countable intersection of regular languages, giving the second counterexample.
Solved by gpt-5.6-sol high.
= 5J
{parent=Paper 3}
{scope}
{title2=Statistical Modelling}
= Solution
{parent=5J}
Assume $X$ has full column rank. The log-likelihood, up to constants, is
$$
-\frac n2\log\sigma^2-\frac1{2\sigma^2}\|Y-X\beta\|^2.
$$
The <ordinary least squares estimators>[ordinary least squares] normal equations give
$$
\widehat\beta=(X^TX)^{-1}X^TY,
$$
and maximizing over the variance gives the maximum-likelihood estimator
$$
\widehat\sigma^2=\frac1n\|(I-H)Y\|^2.
$$
Because $I-H$ is an orthogonal projection of rank $n-p$,
$$
\boxed{\frac{n\widehat\sigma^2}{\sigma^2}\sim\chi^2_{n-p}}.
$$
Writing $\chi^2_{\nu,q}$ for the $q$-quantile, a confidence interval of level $1-\alpha$ is
$$
\boxed{\left[
\frac{n\widehat\sigma^2}{\chi^2_{n-p,1-\alpha/2}},
\frac{n\widehat\sigma^2}{\chi^2_{n-p,\alpha/2}}
\right]}.
$$
Solved by gpt-5.6-sol high.
= 6E
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}
= a
{parent=6e}
{scope}
= Solution
{parent=a}
The boundary condition says that all newborn individuals enter at age zero, and that their influx equals the total birth rate obtained by summing the age-specific births $b(a)n(a,t)$ over the population.
Solved by gpt-5.6-sol high.
= b
{parent=6e}
{scope}
= Solution
{parent=b}
Put $n(a,t)=A(a)T(t)$. Separation gives
$$
\frac{T'}T=-\frac{A'}A-d(a)=r
$$
for a constant $r$. Hence
$$
\boxed{T(t)=T_0e^{rt},\qquad
A(a)=A_0\exp\left(-ra-\int_0^ad(s)ds\right)}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=6e}
{scope}
= Solution
{parent=c}
Substitution into the renewal boundary condition and cancellation of $A_0T(t)$ gives the <Euler-Lotka equation>
$$
\boxed{1=\int_0^\infty b(a)
\exp\left(-ra-\int_0^ad(s)ds\right)da}.
$$
This is the necessary condition selecting the growth exponent $r$.
Solved by gpt-5.6-sol high.
= d
{parent=6e}
{scope}
= Solution
{parent=d}
For $b(a)=\beta e^{-\lambda a}$ and constant death rate $d$,
$$
1=\int_0^\infty\beta e^{-(\lambda+d+r)a}da
=\frac\beta{\lambda+d+r}.
$$
Thus
$$
r=\beta-\lambda-d.
$$
A separable solution exists with this exponent, and it grows precisely when
$$
\boxed{\beta>\beta_c=\lambda+d}.
$$
Solved by gpt-5.6-sol high.
= 7E
{parent=Paper 3}
{scope}
{title2=Further Complex Methods}
= a
{parent=7e}
{scope}
= Solution
{parent=a}
The substitution $t\mapsto1-t$ gives $B(p,q)=B(q,p)$. Direct integration gives
$$
\boxed{B(1,q)=\int_0^1(1-t)^{q-1}dt=\frac1q}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=7e}
{scope}
= Solution
{parent=b}
Integration by parts, or the identities obtained by splitting $1=t+(1-t)$, gives
$$
B(p,z+1)=\frac z{p+z}B(p,z).
$$
Therefore
$$
\boxed{(p+z)B(p,z+1)=zB(p,z)}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=7e}
{scope}
= Solution
{parent=c}
For any $z\notin\{0,-1,-2,\ldots\}$, choose $N$ with $\operatorname{Re}(z+N)>0$ and define
$$
B(p,z)=\frac{(p+z)(p+z+1)\cdots(p+z+N-1)}
{z(z+1)\cdots(z+N-1)}B(p,z+N).
$$
The recurrence makes this independent of $N$ and agrees with the original integral where $\operatorname{Re}z>0$. It is the desired analytic continuation away from the listed points.
Solved by gpt-5.6-sol high.
= d
{parent=7e}
{scope}
= Solution
{parent=d}
Using
$$
B(p,z)=\frac{\Gamma(p)\Gamma(z)}{\Gamma(p+z)},
$$
we see that $z=-n$ is generally a simple pole, inherited from $\Gamma(z)$. A cancellation occurs exactly when $p-n$ is a nonpositive integer. Thus $z=0$ is always a simple pole; for $n\geq1$, $z=-n$ is removable when $p$ is one of $1,2,\ldots,n$, and otherwise is a simple pole.
Solved by gpt-5.6-sol high.
= 8D
{parent=Paper 3}
{scope}
{title2=Classical Dynamics}
= Solution
{parent=8D}
The scalar and vector potentials satisfy
$$
\mathbf E=-\nabla\phi-\partial_t\mathbf A,
\qquad
\mathbf B=\nabla\times\mathbf A.
$$
The canonical momentum and Hamiltonian are
$$
\mathbf p=m\dot{\mathbf r}+q\mathbf A,
\qquad
\boxed{H=\frac{|\mathbf p-q\mathbf A|^2}{2m}+q\phi}.
$$
For the uniform fields choose
$$
\mathbf A=(0,Bx,0),\qquad \phi=-Ex,
$$
which produces the stated Hamiltonian. Since it is time independent and $y,z$ are cyclic coordinates, three independent conserved quantities are
$$
\boxed{H,quad p_y,quad p_z}.
$$
Solved by gpt-5.6-sol high.
= 9B
{parent=Paper 3}
{scope}
{title2=Cosmology}
= a
{parent=9b}
{scope}
= Solution
{parent=a}
Insert
$$
a=(t/t_0)^\gamma,\qquad \phi=\phi_0+\alpha\log t.
$$
Matching powers of $t$ requires $\lambda\alpha=2$, so $\alpha=2/\lambda$. Writing $\overline V=V_0e^{-\lambda\phi_0}$, the scalar equation and Friedmann equation give
$$
\lambda\overline V=\alpha(3\gamma-1),
\qquad
\boxed{\gamma=\frac{16\pi G}{\lambda^2}}.
$$
Thus
$$
\overline V=\frac{2(3\gamma-1)}{\lambda^2},
\qquad
\phi_0=\lambda^{-1}\log(V_0/\overline V).
$$
Existence with positive potential requires $\gamma>1/3$, or $\lambda^2<48\pi G$. Accelerated expansion requires $\gamma>1$, hence
$$
\boxed{\lambda^2<16\pi G}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=9b}
{scope}
= Solution
{parent=b}
Multiplying the inflaton equation by $\dot\phi$ gives
$$
\dot\rho=-3H\dot\phi^2.
$$
Comparison with $\dot\rho+3H(\rho+P)=0$ yields $\rho+P=\dot\phi^2$, and therefore
$$
\boxed{P=\frac12\dot\phi^2-V(\phi)}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=9b}
{scope}
= Solution
{parent=c}
The acceleration equation requires $\rho+3P<0$, so $w=P/\rho<-1/3$. A canonical scalar field with nonnegative potential obeys $P\geq-\rho$. Hence inflation occurs for
$$
\boxed{-1\leq w<-\frac13}.
$$
Solved by gpt-5.6-sol high.
= 10D
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}
= Solution
{parent=10D}
Without Alice's operation, Bob obtains outcome $b$ with probability
$$
p(b)=\langle\psi|I_A\otimes|b\rangle\langle b|\psi\rangle.
$$
If Alice applies $U$ and then measures, but her outcome is not communicated, Bob's probability is
$$
\begin{aligned}
p'(b)
&=\sum_a\langle\psi|
U^\dagger\Pi_aU\otimes|b\rangle\langle b|\psi\rangle\\
&=\langle\psi|I_A\otimes|b\rangle\langle b|\psi\rangle=p(b),
\end{aligned}
$$
because $\sum_a\Pi_a=I_A$. Equivalently, a trace-preserving local operation leaves Bob's <reduced density matrix> unchanged. This is the <quantum no-signalling>[no-communication theorem]: local operations cannot signal without communication of their outcomes.
Solved by gpt-5.6-sol high.
= 11I
{parent=Paper 3}
{scope}
{title2=Number Theory}
= Solution
{parent=11I}
Two integral binary quadratic forms are properly equivalent when one is obtained from the other by a change of variables in $SL_2(\mathbb Z)$. For a negative discriminant $d\equiv0,1\pmod4$, the <class number of a negative discriminant>[class number of binary quadratic forms] $h(d)$ is the number of proper equivalence classes of primitive positive-definite forms of discriminant $d$.
If $f$ properly represents $m$, then $f(r,s)=m$ for coprime $r,s$. Extend $(r,s)^T$ to a matrix in $SL_2(\mathbb Z)$; after the associated variable change, the coefficient of $x^2$ is $m$, so the transformed form is
$$
mx^2+bxy+cy^2.
$$
The converse follows by evaluating this form at $(1,0)$.
Such a form has discriminant $d$ precisely when
$$
b^2-4mc=d,
$$
so it exists exactly when $b^2\equiv d\pmod{4m}$. Thus $m$ is properly represented by some form of discriminant $d$ exactly when $d$ is a square modulo $4m$.
Now let $d=1-4A$. If $n^2+n+A$ is composite for $0\leq n\leq A-2$, it has a prime divisor $p<A$, and
$$
d=(2n+1)^2-4(n^2+n+A)\equiv(2n+1)^2\pmod{4p}.
$$
Conversely, if $d$ is a square modulo $4p$ for a prime $p<A$, choose an odd square root $2n+1$ with $0\leq n<p$. Then $p\mid n^2+n+A$ and $n\leq A-2$, so that value is composite.
Finally, reduction theory says that every nonprincipal positive-definite class of discriminant $1-4A$ has a reduced representative whose leading coefficient has a prime divisor $p<A$ for which $d$ is a square modulo $4p$. Conversely such a prime produces a represented form not equivalent to the principal form $x^2+xy+Ay^2$. Combining this with the preceding equivalence gives
$$
\boxed{h(1-4A)=1\iff n^2+n+A\text{ is prime for }0\leq n\leq A-2}.
$$
Solved by gpt-5.6-sol high.
= 12F
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}
= i
{parent=12f}
{scope}
= Solution
{parent=i}
A <parse tree> has the start variable at its root; the ordered children of a variable spell the right side of the production used at that occurrence, and its leaves, read left to right, spell the derived word.
For a grammar in <Chomsky normal form> with $r$ variables, take pumping length $p=2^r$. A parse tree for a word of length at least $p$ has a root-to-leaf path containing one variable twice. The yield can therefore be written
$$
w=uvxyz,
$$
where $|vxy|\leq p$, $|vy|>0$, and replacing the subtree between the repeated variables zero, one, or several times proves
$$
uv^ixy^iz\in L\qquad(i\geq0).
$$
This is the <pumping lemma for context-free languages>. Without Chomsky normal form but with no epsilon or unit productions, use the maximum production length to choose a larger exponential $p$; bounded branching and the same repeated-variable argument apply.
The language $\{a^ib^ia^i:i\geq0\}$ is not context free. A pumped window of bounded length cannot alter all three long blocks equally, so pumping breaks one of the required equalities.
Solved by gpt-5.6-sol high.
= ii
{parent=12f}
{scope}
= Solution
{parent=ii}
The language $\{a^ib^j:i\geq j\geq0\}$ is context free. For example the grammar
$$
S\to aSb\mid A,
\qquad
A\to aA\mid\epsilon
$$
generates $a^{j+k}b^j$ with $j,k\geq0$, and every required word has this form.
Solved by gpt-5.6-sol high.
= iii
{parent=12f}
{scope}
= Solution
{parent=iii}
The language $\{wabw:w\in\{a,b\}^*\}$ is not context free. Intersecting with a suitable regular language that forces both copies into long homogeneous blocks reduces to a marked copy language; the context-free pumping lemma then places the two pumped pieces within only finitely many local blocks and cannot change the two copies identically. Equivalently, the standard copy-language proof applies with the fixed marker $ab$.
Solved by gpt-5.6-sol high.
= 13E
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}
= a
{parent=13e}
{scope}
= Solution
{parent=a}
The term $\lambda_iS_i$ moves susceptible people in group $i$ into its infective class, while $\gamma I_i$ moves infectives into recovery. In
$$
\lambda_i=\beta\sum_jC_{ij}I_j/N_j,
$$
$C_{ij}$ is the contact rate of a person in group $i$ with group $j$, and $I_j/N_j$ is the infectious fraction there. The reciprocity condition $N_iC_{ij}=N_jC_{ji}$ says that the total number of $i$--$j$ contacts agrees when counted from either group. Common $\beta$ and $\gamma$ mean age-independent transmission per infectious contact and recovery rate. There is no transmission between distinct age groups exactly when
$$
C_{ij}=0\quad(i\ne j).
$$
Solved by gpt-5.6-sol high.
= b
{parent=13e}
{scope}
= Solution
{parent=b}
With $S_i\simeq N_i$, the infective equations become
$$
\dot I_i=\beta\sum_j\frac{N_iC_{ij}}{N_j}I_j-\gamma I_i
=\beta\sum_jC_{ji}I_j-\gamma I_i,
$$
where reciprocity was used. For a column vector $\mathbf I$,
$$
\dot{\mathbf I}=\gamma(\mathbf L-\mathbf1)\mathbf I,
\qquad
\boxed{\mathbf L=\frac\beta\gamma\mathbf C^T}.
$$
Therefore
$$
\boxed{\mathbf I(t)=e^{\gamma(\mathbf L-\mathbf1)t}\mathbf I(0)}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=13e}
{scope}
= Solution
{parent=c}
Since $C$ and $C^T$ have the same eigenvalues, an exponentially growing mode exists precisely when $L$ has an eigenvalue with real part greater than one. For the nonnegative contact matrix this is
$$
\boxed{\frac\beta\gamma\rho(C)>1},
$$
where $\rho(C)$ is the <spectral radius>.
Solved by gpt-5.6-sol high.
= 14B
{parent=Paper 3}
{scope}
{title2=Cosmology}
= a
{parent=14b}
{scope}
= Solution
{parent=a}
Substituting $P=\rho/3$ into the continuity equation gives
$$
\dot\rho+4\frac{\dot a}{a}\rho=0,
$$
so
$$
\boxed{\rho=\rho_0a^{-4}}.
$$
Three powers arise from dilution of photon number in physical volume and one from the cosmological redshift of each photon's energy.
Solved by gpt-5.6-sol high.
= b
{parent=14b}
{scope}
= Solution
{parent=b}
The positive-curvature term identifies $R$ as the curvature radius of the spatial three-sphere when the dimensionless scale factor is $a=1$; the physical curvature radius is $Ra(t)$.
Solved by gpt-5.6-sol high.
= c
{parent=14b}
{scope}
= Solution
{parent=c}
Put $a=\alpha\widetilde a$ with
$$
\boxed{\alpha=\frac cR},
\qquad
\boxed{\Gamma=\frac{8\pi G\rho_0R^4}{3c^6}}.
$$
Then
$$
\left(\frac{\dot{\widetilde a}}{\widetilde a}\right)^2
=\frac\Gamma{\widetilde a^4}-\frac1{\widetilde a^2}+\frac\Lambda3.
$$
For $y=\widetilde a^2$ this becomes
$$
\dot y^2=4\Gamma-4y+\frac{4\Lambda}{3}y^2.
$$
The expanding big-bang branch has
$$
y(0)=0,\qquad \dot y(0)=2\sqrt\Gamma.
$$
Differentiation gives
$$
\ddot y=\frac{4\Lambda}{3}y-2.
$$
Thus, with $\omega=\sqrt{4\Lambda/3}$,
$$
\boxed{y(t)=\frac3{2\Lambda}\left[1-\cosh(\omega t)+\lambda\sinh(\omega t)\right]},
$$
where the initial derivative gives
$$
\boxed{\lambda=2\sqrt{\frac{\Lambda\Gamma}{3}}}.
$$
For $\lambda>1$, $\widetilde a$ expands forever and becomes asymptotically exponential. For $\lambda<1$, it reaches a maximum and recollapses to zero. At $\lambda=1$, $y$ increases monotonically to $3/(2\Lambda)$, the limiting static radius.
Solved by gpt-5.6-sol high.
= 15D
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}
= a
{parent=15d}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
Write
$$
|\psi_0\rangle=\frac12\sum_{x\in\mathcal B_2}|x\rangle
=\sin\theta|x_0\rangle+\cos\theta|\beta\rangle,
\qquad \theta=\frac\pi6,
$$
where $|\beta\rangle$ is the normalized component orthogonal to $|x_0\rangle$. On this real plane, $I_{x_0}$ is reflection in the line perpendicular to $|x_0\rangle$, while
$$
-J=2|\psi_0\rangle\langle\psi_0|-I
$$
is reflection in the line spanned by $|\psi_0\rangle$. Their product $Q=-JI_{x_0}$ is a rotation through $2\theta=\pi/3$ toward $|x_0\rangle$. This is the geometric core of <Grover search algorithm>[Grover search].
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
Prepare $|\psi_0\rangle=H^{\otimes2}|00\rangle$. One query to $U_f$, with phase kickback, implements $I_{x_0}$. The reflection $-J$ is independent of $f$ and can be implemented by conjugating the phase flip of $|00\rangle$ with $H^{\otimes2}$. Since $|\psi_0\rangle$ begins at angle $\theta=\pi/6$ from the unmarked direction, one Grover iterate rotates it by $2\theta$ and gives
$$
Q|\psi_0\rangle=|x_0\rangle.
$$
A computational-basis measurement therefore determines $x_0$ with certainty after one oracle query.
Solved by gpt-5.6-sol high.
= b
{parent=15d}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Let $a=\|\Pi_A\xi\|$ and $b=\|\Pi_{A^\perp}\xi\|$. Since $a^2+b^2=1$, choose $\theta\in[0,\pi/2]$ with $a=\sin\theta$, $b=\cos\theta$. Normalizing the two nonzero projections gives
$$
|\xi\rangle=\sin\theta|\alpha\rangle+\cos\theta|\beta\rangle.
$$
At an endpoint, the unused normalized vector may be chosen arbitrarily in its subspace.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Since $I_A|\alpha\rangle=-|\alpha\rangle$,
$$
Q|\alpha\rangle=I_\xi|\alpha\rangle
=|\alpha\rangle-2\sin\theta|\xi\rangle.
$$
Substituting the decomposition of $|\xi\rangle$ and using the double-angle identities gives
$$
\boxed{Q|\alpha\rangle=\cos2\theta|\alpha\rangle-sin2\theta|\beta\rangle}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
The two stated formulas show that $Q$ rotates the $(|\beta\rangle,|\alpha\rangle)$ plane by $2\theta$. Starting from $|\xi\rangle$, whose angle from $|\beta\rangle$ is $\theta$, two iterations give angle
$$
\theta+2(2\theta)=5\theta=\frac\pi2.
$$
Hence $Q^2|\xi\rangle=|\alpha\rangle$. The reflection about $|\xi\rangle$ is implementable as
$$
I_\xi=U(I-2|0\cdots0\rangle\langle0\cdots0|)U^\dagger.
$$
Thus apply $U$ to $|0\cdots0\rangle$, followed by two applications of $-I_\xi I_A$. This prepares $|\alpha\rangle$ exactly and deterministically.
Solved by gpt-5.6-sol high.
= 16G
{parent=Paper 3}
{scope}
{title2=Logic and Set Theory}
= a
{parent=16g}
{scope}
= Solution
{parent=a}
For cardinals, $\kappa<\lambda$ means that there is an injection $\kappa\to\lambda$ but no bijection. Under the <axiom of choice>, every set is well-orderable, so every infinite cardinal is the initial ordinal $\aleph_\alpha$ for a unique ordinal $\alpha$. <Cantor diagonal argument>[Cantor theorem] gives
$$
\aleph_\alpha<2^{\aleph_\alpha},
$$
and minimality of the successor cardinal gives
$$
\aleph_{\alpha+1}\leq2^{\aleph_\alpha}.
$$
The strict inequality $\aleph_{\alpha+1}<2^{\aleph_\alpha}$ printed in the conversion is not provable and can be false under the generalized continuum hypothesis; the intended symbol must be $\leq$.
Solved by gpt-5.6-sol high.
= b
{parent=16g}
{scope}
= Solution
{parent=b}
The map $a\mapsto\{a\}$ injects $x$ into $\mathcal P(x)$. Cantor's diagonal argument shows that no map $x\to\mathcal P(x)$ is surjective: for $f:x\to\mathcal P(x)$, the set $D=\{a:a\notin f(a)\}$ differs from every $f(a)$. Hence $\kappa<2^\kappa$.
If $f:x\to y$ is surjective, inverse image gives an injection
$$
\mathcal P(y)\to\mathcal P(x),
\qquad A\mapsto f^{-1}(A).
$$
For every infinite initial ordinal $\omega_\alpha$, transfinite recursion constructs a pairing of $\omega_\alpha\times\omega_\alpha$ with $\omega_\alpha$: at each stage fewer than $\aleph_\alpha$ earlier pairs have been used. Thus
$$
\aleph_\alpha\aleph_\alpha=\aleph_\alpha
$$
without invoking choice for arbitrary families.
Every ordinal below $\omega_{\alpha+1}$ has cardinal at most $\aleph_\alpha$ and can be coded by a well-ordering relation on a subset of $\omega_\alpha$. Sending such a relation to its order type, and all non-well-orders to zero, gives a surjection
$$
\mathcal P(\omega_\alpha\times\omega_\alpha)\to\omega_{\alpha+1}.
$$
Using the pairing above to code relations as subsets of $\omega_\alpha$ yields
$$
\boxed{\aleph_{\alpha+1}\leq2^{\aleph_\alpha}}.
$$
Again, strictness beyond this is independent of the usual axioms.
Solved by gpt-5.6-sol high.
= 17G
{parent=Paper 3}
{scope}
{title2=Graph Theory}
= a
{parent=17g}
{scope}
= Solution
{parent=a}
The diagonal <diagonal Ramsey number>[Ramsey number] $R(k)$ is the least $n$ such that every red-blue coloring of $K_n$ contains a monochromatic $K_k$. The standard recursion
$$
R(s,t)\leq R(s-1,t)+R(s,t-1)
$$
gives
$$
R(k)\leq\binom{2k-2}{k-1}<4^k.
$$
Thus the converted $k^4$ is a superscript-order error; that bound is false for large $k$.
In a red-blue coloring, if the red spanning graph is connected it has a red spanning tree. Otherwise its red components are joined pairwise by blue edges, making the blue graph connected, so it has a blue spanning tree.
The result fails for three colors. Color the six edges of $K_4$ by its three perfect matchings, one color per matching. Every monochromatic graph then consists of two disjoint edges and has no spanning tree.
Solved by gpt-5.6-sol high.
= b
{parent=17g}
{scope}
= Solution
{parent=b}
A directed path of length two is determined by its middle vertex $x$ and an ordered pair of distinct neighbors, giving
$$
\sum_xd(x)(d(x)-1).
$$
For the colored complete graph, let
$$
S=\sum_x\left\{\binom{d_r(x)}2+\binom{d_b(x)}2\right\}.
$$
A monochromatic triangle contributes three monochromatic two-edge angles to $S$, while a nonmonochromatic triangle contributes one. If $M$ and $N$ are their respective numbers, then
$$
S=3M+N,
\qquad
\binom n3=M+N.
$$
Eliminating $N$ gives
$$
\boxed{M=\frac12S-\frac12\binom n3}.
$$
Solved by gpt-5.6-sol high.
= 18I
{parent=Paper 3}
{scope}
{title2=Galois Theory}
= Solution
{parent=18I}
The elementary symmetric functions are
$$
e_r(x_1,\ldots,x_n)=\sum_{1\leq i_1<\cdots<i_r\leq n}x_{i_1}\cdots x_{i_r}.
$$
The <fundamental theorem of symmetric polynomials> says that every symmetric polynomial is uniquely a polynomial in $e_1,\ldots,e_n$.
If $r_1,\ldots,r_n$ are the roots of a monic polynomial, its <polynomial discriminant>[discriminant] is
$$
\operatorname{Disc}(f)=\prod_{i<j}(r_i-r_j)^2.
$$
This is symmetric in the roots, hence the theorem and Vieta's formulas express it polynomially in the coefficients. Equivalently,
$$
\operatorname{Disc}(f)=(-1)^{n(n-1)/2}\operatorname{Res}(f,f').
$$
For $x^5+q$ this gives
$$
\boxed{\operatorname{Disc}(x^5+q)=5^5q^4=3125q^4}.
$$
For $f=x^5+px^2+q$, a repeated root solves $f=f'=0$. Elimination gives
$$
\boxed{\operatorname{Disc}(f)=q(108p^5+3125q^3)}.
$$
Thus the discriminant vanishes exactly when $q=0$ or $108p^5+3125q^3=0$.
Solved by gpt-5.6-sol high.
= 19I
{parent=Paper 3}
{scope}
{title2=Representation Theory}
= a
{parent=19i}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
For an $H$-module $W$,
$$
\operatorname{Ind}_H^GW=\mathbb C[G]\otimes_{\mathbb C[H]}W.
$$
The action is left multiplication on the first factor:
$$
g_0\cdot(g\otimes w)=g_0g\otimes w.
$$
It is well defined because the tensor relation $gh\otimes w=g\otimes hw$ is preserved by left multiplication.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
If the $H$-classes in $H\cap\mathcal C(g)$ have representatives $x_i$, the induced-character formula is
$$
\boxed{(\operatorname{Ind}_H^G\psi)(g)
=|C_G(g)|\sum_{i=1}^s\frac{\psi(x_i)}{|C_H(x_i)|}}.
$$
It follows by grouping the standard sum over $x\in G$ satisfying $x^{-1}gx\in H$.
Solved by gpt-5.6-sol high.
= b
{parent=19i}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
The subgroup $\langle a\rangle$ is the unique Sylow-seven subgroup and is normal. No element outside it centralizes $a$, so $|C_G(a)|=7$. Similarly $C_G(b)=\langle b\rangle$ has order three. Thus nonidentity powers of $a$ form two classes of size three, while the elements of orders three form two classes of size seven. Together with the identity, there are five conjugacy classes.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
The commutator subgroup is $\langle a\rangle$, so
$$
G_{\rm ab}\cong G/\langle a\rangle\cong C_3.
$$
Hence the three linear characters are
$$
\chi_j(a)=1,
\qquad
\chi_j(b)=\omega^j,
\qquad j=0,1,2,quad \omega=e^{2\pi i/3}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
Let $\zeta=e^{2\pi i/7}$ and let $\lambda_k(a)=\zeta^k$. Conjugation by $b$ partitions the six nontrivial characters of $H=\langle a\rangle$ into the two orbits
$$
\{1,2,4\},
\qquad
\{3,5,6\}.
$$
Inducing one representative from each orbit gives two irreducible degree-three characters $\psi,\overline\psi$. On the five classes represented by $1,a,a^3,b,b^2$ their values are
$$
\psi=(3,\zeta+\zeta^2+\zeta^4,
\zeta^3+\zeta^5+\zeta^6,0,0),
$$
with the two cyclotomic sums interchanged for $\overline\psi$. Their squared degrees, together with the three linear characters, sum to
$$
3\cdot1^2+2\cdot3^2=21,
$$
so these are all irreducible characters.
Solved by gpt-5.6-sol high.
= 20F
{parent=Paper 3}
{scope}
{title2=Algebraic Topology}
= a
{parent=20f}
{scope}
= Solution
{parent=a}
Triangulate $X$ and adjoin one new vertex joined to every simplex; the resulting simplicial complex realizes the <simplicial cone>[cone] $CX$. The cone contracts to its apex, so
$$
\boxed{H_0(CX)\cong\mathbb Z,\qquad H_i(CX)=0\ (i>0)}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=20f}
{scope}
= Solution
{parent=b}
Write $X'=X\cup_A CA$. After replacing the two pieces by small open neighborhoods, their intersection deformation retracts to $A$. The <Mayer-Vietoris theorem>[Mayer-Vietoris sequence], together with $H_i(CA)=0$ for $i>0$ and $H_0(CA)=\mathbb Z$, becomes
$$
\cdots\to H_{i+1}(X')\to H_i(A)\to H_i(X)\to H_i(X')\to H_{i-1}(A)\to\cdots
$$
and at degree zero ends as
$$
\cdots\to H_1(X')\to H_0(A)\to\mathbb Z\oplus H_0(X)\to H_0(X')\to0,
$$
which is the required sequence.
Solved by gpt-5.6-sol high.
= c
{parent=20f}
{scope}
= Solution
{parent=c}
Attaching the cone collapses the circle $A=S^1\times\{x\}$ in the torus. The quotient has the homotopy type
$$
X'\simeq S^1\vee S^2.
$$
Equivalently, the exact sequence in part (b) kills one of the two generators of $H_1(T^2)$ and preserves the fundamental two-cycle. Therefore
$$
\boxed{H_i(X')\cong
\begin{cases}
\mathbb Z,&i=0,1,2,\\
0,&i>2.
\end{cases}}
$$
Solved by gpt-5.6-sol high.
= 21H
{parent=Paper 3}
{scope}
{title2=Linear Analysis}
= a
{parent=21h}
{scope}
= Solution
{parent=a}
A family $F$ of <continuous functions> on a <compact space> $K$ is <equicontinuous family of functions>[equicontinuous] if, for every $x$ and every $ε>0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<ε$ for every $y$ in $U$ and every $f$ in $F$. The <Arzela-Ascoli theorem> says that $F$ has compact closure in $C(K)$ with the <uniform norm> if and only if it is equicontinuous and pointwise bounded.
Solved by gpt-5.6-sol high.
= b
{parent=21h}
{scope}
= Solution
{parent=b}
Apply <Arzela-Ascoli theorem> on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging <uniform convergence>[uniformly] there. Successively extract subsequences for $m=1,2,\ldots$, and take the <diagonal argument>[diagonal subsequence]. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the <real numbers>; the locally uniform limit of <continuous functions> is continuous.
Solved by gpt-5.6-sol high.
= c
{parent=21h}
{scope}
= Solution
{parent=c}
Fix $x$ and $ε>0$. Since $\mathcal K$ is compact in the <uniform norm>, choose a finite $ε/3$-net $f_1,\ldots,f_r$. By <continuous function>[continuity], there is a neighbourhood $U$ of $x$ on which $|f_i(y)-f_i(x)|<ε/3$ for every $i$. For $f$ in $\mathcal K$, choose $f_i$ with $\|f-f_i\|_∞<ε/3$. The <triangle inequality> gives $|f(y)-f(x)|<ε$ on $U$, uniformly in $f$. Thus $\mathcal K$ is <equicontinuous family of functions>[equicontinuous].
Solved by gpt-5.6-sol high.
= d
{parent=21h}
{scope}
= Solution
{parent=d}
A <locally compact space>[locally compact] Hausdorff space is regular. Cover $K$ by finitely many open sets whose compact closures lie in $U$; their union $V$ satisfies $K\subset V\subset\overline V\subset U$, with $\overline V$ compact. The compact Hausdorff space $\overline V$ is normal, so the <Urysohn lemma> gives a <continuous function> $g:\overline V\to[0,1]$ equal to $1$ on $K$ and $0$ on $\overline V\setminus V$. Extend $g$ by zero outside $\overline V$. The boundary values make the extension continuous, and its support is a compact subset of $U$.
Solved by gpt-5.6-sol high.
= 22H
{parent=Paper 3}
{scope}
{title2=Analysis of Functions}
= a
{parent=22h}
{scope}
= Solution
{parent=a}
The <Riemann-Lebesgue lemma> states that the <Fourier transform> of an $L^1$ function is continuous and tends to zero at infinity.
For a <Schwartz space>[Schwartz function] $f$, differentiation under the integral and <integration by parts> give
$$
ξ^α D^β \widehat f(ξ)=C_{α,β}\widehat{D^α(x^βf)}(ξ).
$$
The <Riemann-Lebesgue lemma> bounds the right side by an $L^1$ seminorm of $D^α(x^βf)$. Every Schwartz seminorm of $\widehat f$ is therefore bounded by finitely many Schwartz seminorms of $f$. Hence the Fourier transform maps $\mathcal S(\mathbb R^n)$ continuously into itself.
Solved by gpt-5.6-sol high.
= b
{parent=22h}
{scope}
= Solution
{parent=b}
Taking the <Fourier transform of a derivative> gives
$$
iξ\mathbin\cdot\widehat B=\widehat f,\qquad ξ\mathbin\times\widehat B=0.
$$
Thus, away from $ξ=0$,
$$
\widehat B_j(ξ)=-iξ_j|ξ|^{-2}\widehat f(ξ).
$$
This formula determines the $L^2$ solution uniquely; a distribution supported only at $ξ=0$ cannot belong to $L^2$ unless it is zero.
For $|ξ|\leq1$, the <Fourier transform> estimate $|\widehat f(ξ)|\leq C\|f\|_1$ and local integrability of $|ξ|^{-2}$ in three dimensions control the squared norm. For $|ξ|>1$,
$$
(1+|ξ|^2)^{s+1}|\widehat B_j|^2\leq C(1+|ξ|^2)^s|\widehat f|^2.
$$
The <Plancherel theorem> therefore yields $\|B_j\|_{H^{s+1}}\leq C(\|f\|_1+\|f\|_{H^s})$. Finally the <Sobolev embedding theorem> in three dimensions gives $H^{s+1}\subset C^1$ when $s+1>1+3/2$, namely when $s>3/2$.
Solved by gpt-5.6-sol high.
= 23F
{parent=Paper 3}
{scope}
{title2=Riemann Surfaces}
= a
{parent=23f}
{scope}
= Solution
{parent=a}
A point is ramified with multiplicity $m_i$ exactly when it is a zero of $f'$ of multiplicity $m_i-1$. Since $f'$ has degree $d-1$, the <fundamental theorem of algebra> gives $\sum_i(m_i-1)=d-1$.
Conversely, choose distinct complex numbers $a_i$ and put $g(z)=c\prod_i(z-a_i)^{m_i-1}$ with $c\ne0$. The assumption makes $g$ a polynomial of degree $d-1$. Any <antiderivative> $f$ of $g$ has degree $d$ and precisely the prescribed <ramification divisor>[ramification] multiplicities.
Solved by gpt-5.6-sol high.
= b
{parent=23f}
{scope}
= Solution
{parent=b}
Outside the finite set of <branch points>, the derivative is nonzero. The <inverse function theorem> supplies evenly covered neighbourhoods, and compactness of the <Riemann sphere> makes every fibre finite, so the restriction is a covering map.
Lift a loop $γ$ from each point of the fibre over $z_0$; its endpoints define a <permutation> $σ_γ$. The homotopy lifting property shows that this permutation depends only on the based <homotopy> class. The punctured source is connected, so a path between any two points in the fibre projects to a loop whose lift joins them. Hence the monodromy group acts transitively.
Solved by gpt-5.6-sol high.
= c
{parent=23f}
{scope}
= Solution
{parent=c}
The monodromy around both $0$ and $1$ must be a product of two disjoint transpositions, hence a $(2,2)$ cycle type in $S_4$. The product relation for loops around the three punctures says $σ_0σ_1σ_∞=1$. By the stated fact, $σ_0σ_1$ is either the identity or another $(2,2)$ permutation. Consequently $σ_∞=(σ_0σ_1)^{-1}$ cannot contain a $3$-cycle. But ramification index $3$ over infinity would force exactly such a cycle, a contradiction.
Solved by gpt-5.6-sol high.
= 24I
{parent=Paper 3}
{scope}
{title2=Algebraic Geometry}
= i
{parent=24i}
{scope}
= Solution
{parent=i}
A <projective variety> is smooth when every local ring is regular, equivalently when its Jacobian has the expected rank at every point. The affine curve $y=x^3$ is smooth because the derivative of $y-x^3$ with respect to $y$ is $1$, while its projective closure $YZ^2=X^3$ is singular at $[0:1:0]$.
The genus is $\dim_k H^1(C,\mathcal O_C)$. Intersect the Fermat cubic threefold with the plane $X_3=X_4=0$. This gives the smooth plane cubic $X_0^3+X_1^3+X_2^3=0$, whose <genus of a smooth plane curve> is $(3-1)(3-2)/2=1$.
An irreducible plane conic is smooth: a singular quadratic form in three variables has rank at most two and factors over the algebraically closed field, contradicting irreducibility. Projection from any point of the conic, or its degree-two complete linear system, then identifies it with the <projective line>.
Finally, a nonzero ternary quadratic form is determined up to nonzero scalar by its six coefficients. Thus generalized conics are parametrized bijectively by $\mathbb P^5$.
Solved by gpt-5.6-sol high.
= ii
{parent=24i}
{scope}
= Solution
{parent=ii}
Write a conic as $x^TAx=0$ for a symmetric $3\times3$ matrix $A$, with off-diagonal entries equal to half the corresponding coefficients. A union of two distinct lines has matrix rank two. Thus $R^\circ$ is the rank-two part of the determinant hypersurface $\det A=0$. Its closure also contains rank-one matrices, representing double lines, so it is not <Zariski-closed set>[Zariski closed]. Since the determinant is one nonzero homogeneous equation in $\mathbb P^5$, its dense rank-two locus has dimension $4$.
The final sentence of the converted question writes $R^c$, but the natural intended set is $R^\circ$. The homogeneous ideal of its closure is the principal ideal generated by the cubic polynomial $\det A$: the determinant hypersurface is irreducible, and $R^\circ$ is dense in it.
Solved by gpt-5.6-sol high.
= 25F
{parent=Paper 3}
{scope}
{title2=Differential Geometry}
= Solution
{parent=25F}
A point $y$ is a <regular value> of $f$ when every derivative $D_xf$ for $x$ in $f^{-1}(y)$ is surjective. <Sard theorem> says that the critical values of a smooth map have measure zero. The stack-of-records theorem says that the inverse image of a regular value is a smooth submanifold of codimension $\dim Y$.
Maps $f$ and $g$ are smoothly <homotopy>[homotopic] when a smooth $F:X\times[0,1]\to Y$ has endpoint maps $f$ and $g$. Perturb such a homotopy relative to its ends so that it is transverse to $y$. Then $F^{-1}(y)$ is a compact one-dimensional manifold whose boundary is $f^{-1}(y)\sqcup g^{-1}(y)$. Every compact one-dimensional manifold has an even number of boundary points, proving the parity formula.
The hypothesis on $U$ says that the <degree modulo two> of $h$ is one. Suppose no $h(x)$ and $h(-x)$ were antipodal. Then
$$
k(x)={h(x)+h(-x)\over |h(x)+h(-x)|}
$$
would be defined, even, and homotopic to $h$ by normalized straight-line interpolation. For a regular value, the fibres of the even map $k$ occur in antipodal pairs, so its degree modulo two is zero. Homotopy invariance gives a contradiction. Hence some antipodal pair is mapped to an antipodal pair.
Solved by gpt-5.6-sol high.
= 26H
{parent=Paper 3}
{scope}
{title2=Probability and Measure}
= Solution
{parent=26H}
If the variables are <independent random variables>, apply the defining product rule first to indicator functions, then to simple functions, and finally use bounded measurable approximation to obtain the displayed identity. Conversely, choosing $f_n=1_{A_n}$ gives
$$
\mathbb P(X_1\in A_1,\ldots,X_N\in A_N)=\prod_n\mathbb P(X_n\in A_n),
$$
which is independence.
Let $S_m=\sum_{n\leq m}X_n$. Independence and zero means imply orthogonality in $L^2$, so
$$
\|S_m-S_l\|_2^2=\sum_{n=l+1}^mσ_n^2.
$$
If the variance series converges, $(S_m)$ is Cauchy in the complete space $L^2$ and hence converges there. Conversely, $L^2$ convergence makes $(S_m)$ Cauchy, so every tail of the nonnegative variance series tends to zero; therefore $\sum_nσ_n^2<∞$.
Solved by gpt-5.6-sol high.
= 27K
{parent=Paper 3}
{scope}
{title2=Applied Probability}
= a
{parent=27k}
{scope}
= Solution
{parent=a}
The queue length is a <birth-death process> with birth rate $λr^n$ in state $n$ and death rate $μ\min(n,2)$. Its reversible weights satisfy
$$
π_n=π_0\prod_{j=0}^{n-1}{λr^j\over μ\min(j+1,2)}.
$$
For $0<r<1$, the factor $r^{n(n-1)/2}$ makes their sum finite for every $λ,μ>0$. For $r=1$, the tail is geometric with ratio $λ/(2μ)$, so a <stationary distribution> exists exactly when $λ<2μ$.
Solved by gpt-5.6-sol high.
= b
{parent=27k}
{scope}
= Solution
{parent=b}
On states $0,\ldots,N$, births have rate $1$ below $N$ and deaths have rate $μ$ above $0$. For $μ>0$ this finite <continuous-time Markov chain> is irreducible, so its unique stationary law is
$$
π_n={μ^{-n}\over\sum_{j=0}^N μ^{-j}},\qquad0\leq n\leq N.
$$
For $μ=1$ this is uniform. If the parameter value $μ=0$ is admitted, state $N$ is absorbing and the unique stationary law is instead the point mass at $N$.
Solved by gpt-5.6-sol high.
= c
{parent=27k}
{scope}
= Solution
{parent=c}
In equilibrium the chain is reversible by <detailed balance>. Reverse time. Forward departures become reverse accepted arrivals; reversibility identifies the reversed chain with the original one, whose attempted arrivals form a <Poisson process>. The accepted points are exactly those occurring while the state is below $N$. The standard quasi-reversibility argument therefore makes the equilibrium departure process Poisson, with rate equal to the mean departure rate
$$
μ\mathbb P(X>0)=1-π_N.
$$
Solved by gpt-5.6-sol high.
= 28J
{parent=Paper 3}
{scope}
{title2=Principles of Statistics}
= a
{parent=28j}
{scope}
= Solution
{parent=a}
The <log-likelihood> is, up to constants, $nα\log β-β\sum_iX_i$, so the <maximum likelihood estimator> is $\widehat β=α/\overline X$. By the <central limit theorem>,
$$
\sqrt n(\overline X-α/β)\Rightarrow N(0,α/β^2).
$$
Applying the <delta method> to $g(x)=α/x$ gives
$$
\sqrt n(\widehat β-β)\Rightarrow N(0,β^2/α).
$$
Solved by gpt-5.6-sol high.
= b
{parent=28j}
{scope}
= Solution
{parent=b}
Let $z_{1-γ/2}$ be the corresponding <standard normal distribution> quantile. <Slutsky theorem> permits replacing $β$ by the consistent estimator $\widehat β$, giving the confidence interval
$$
\left[\widehat β-z_{1-γ/2}{\widehat β\over\sqrt{nα}},\
\widehat β+z_{1-γ/2}{\widehat β\over\sqrt{nα}}\right].
$$
After standardization, the preceding convergence in distribution and continuity of the normal distribution show that its coverage tends to $1-γ$.
Solved by gpt-5.6-sol high.
= c
{parent=28j}
{scope}
= Solution
{parent=c}
For <bootstrapping (statistics)>, repeatedly sample $X_1^*,\ldots,X_n^*$ with replacement from the empirical distribution of the observations and compute $\widehat β^*=α/\overline X^*$. Conditional on the data, estimate the $γ/2$ and $1-γ/2$ quantiles $q_l,q_u$ of $\sqrt n(\widehat β^*-\widehat β)$. The basic bootstrap interval is
$$
[\widehat β-q_u/\sqrt n,\widehat β-q_l/\sqrt n].
$$
As the number of resamples tends to infinity it estimates the conditional quantiles; bootstrap consistency and the continuous mapping theorem then give asymptotic coverage $1-γ$.
Solved by gpt-5.6-sol high.
= 29K
{parent=Paper 3}
{scope}
{title2=Stochastic Financial Models}
= a
{parent=29k}
{scope}
= Solution
{parent=a}
For a bounded <stopping time> $T\leq N$, write
$$
\widehat M_T=\widehat M_0+\sum_{n=1}^N1_{\{T\geq n\}}(\widehat M_n-\widehat M_{n-1}).
$$
The indicator is measurable at time $n-1$. Taking conditional expectations term by term and using the <supermartingale> inequality gives $\mathbb E\widehat M_T\leq\mathbb E\widehat M_0$. The same calculation gives equality for the <martingale> $M$. Since their initial values agree, $\mathbb E M_T\geq\mathbb E\widehat M_T$.
Solved by gpt-5.6-sol high.
= b
{parent=29k}
{scope}
= Solution
{parent=b}
If $θ_n$ shares are held during period $n$, the remaining wealth $X_{n-1}-θ_nS_{n-1}$ is in the bank account. <Self-financing portfolio>[Self-financing] therefore gives
$$
X_n=θ_nS_n+(1+r)(X_{n-1}-θ_nS_{n-1}),
$$
which rearranges to the required equation.
Solved by gpt-5.6-sol high.
= c
{parent=29k}
{scope}
= Solution
{parent=c}
Condition on the information through time $n-1$. Independence of $ξ_n$ and the definition of the <Bellman equation for terminal-wealth utility> imply
$$
\mathbb E[V(n,X_n)\mid\mathcal F_{n-1}]\leq V(n-1,X_{n-1}),
$$
because the actual holding is one candidate in the supremum. Thus $V(n,X_n)$ is a <supermartingale> for every strategy.
For the stated strategy it is a <martingale>, so
$$
\mathbb E[U(X_N^*)]=V(0,X_0).
$$
Every competing strategy has expected terminal value at most $V(0,X_0)$ by the supermartingale inequality from part (a), evaluated at the bounded time $N$. Hence $θ^*$ is optimal.
Solved by gpt-5.6-sol high.
= 30A
{parent=Paper 3}
{scope}
{title2=Asymptotic Methods}
= a
{parent=30a}
{scope}
= Solution
{parent=a}
One form of <Watson lemma> is this. If $f(t)$ has the asymptotic expansion $f(t)\sim\sum_{n\geq0}a_nt^{λ_n-1}$ as $t\downarrow0$, with $0<λ_0<λ_1<\cdots$, and the Laplace integral has suitable growth control away from zero, then
$$
\int_0^b e^{-xt}f(t)\,dt\sim\sum_{n\geq0}a_nΓ(λ_n)x^{-λ_n}
$$
as $x\to∞$, with truncation after any fixed number of terms justified by the corresponding remainder estimate.
Solved by gpt-5.6-sol high.
= b
{parent=30a}
{scope}
= Solution
{parent=b}
Complete the square: $z^2-2iz=(z-i)^2+1$. The pole of $(1-iz)^{-1}$ is at $z=-i$, so the contour may be shifted upward through the <saddle point> $z=i$ without crossing a singularity. Setting $z=t+i$ gives the exact steepest-descent representation
$$
I(x)=e^{-x}\int_{-∞}^{∞}{e^{-xt^2}\over2-it}\,dt.
$$
Near the saddle,
$$
{1\over2-it}=\sum_{n\geq0}{(it)^n\over2^{n+1}}.
$$
Odd terms integrate to zero, while
$$
\int_{-∞}^{∞}t^{2m}e^{-xt^2}\,dt=Γ(m+1/2)x^{-m-1/2}.
$$
The <Watson lemma> therefore yields the full <asymptotic expansion>
$$
I(x)\sim e^{-x}\sum_{m=0}^{∞}{(-1)^mΓ(m+1/2)\over2^{2m+1}x^{m+1/2}}.
$$
Solved by gpt-5.6-sol high.
= 31A
{parent=Paper 3}
{scope}
{title2=Dynamical Systems}
= a
{parent=31a}
{scope}
= Solution
{parent=a}
The second equation gives $y=x(1-x)$. The first then says either $y=0$, giving the persistent equilibrium points $(0,0)$ and $(1,0)$, or
$$
x^2-(β+1)x-μ=0,
$$
giving $x_±=((β+1)±\sqrt{(β+1)^2+4μ})/2$ and $y_±=x_±(1-x_±)$. These two points exist for $μ\geq-(β+1)^2/4$.
At the origin the <Jacobian matrix> has trace $-1$ and determinant $-μ$, so it is stable for $μ<0$ and a <saddle point> for $μ>0$. At $(1,0)$ the trace is $-1$ and determinant $μ+β$, so it is stable for $μ>-β$ and a saddle for $μ<-β$. Stationary bifurcations occur at
$$
μ=-{(β+1)^2\over4},\qquad μ=-β,\qquad μ=0,
$$
with the first two coinciding when $β=1$.
Solved by gpt-5.6-sol high.
= b
{parent=31a}
{scope}
= Solution
{parent=b}
At a non-axis fixed point the determinant and trace are
$$
\det J=x(1-x)(2x-β-1),\qquad \operatorname{tr}J=βx(1-x)-1.
$$
These formulas determine the solid stable and dashed unstable branches in the <bifurcation diagram>. The pair $x_±$ is born in a <saddle-node bifurcation> at $μ=-(β+1)^2/4$. For $β\ne1$, it exchanges branches with $x=1$ at $μ=-β$ and with $x=0$ at $μ=0$, both <transcritical bifurcations>. If $0<β<1$, the saddle-node lies to the left of $x=1$; if $β>1$, it lies to the right. For $β=1$, the saddle-node and the collision at $x=1$ combine into a <pitchfork bifurcation normal form>[pitchfork bifurcation] at $μ=-1$.
Solved by gpt-5.6-sol high.
= c
{parent=31a}
{scope}
= Solution
{parent=c}
Put $x=1+u$, $y=v$, and $μ=-1+ν$, adjoining $\dot ν=0$. Then
$$
\dot u=v(ν+u+v),\qquad \dot v=-u-v-u^2.
$$
The zero <eigenvalue> has centre direction $v=-u$. Solving the <center manifold> invariance equation gives
$$
v=-u-u^2-uν+O(|(u,ν)|^3).
$$
Substitution yields the reduced equation
$$
\dot u=u(u^2-ν)+O(|(u,ν)|^4).
$$
Its equilibria are $u=0$ and $u=±\sqrt ν+O(ν)$ for $ν>0$, exactly matching the three branches meeting at the pitchfork in part (b).
Solved by gpt-5.6-sol high.
= d
{parent=31a}
{scope}
= Solution
{parent=d}
For $ν=μ+1>0$ small, $(1,0)$ has two negative <eigenvalues> and is a <stable node>. The two nearby equilibria satisfy $u=±\sqrt ν+O(ν)$ and are saddles. Their stable and unstable separatrices form the local phase portrait; trajectories in the wedge between the stable manifolds approach $(1,0)$, while trajectories crossing an unstable direction leave the neighbourhood.
Solved by gpt-5.6-sol high.
= 32D
{parent=Paper 3}
{scope}
{title2=Integrable Systems}
= a
{parent=32d}
{scope}
= Solution
{parent=a}
The transformations obey $g_1^s g_1^r=g_1^{s+r}$ and $g_1^0$ is the identity. If $\widetilde x(\widetilde t)=x(t)+st$, then $d^2\widetilde x/d\widetilde t^2=d^2x/dt^2$. Thus solutions of the free-particle equation are mapped to solutions, so this is a one-parameter group of Lie symmetries.
Solved by gpt-5.6-sol high.
= b
{parent=32d}
{scope}
= Solution
{parent=b}
The flow equations of the displayed <vector field> are $\dot ψ_1=-ψ_2$ and $\dot ψ_2=ψ_1$. Hence $dψ/ds=iψ$, whose solution is $ψ(s)=e^{is}ψ(0)$. The vector field therefore generates the stated phase rotation.
Solved by gpt-5.6-sol high.
= c
{parent=32d}
{scope}
= Solution
{parent=c}
Direct substitution gives $g^sg^r=g^{s+r}$. Differentiating at $s=0$ gives $t\partial_x+ix(ψ\partial_ψ-ψ^*\partial_{ψ^*})$.
Writing the transformed field in transformed coordinates gives
$$
\widetilde ψ(t,\widetilde x)=e^{i(s\widetilde x-s^2t/2)}ψ(t,\widetilde x-st).
$$
Consequently
$$
\widetilde ψ_t=e^{i\Phi}(ψ_t-sψ_x-is^2ψ/2),
$$
$$
\widetilde ψ_x=e^{i\Phi}(ψ_x+isψ),\qquad
\widetilde ψ_{xx}=e^{i\Phi}(ψ_{xx}+2isψ_x-s^2ψ),
$$
where $\Phi=s\widetilde x-s^2t/2$. These formulas, together with their complex conjugates and the analogous mixed derivatives, give the second prolongation. Differentiating at zero yields
$$
η^0=ixψ_t-ψ_x,\qquad η^{11}=ixψ_{xx}+2iψ_x.
$$
Solved by gpt-5.6-sol high.
= d
{parent=32d}
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= Solution
{parent=d}
Substituting the prolonged formulas shows
$$
i\widetilde ψ_t+\widetilde ψ_{xx}/2+|\widetilde ψ|^2\widetilde ψ
=e^{i\Phi}(iψ_t+ψ_{xx}/2+|ψ|^2ψ),
$$
so the nonlinear Schrödinger equation is invariant. Applying the symmetry to the given solitary wave produces
$$
ψ(t,x)=a\exp\{i[sx+(a^2-s^2)t/2]\}\operatorname{sech}(a(x-st)),
$$
which travels with speed $s$.
Solved by gpt-5.6-sol high.
= 33B
{parent=Paper 3}
{scope}
{title2=Principles of Quantum Mechanics}
= a
{parent=33b}
{scope}
= Solution
{parent=a}
The <addition of angular momentum> rule gives
$$
j=|j_1-j_2|,|j_1-j_2|+1,\ldots,j_1+j_2.
$$
For each such $j$, measurement along any axis has magnetic quantum number $m=-j,-j+1,\ldots,j$.
Solved by gpt-5.6-sol high.
= b
{parent=33b}
{scope}
= Solution
{parent=b}
For two equal spins, exchange acts on the coupled state $|j,m\rangle$ by the <Clebsch-Gordan decomposition>[Clebsch-Gordan coefficient] symmetry factor $(-1)^{2j_1-j}$. It is therefore $-1$ when $j=2j_1-1$, making every state in that multiplet antisymmetric, and $+1$ when $j=2j_1-2$, making the next multiplet symmetric.
Solved by gpt-5.6-sol high.
= c
{parent=33b}
{scope}
= Solution
{parent=c}
Parity conservation and the negative parity of $X$ require odd <orbital angular momentum> $L$. Since the two $Y$ particles are identical bosons, their spin state must then be antisymmetric, so their total spin is $S=1$. Total angular momentum zero forces $L=S=1$.
Conditioning on spin projection $m_S=1$ leaves orbital projection $m_L=-1$. The angular density is therefore proportional to $|Y_{1,-1}(θ,φ)|^2\propto\sin^2θ$. Including the solid-angle measure, the requested probability is
$$
{\int_{π/4}^{3π/4}\sin^3θ\,dθ\over\int_0^π\sin^3θ\,dθ}
= {5\sqrt2\over8}.
$$
Solved by gpt-5.6-sol high.
= 34B
{parent=Paper 3}
{scope}
{title2=Applications of Quantum Mechanics}
= a
{parent=34b}
{scope}
= Solution
{parent=a}
A <Bravais lattice> is the integer span of three linearly independent primitive vectors. Its <reciprocal lattice> consists of vectors $G$ satisfying $G\mathbin\cdot r\in2π\mathbb Z$ for every lattice vector $r$.
Translations by lattice vectors commute with the <Hamiltonian>. Their simultaneous eigenvalues have the form $e^{ik\cdot r}$, so an energy eigenfunction may be chosen with
$$
ψ_k(x+r)=e^{ik\cdot r}ψ_k(x).
$$
Thus $ψ_k=e^{ik\cdot x}u_k$ with lattice-periodic $u_k$, which is <Bloch theorem>. Wavevectors differing by a reciprocal-lattice vector describe the same translation character, and the <Brillouin zone> is a fundamental Wigner-Seitz cell of the reciprocal lattice.
Solved by gpt-5.6-sol high.
= b
{parent=34b}
{scope}
= Solution
{parent=b}
Take
$$
a_1=(-1,1,1)/2,\quad a_2=(1,-1,1)/2,\quad a_3=(1,1,-1)/2.
$$
Their integer span consists exactly of the cubic points together with the body-centre points, so the BCC set is a <Bravais lattice>. Using the inverse matrix supplied in the question, reciprocal primitive vectors are
$$
b_1=2π(0,1,1),\quad b_2=2π(1,0,1),\quad b_3=2π(1,1,0).
$$
These generate a face-centred cubic lattice; thus the <reciprocal lattice> of BCC is FCC, with the displayed scale.
Solved by gpt-5.6-sol high.
= 35C
{parent=Paper 3}
{scope}
{title2=Statistical Physics}
= a
{parent=35c}
{scope}
= Solution
{parent=a}
One quantum state occupies volume $(2π)^d/V$ in wavevector space. Counting the shell of radius $k$ and including spin degeneracy gives
$$
g(E)dE={g_sV S_{d-1}\over(2π)^d}k^{d-1}dk.
$$
Using $k=(E/A)^{1/α}/\hbar$ gives the stated power with
$$
B={g_sS_{d-1}\over α(2π)^d\hbar^dA^{d/α}}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=35c}
{scope}
= Solution
{parent=b}
The <Bose-Einstein distribution> is
$$
\overline n_r={1\over e^{β(E_r-μ)}-1}.
$$
Positivity requires $μ$ not to exceed the ground-state energy, here zero. For a finite system above the condensation limit one has $μ<0$; it approaches zero from below at <Bose-Einstein condensation>.
Solved by gpt-5.6-sol high.
= c
{parent=35c}
{scope}
= Solution
{parent=c}
For a homogeneous dispersion $E(p)=A|p|^α$, the kinetic expression for pressure gives
$$
pV={1\over d}\sum_r\overline n_r,p_r\mathbin\cdot\nabla_pE(p_r)
={α\over d}\sum_r\overline n_rE_r.
$$
Thus $pV=DE$ with $D=α/d$.
Solved by gpt-5.6-sol high.
= d
{parent=35c}
{scope}
= Solution
{parent=d}
In the continuum limit the number outside the ground state is
$$
N_{\rm ex}=BV\int_0^∞{E^{d/α-1}\over e^{β(E-μ)}-1}\,dE.
$$
At the largest allowed chemical potential, $μ=0$, the integrand behaves near zero as $E^{d/α-2}$. The integral is finite exactly when $d/α>1$. Excess particles must then occupy the ground state, so a <Bose-Einstein condensation>[Bose-Einstein condensate] exists precisely for $α<d$.
Solved by gpt-5.6-sol high.
= 36C
{parent=Paper 3}
{scope}
{title2=Electrodynamics}
= a
{parent=36c}
{scope}
= Solution
{parent=a}
For a point charge, $p=q x$, so the radiation fields are transverse and satisfy $|E_{\rm Rad}|=c|B_{\rm Rad}|$. The radial <Poynting vector> is
$$
S_r={μ_0q^2\over16π^2cr^2}|\widehat x\times\ddot x|^2.
$$
Choosing the acceleration as polar axis and integrating $\sin^2θ$ over the sphere gives $\int\sin^2θ\,dΩ=8π/3$. Hence the Larmor formula is
$$
P={μ_0q^2|\ddot x|^2\over6πc}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=36c}
{scope}
= Solution
{parent=b}
Ignoring the small radiative back-reaction, <conservation of energy> gives $\dot x=\sqrt{2(E-f(x))/m}$ and Newton's equation gives $\ddot x=-f'(x)/m$. Integrating the Larmor formula with $dt=dx/\dot x$ yields, for $E>V_0$,
$$
ΔE_{\rm Rad}={q^2μ_0\over6πm^2c}\sqrt{m/2}
\int_0^L{[f'(x)]^2\over\sqrt{E-f(x)}}\,dx.
$$
If $E<V_0$, the particle turns at $x_*$ defined by $f(x_*)=E$ and traverses the accelerated region twice, so the same formula holds with twice the integral from $0$ to $x_*$.
Solved by gpt-5.6-sol high.
= c
{parent=36c}
{scope}
= Solution
{parent=c}
For $f(x)=V_0x/L$, direct integration gives
$$
ΔE_{\rm Rad}={q^2μ_0\over3πm^2c}\sqrt{m/2}{V_0\over L}
(\sqrt E-\sqrt{E-V_0}),\qquad E>V_0,
$$
and
$$
ΔE_{\rm Rad}={2q^2μ_0\over3πm^2c}\sqrt{m/2}{V_0\over L}\sqrt E,
\qquad E<V_0.
$$
The reflected expression increases up to threshold and the transmitted expression decreases above it. The supremum is therefore approached from below at $E=V_0$ and equals
$$
{2q^2μ_0\over3πm^2c}\sqrt{m/2}{V_0^{3/2}\over L}.
$$
Solved by gpt-5.6-sol high.
= 37C
{parent=Paper 3}
{scope}
{title2=General Relativity}
= a
{parent=37c}
{scope}
= Solution
{parent=a}
The upper-left block has <eigenvalues> $1$ and $-1$, while the remaining two eigenvalues are both $-1$. The metric signature is therefore $(1,3)$ up to sign convention. It is Lorentzian.
Solved by gpt-5.6-sol high.
= b
{parent=37c}
{scope}
= Solution
{parent=b}
Let $F=1-2M/r$, define the tortoise coordinate by $dr_*/dr=F^{-1}$, and set the advanced time $v=t+r_*$. Since $dt=dv-F^{-1}dr$, substitution gives
$$
ds^2=-Fdv^2+2\,dv\,dr+r^2dΩ^2.
$$
These are <ingoing Eddington-Finkelstein coordinates>; all coefficients are regular at the <Schwarzschild event horizon>[event horizon] $r=2M$.
Solved by gpt-5.6-sol high.
= c
{parent=37c}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
Outside the ingoing null shell the enclosed mass is $E$, so use the ingoing <ingoing Eddington-Finkelstein coordinates>
$$
ds^2=-(1-2E/r)dv^2+2\,dv\,dr+r^2dΩ^2.
$$
This is the <Schwarzschild spacetime>[Schwarzschild metric] of mass $E$, regular across $r=2E$ and valid for $r>0$ on the exterior side of the shell.
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
Inside the spherical shell the enclosed mass vanishes by the spherical vacuum solution, so spacetime is flat. In the same ingoing coordinates,
$$
ds^2=-dv^2+2\,dv\,dr+r^2dΩ^2,
$$
valid for $r\geq0$ on the interior side before the shell reaches the centre.
Solved by gpt-5.6-sol high.
= iii
{parent=c}
{scope}
= Solution
{parent=iii}
Because the shell lies well inside its Schwarzschild radius once it reaches $r=2E$, an <Schwarzschild event horizon>[event horizon] forms and the final exterior is a <Schwarzschild spacetime>[black hole] of mass $E$. In the ideal instantaneous-shell model the collapse ends at a spacetime singularity.
Solved by gpt-5.6-sol high.
= 38A
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics II}
= a
{parent=38a}
{scope}
= Solution
{parent=a}
The sink speed scales as $U\sim A/R^2$, so the <Reynolds number> is $Re\sim UR/ν=A/(νR)$. The approximation applies in the high-Reynolds-number region $R\ll A/ν$. Radial inertia scales as $U\,dU/dR\sim A^2/R^5$, so the radial pressure gradient has the same $R^{-5}$ dependence.
Solved by gpt-5.6-sol high.
= b
{parent=38a}
{scope}
= Solution
{parent=b}
The definitions through the Stokes stream function give
$$
{1\over r}\partial_r(ru)+\partial_zw
=-{1\over r}\partial_r\partial_zΨ+{1\over r}\partial_z\partial_rΨ=0,
$$
so <incompressible flow>[incompressibility] is automatic.
Balancing inertia $A^2/r^5$ with viscosity $νU/δ^2$ gives
$$
δ(r)\sim(νr^3/A)^{1/2}.
$$
The streamfunction scale is $rUδ\sim(Aνr)^{1/2}$, motivating
$$
Ψ=(Aνr)^{1/2}F(η),\qquad η=z(A/(νr^3))^{1/2}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=38a}
{scope}
= Solution
{parent=c}
The similarity form gives
$$
u=-{A\over r^2}F',\qquad
w={(Aν)^{1/2}\over2r^{3/2}}(F-3ηF').
$$
Using the outer pressure gradient $ρ^{-1}dP/dr=2A^2/r^5$ in the correctly written radial boundary-layer equation $u u_r+w u_z=-ρ^{-1}P_r+νu_{zz}$ gives
$$
F^{(3)}+2(1-F'^2)-{1\over2}FF''=0.
$$
No slip and matching to the sink require $F(0)=F'(0)=0$ and $F'(∞)=1$. The boundary-layer flux is of order $rUδ$, relative to the sink flux $A$ it is
$$
{rUδ\over A}\sim(νr/A)^{1/2}.
$$
Mass conservation therefore requires an outer correction of this relative size, which is small in the assumed region.
Solved by gpt-5.6-sol high.
= 39A
{parent=Paper 3}
{scope}
{title2=Waves}
= a
{parent=39a}
{scope}
= Solution
{parent=a}
Substitution of the monochromatic ansatz gives
$$
\widehat w''+k^2(N^2/ω^2-1)\widehat w=0.
$$
Thus the outer solutions oscillate with $m=k\sqrt{N_0^2/ω^2-1}$ and the middle solutions are combinations of $\cosh kz$ and $\sinh kz$. The incident wave has phase $kx-mz-ωt$; its group velocity is upward, whereas the reflected sign has downward group velocity. The transmitted wave must again use the upward-energy sign. This gives exactly the three forms stated.
Solved by gpt-5.6-sol high.
= b
{parent=39a}
{scope}
= Solution
{parent=b}
Continuity of normal velocity and pressure at each density interface requires continuity of $\widehat w$ and $d\widehat w/dz$ at $z=±L$. These four conditions determine the four unknown amplitudes.
Solved by gpt-5.6-sol high.
= c
{parent=39a}
{scope}
= Solution
{parent=c}
Matching at $z=L$ first expresses the middle solution in terms of $A_T$. Propagating its value and derivative across the layer of thickness $2L$ uses the transfer matrix
$$
\begin{pmatrix}\cosh2α&k^{-1}\sinh2α\\k\sinh2α&\cosh2α\end{pmatrix},
\qquad α=kL.
$$
Matching this result to $A_I+A_R$ and $-imA_I+imA_R$ at $z=-L$, then eliminating $A_R$, gives
$$
{A_T\over A_I}={2imk\over2imk\cosh2α+(k^2-m^2)\sinh2α}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=39a}
{scope}
= Solution
{parent=d}
Taking the squared <modulus> and using $\cosh^2u=1+\sinh^2u$ gives
$$
\left|{A_T\over A_I}\right|^2
=\left[1+{(k^2+m^2)^2\over4k^2m^2}\sinh^22α\right]^{-1}.
$$
The incident wavevector makes an angle $ψ$ with the downward vertical, so $\sin2ψ=2km/(k^2+m^2)$. Substitution gives the required formula.
Solved by gpt-5.6-sol high.
= 40E
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}
= a
{parent=40e}
{scope}
= Solution
{parent=a}
For the discrete <Fourier mode> $u_m^n=G^ne^{imθ}$, the centred second difference has multiplier $-4\sin^2(θ/2)$. Hence the <amplification factor> is
$$
G(θ)={1-2μ\sin^2(θ/2)\over1+2μ\sin^2(θ/2)}.
$$
For every $μ>0$, $|G(θ)|\leq1$. The discrete <Parseval identity> then shows directly that the discrete $L^2$ norm cannot increase at any time step, proving unconditional stability for the Cauchy problem.
Solved by gpt-5.6-sol high.
= b
{parent=40e}
{scope}
= Solution
{parent=b}
Let $A_h$ be the symmetric Dirichlet second-difference matrix. Its <eigenvalues> are nonpositive, so every eigenvalue of
$$
R_h=(I-kA_h/2)^{-1}(I+kA_h/2)
$$
has modulus at most one. Thus $\|R_h\|_{2,h}\leq1$.
Insert the exact smooth solution into the scheme. The <local truncation error> per step has discrete norm $O(k^3+kh^2)$, from the order-three trapezoidal time residual and the order-two centred spatial residual. If $e^n$ is the grid error, stability and iteration give
$$
\|e^{n+1}\|_{2,h}\leq\|e^n\|_{2,h}+C(k^3+kh^2).
$$
There are at most $1/k$ steps and $e^0=0$, hence
$$
\max_n\|e^n\|_{2,h}\leq C(k^2+h^2)\to0.
$$
This proves convergence for every $μ=k/h^2>0$ without invoking the Lax equivalence theorem.
Solved by gpt-5.6-sol high.