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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1.bigb
= Paper 1
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperii_1_2022.pdf
= 1I
{parent=Paper 1}
{scope}
{title2=Number Theory}
= Solution
{parent=1I}
If $m$ and $n$ are <coprime integers>[coprime], every divisor of $mn$ is uniquely $ab$ with $a\mid m$ and $b\mid n$. Hence, for a <multiplicative arithmetic function> $f$,
$$
g(mn)=\sum_{a\mid m}\sum_{b\mid n}f(ab)
=\left(\sum_{a\mid m}f(a)\right)
\left(\sum_{b\mid n}f(b)\right)=g(m)g(n).
$$
The <Mobius function> is $\mu(n)=0$ if a prime square divides $n$, and $\mu(n)=(-1)^r$ if $n$ is a product of $r$ distinct primes. The <Euler totient function> $\phi(n)$ counts residues modulo $n$ coprime to $n$. From the prime factorizations,
$$
\frac{\phi(n)}n=\prod_{p\mid n}\left(1-\frac1p\right)
=\sum_{d\mid n}\frac{\mu(d)}d.
$$
Both sides of the second identity are multiplicative, and at a prime power $p^a$,
$$
\sum_{d\mid p^a}\frac{\mu(d)^2}{\phi(d)}
=1+\frac1{p-1}=\frac p{p-1}
=\frac{p^a}{\phi(p^a)}.
$$
Therefore
$$
\boxed{\frac{n}{\phi(n)}=
\sum_{d\mid n}\frac{\mu(d)^2}{\phi(d)}}.
$$
Solved by gpt-5.6-sol high.
= 2G
{parent=Paper 1}
{scope}
{title2=Topics in Analysis}
= Solution
{parent=2G}
Directly,
$$
a+\frac{At+B}{Ct+D}
=\frac{(aC+A)t+(aD+B)}{Ct+D}.
$$
Thus take $\alpha=aC+A$, $\beta=aD+B$, $\gamma=C$, and $\delta=D$. They are nonnegative and
$$
\alpha\delta-\beta\gamma=AD-BC=1.
$$
Each finite block of a <continued fraction> acts by a fractional linear transformation with an integral matrix of determinant $\pm1$. If the block $a_1,\ldots,a_N$ repeats, removing one period leaves the same tail $x$, so
$$
x=\frac{Ax+B}{Cx+D}
$$
for integers $A,B,C,D$. Hence $Cx^2+(D-A)x-B=0$, with the linear case allowed when $C=0$.
For period $(a,b)$,
$$
x=a+\frac1{b+1/x},
$$
so
$$
\boxed{bx^2-abx-a=0}.
$$
The continued fraction is positive, so it is the positive root
$$
\boxed{x=\frac{ab+\sqrt{a^2b^2+4ab}}{2b}},
$$
while the other root is negative.
Solved by gpt-5.6-sol high.
= 3K
{parent=Paper 1}
{scope}
{title2=Coding and Cryptography}
= a
{parent=3k}
{scope}
= Solution
{parent=a}
For a $q$-ary code with codeword lengths $l_1,\ldots,l_m$, <Kraft inequality> is
$$
\boxed{\sum_{i=1}^m q^{-l_i}\leq1}.
$$
Choose an infinite random $q$-ary string with independent uniform symbols. The event that it begins with codeword $i$ has probability $q^{-l_i}$. For a <prefix code>[prefix-free code], these events are disjoint, so their probabilities sum to at most one. This proves necessity.
Solved by gpt-5.6-sol high.
= b
{parent=3k}
{scope}
= Solution
{parent=b}
If one comma codeword were a prefix of another, its terminal comma would occur inside the longer codeword, contrary to the definition. Thus every comma code is prefix-free. Directly, the disjoint cylinder sets of infinite strings beginning with the respective comma-terminated words have measures $q^{-l_i}$, so their total measure gives Kraft's inequality.
Kraft's inequality does not imply unique decipherability. Over the binary alphabet,
$$
C=\{0,01,10\}
$$
has Kraft sum $1/2+1/4+1/4=1$, but
$$
010=0\,|\,10=01\,|\,0.
$$
Thus it is not <prefix code>[uniquely decipherable].
Solved by gpt-5.6-sol high.
= 4I
{parent=Paper 1}
{scope}
{title2=Automata and Formal Languages}
= Solution
{parent=4I}
Fix an effective enumeration of register-machine programs. The $n$th register machine is $P_n$, and its domain
$$
W_n=\{x:P_n\text{ halts on input }x\}
$$
is the $n$th <recursively enumerable set>. A <many-one reduction> $A\leq_mB$ is a total computable function $h$ satisfying $x\in A\iff h(x)\in B$. <Rice theorem> says that every nontrivial property depending only on the computed partial function, or equivalently on an r.e. set in its extensional form, has an undecidable index set.
There is no total equality algorithm: it would decide whether $W_n$ is empty by comparing it with a fixed index for the empty set, contradicting Rice's theorem.
There is no partial algorithm that halts exactly when $W_m=W_n$ either. Given $e$, effectively construct an index $h(e)$ whose machine enumerates nothing unless $P_e(e)$ halts, after which it enumerates $0$. Then
$$
W_{h(e)}=\varnothing\iff e\notin K.
$$
A semialgorithm for equality with a fixed empty index would enumerate the complement of the <halting problem> $K$. Since $K$ is r.e., both it and its complement would then be r.e., making $K$ recursive, a contradiction.
Solved by gpt-5.6-sol high.
= 5J
{parent=Paper 1}
{scope}
{title2=Statistical Modelling}
= Solution
{parent=5J}
For $Y_\mu\sim\operatorname{Poisson}(\mu)$, the <delta method> with $g(y)=2\sqrt y$ gives
$$
\operatorname{Var}(g(Y_\mu))
\approx g'(\mu)^2\operatorname{Var}(Y_\mu)
=\frac1\mu\mu=1.
$$
Thus the transformation stabilizes variance at large $\mu$.
A Gaussian linear model for $\sqrt Y$ treats the transformed observations as having additive constant-variance errors and models $\mathbb E(\sqrt Y\mid X)$ as linear. A <Poisson regression> with square-root link retains the Poisson likelihood and models
$$
\sqrt{\mathbb E(Y\mid X)}=X\beta.
$$
The two procedures therefore have different likelihoods, fitted means, and constraints; transforming the response is not the same as transforming its conditional mean.
Solved by gpt-5.6-sol high.
= 6C
{parent=Paper 1}
{scope}
{title2=Mathematical Biology}
= a
{parent=6c}
{scope}
= Solution
{parent=a}
The positive fixed point satisfies $x_*=x_*e^{r(1-x_*)}$, hence $x_*=1$. Put $x_n=1+u_n$ and retain first-order terms:
$$
1+u_{n+1}=(1+u_n)e^{-ru_{n-1}}
=1+u_n-ru_{n-1}+O(u^2).
$$
Thus
$$
\boxed{u_{n+1}=u_n-ru_{n-1}},
$$
with characteristic polynomial $\lambda^2-\lambda+r$.
Solved by gpt-5.6-sol high.
= b
{parent=6c}
{scope}
= Solution
{parent=b}
The <Jury stability criterion> for $\lambda^2-\lambda+r$ gives both roots inside the unit disc exactly when
$$
\boxed{0<r<1}.
$$
The decay is monotone when both roots are real and positive. Since the discriminant is $1-4r$, this occurs for
$$
\boxed{0<r\leq\frac14}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=6c}
{scope}
= Solution
{parent=c}
For $1/4<r<1$, the roots are a complex-conjugate pair with modulus $\sqrt r<1$. The perturbation therefore changes sign or phase while its envelope decays:
$$
\boxed{\frac14<r<1\quad\text{gives oscillatory decay}.}
$$
Solved by gpt-5.6-sol high.
= d
{parent=6c}
{scope}
= Solution
{parent=d}
At the stability boundary $r_*=1$, the characteristic roots are
$$
\lambda=\frac{1\pm i\sqrt3}{2}=e^{\pm i\pi/3}.
$$
They lie on the unit circle and satisfy $\lambda^6=1$, with no smaller positive common period. Hence
$$
\boxed{r_*=1,\qquad p=6}.
$$
Solved by gpt-5.6-sol high.
= 7E
{parent=Paper 1}
{scope}
{title2=Further Complex Methods}
= Solution
{parent=7E}
Use a keyhole contour about the negative real axis, indented symmetrically around the pole $s=t$. The jump of $s^{z-1}$ across the cut is determined by the chosen branch, and the small and large circular contributions vanish for $0<\operatorname{Re}z<1$. The <residue theorem>, with the symmetric indentation interpreted as a <Cauchy principal value>, gives
$$
\boxed{\operatorname{PV}\int_{-\infty}^{\infty}
\frac{s^{z-1}}{s-t}\,ds=\pi i\,t^{z-1}}.
$$
Splitting the real axis at zero and substituting $s\mapsto-s$ in the negative part gives two linear relations. Solving them yields, for real $0<z<1$,
$$
\boxed{\int_0^\infty\frac{s^{z-1}}{s+t}\,ds
=\pi t^{z-1}\csc(\pi z)},
$$
$$
\boxed{\operatorname{PV}\int_0^\infty\frac{s^{z-1}}{s-t}\,ds
=-\pi t^{z-1}\cot(\pi z)}.
$$
Both sides are holomorphic functions of $z$ throughout the vertical strip $0<\operatorname{Re}z<1$: convergence is locally uniform there, and the trigonometric expressions are holomorphic away from integer poles. The <identity theorem> therefore extends the identities from real $z$ to the whole strip.
Solved by gpt-5.6-sol high.
= 8B
{parent=Paper 1}
{scope}
{title2=Classical Dynamics}
= a
{parent=8b}
{scope}
= Solution
{parent=a}
For a type-two generating function, compare
$$
dF_2=p\,dq+Q\,dP
$$
with
$$
dF_2=\frac{\partial F_2}{\partial q},dq
+\frac{\partial F_2}{\partial P},dP.
$$
The coefficients give the canonical-transformation equations
$$
\boxed{p=\frac{\partial F_2}{\partial q},
\qquad Q=\frac{\partial F_2}{\partial P}}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=8b}
{scope}
= Solution
{parent=b}
Hamilton's equations give
$$
\dot p_i=n_iH_1(p_1,p_2)\sin\theta,
$$
so
$$
\boxed{n_2p_1-n_1p_2\text{ is conserved}}.
$$
Use
$$
F_2=(n_1q_1+n_2q_2)P_1+q_2P_2.
$$
Then
$$
Q_1=\theta,\quad Q_2=q_2,\quad
p_1=n_1P_1,\quad p_2=n_2P_1+P_2,
$$
and $P_2=p_2-(n_2/n_1)p_1$ is the conserved combination. The transformed Hamiltonian depends on $Q_1$ but not $Q_2$. Thus $H$ and $P_2$ are two independent first integrals in involution for two degrees of freedom, proving <Integrable systems>[Liouville integrability].
Solved by gpt-5.6-sol high.
= 9A
{parent=Paper 1}
{scope}
{title2=Cosmology}
= Solution
{parent=9A}
Chemical equilibrium for $p+e\leftrightarrow H$ gives
$$
\mu_H=\mu_p+\mu_e,
$$
and <charge neutrality> gives $n_p=n_e$. Insert the nonrelativistic Maxwell-Boltzmann number densities, use $m_Hc^2=m_pc^2+m_ec^2-E_{\rm bind}$, and neglect the proton correction in the reduced mass. The spin degeneracies cancel in the ground-state approximation. One obtains <Saha ionization equation>[Saha equation]
$$
\boxed{
\frac{n_H}{n_e^2}
=\left(\frac{2\pi\hbar^2}{m_ek_BT}\right)^{3/2}
\exp\!\left(\frac{E_{\rm bind}}{k_BT}\right)}.
$$
Equivalently, its reciprocal is the ionization ratio. The exponential favors neutral hydrogen as the temperature falls, while the translational phase-space factor favors ionization at high temperature.
Solved by gpt-5.6-sol high.
= 10D
{parent=Paper 1}
{scope}
{title2=Quantum Information and Computation}
= a
{parent=10d}
{scope}
= Solution
{parent=a}
The <Quantum no-signalling>[no-signalling theorem] states that any trace-preserving quantum operation performed locally by Bob leaves Alice's reduced density matrix unchanged when Bob's outcome is not communicated:
$$
\operatorname{Tr}_B[(I\otimes\mathcal E_B)(\rho_{AB})]
=\operatorname{Tr}_B\rho_{AB}.
$$
Thus Bob's choice cannot change the statistics of any measurement made by Alice.
Solved by gpt-5.6-sol high.
= b
{parent=10d}
{scope}
= Solution
{parent=b}
A unitary cloner with blank state $|0\rangle$ would satisfy
$$
U(|\psi\rangle|0\rangle)=|\psi\rangle|\psi\rangle
$$
for every $|\psi\rangle\in S$. For $|\psi\rangle,|\phi\rangle\in S$, preservation of inner products gives
$$
\langle\psi|\phi\rangle
=\langle\psi|\phi\rangle^2.
$$
Hence every pair is either orthogonal or represents the same state. Conversely, an orthonormal set can be cloned by defining the map on its basis states and extending it to a unitary. Therefore the <Quantum cloning>[no-cloning theorem] says that a set of pure states is unitarily clonable exactly when its distinct members are mutually orthogonal.
Solved by gpt-5.6-sol high.
= c
{parent=10d}
{scope}
= Solution
{parent=c}
Let Alice and Bob share the <Bell state>
$$
|\Phi^+\rangle
=\frac{|00\rangle+|11\rangle}{\sqrt2}
=\frac{|++\rangle+|--\rangle}{\sqrt2}.
$$
Alice may measure in the computational basis, remotely preparing Bob's ensemble $\{|0\rangle,|1\rangle\}$, or in the Hadamard basis, preparing $\{|+\rangle,|-\rangle\}$. Without her outcome, both ensembles have density matrix $I/2$, as no-signalling requires.
If Bob could clone all four states, the resulting two-qubit ensembles would be
$$
\frac12(|00\rangle\langle00|+|11\rangle\langle11|)
$$
and
$$
\frac12(|++\rangle\langle++|+|--\rangle\langle--|).
$$
They are distinguishable: a $Z\otimes Z$ measurement is perfectly correlated for the first and only equally likely correlated or anticorrelated for the second. Bob could infer Alice's basis choice instantaneously, violating no-signalling. Thus no-signalling implies no-cloning for this set.
Solved by gpt-5.6-sol high.
= 11K
{parent=Paper 1}
{scope}
{title2=Coding and Cryptography}
= a
{parent=11k}
{scope}
= Solution
{parent=a}
A binary <cyclic code> of odd length $n$ is an ideal of $\mathbb F_2[X]/(X^n-1)$. For a primitive $n$th root $\alpha$, its defining set is the set of powers $\alpha^i$ at which every code polynomial vanishes. A <BCH code> of design distance $\delta$ has $\delta-1$ consecutive powers
$$
\alpha^b,\alpha^{b+1},\ldots,\alpha^{b+\delta-2}
$$
in its defining set.
If a nonzero codeword had weight $w<\delta$, write it as $c(X)=\sum_{j=1}^wc_jX^{i_j}$. Evaluation at $w$ consecutive defining roots gives a homogeneous Vandermonde system in the nonzero values $c_j\alpha^{bi_j}$. Its determinant is nonzero because the support elements $\alpha^{i_j}$ are distinct. Thus every $c_j$ would vanish, a contradiction. This proves the <BCH bound>
$$
\boxed{d_{\min}\geq\delta}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=11k}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
The defining roots are $\alpha,\alpha^2,\alpha^3,\alpha^4$. Frobenius conjugacy gives the two binary cyclotomic classes
$$
\{1,2,4,8\},
\qquad
\{3,6,12,9\}.
$$
Hence
$$
m_\alpha=m_{\alpha^2}=m_{\alpha^4}=X^4+X+1,
$$
while $\alpha^3$ has order five and
$$
m_{\alpha^3}=X^4+X^3+X^2+X+1.
$$
The generator is their least common multiple:
$$
\boxed{g(X)=X^8+X^7+X^6+X^4+1}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
For received polynomial $r$, define the syndromes $S_j=r(\alpha^j)$. If the error positions are $i_1,\ldots,i_\nu$, the <error locator polynomial> is
$$
\boxed{\sigma(Z)=\prod_{h=1}^{\nu}(1-\alpha^{i_h}Z)}.
$$
Its roots are $\alpha^{-i_h}$, and its coefficients satisfy the Newton syndrome recurrences. For this design-distance-five code, bounded-distance decoding seeks degree at most two.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
For $r(X)=1+X+X^7$, reduction using $\alpha^4=\alpha+1$ gives
$$
(S_1,S_2,S_3,S_4)=(\alpha^3,\alpha^6,\alpha^8,\alpha^{12}).
$$
The degree-two syndrome recurrence yields
$$
\boxed{\sigma(Z)=1+\alpha^3Z+\alpha^9Z^2}.
$$
Checking the fifteen nonzero field elements shows that this polynomial has no root in $\mathbb F_{16}$. It therefore cannot be expressed as $(1-\alpha^iZ)(1-\alpha^jZ)$ for error positions $i,j$. The received word lies outside the code's two-error decoding radius, so the error positions cannot be determined uniquely by this decoder.
Solved by gpt-5.6-sol high.
= 12I
{parent=Paper 1}
{scope}
{title2=Automata and Formal Languages}
= Solution
{parent=12I}
The <primitive recursive function>[primitive recursive functions] are the smallest class containing zero, successor, and projections and closed under composition and primitive recursion. Addition is defined by
$$
P(m,0)=m,
\qquad P(m,n+1)=S(P(m,n)),
$$
and multiplication by
$$
T(m,0)=0,
\qquad T(m,n+1)=P(T(m,n),m).
$$
Thus both are primitive recursive directly from the definition. Inductively,
$$
T_{k+1}(n_1,\ldots,n_{k+1})
=T(T_k(n_1,\ldots,n_k),n_{k+1})
$$
is primitive recursive. For fixed $a$,
$$
E_a(0)=1,
\qquad E_a(n+1)=T(a,E_a(n)),
$$
so $E_a(n)=a^n$ is primitive recursive.
Encode $(x_0,\ldots,x_{k-1})$ by $2^{x_0}3^{x_1}\cdots p_k^{x_{k-1}}$. The exponent functions $M_{p_i}$ decode the coordinates. Therefore
$$
\overline F(N)=
\prod_{i=0}^{k-1}p_{i+1}^{
f_i(M_2(N),M_3(N),\ldots,M_{p_k}(N))}
$$
is a primitive recursive one-number encoding of $F$; the finite product is built from the primitive recursive multiplication and exponentiation just established.
The Fibonacci function is primitive recursive. Encode the pair $(B(n),B(n+1))$ by
$$
C(n)=2^{B(n)}3^{B(n+1)}.
$$
Starting with $C(0)=3$, primitive recursion using
$$
C(n+1)=2^{M_3(C(n))}
3^{M_2(C(n))+M_3(C(n))}
$$
constructs $C$, and $B(n)=M_2(C(n))$.
There is no universal exponential bound of the stated form. The primitive recursive function
$$
f(n)=2^{n^2}
$$
exceeds $R^n$ for every fixed $R>0$ once $n$ is sufficiently large.
Solved by gpt-5.6-sol high.
= 13J
{parent=Paper 1}
{scope}
{title2=Statistical Modelling}
= a
{parent=13j}
{scope}
= Solution
{parent=a}
The first command counts missing values in every column: there are $177$ missing ages and $687$ missing cabin entries, with none in the other variables. The next command deletes `Cabin`. The final command replaces every missing age by the mean of the observed ages, using `na.rm = TRUE` so missing entries do not contaminate that mean.
Solved by gpt-5.6-sol high.
= b
{parent=13j}
{scope}
= Solution
{parent=b}
The code fits a <Bernoulli logistic-regression model>. Conditionally on the covariates, the responses are independent with
$$
Y_i\sim\operatorname{Bernoulli}(p_i),
\qquad
\log\frac{p_i}{1-p_i}=x_i^T\beta,
$$
where the design includes the displayed numerical predictors and indicator columns for factor levels. It maximizes
$$
L(\beta)=\prod_{i=1}^{889}p_i^{Y_i}(1-p_i)^{1-Y_i}.
$$
<Akaike information criterion> is
$$
\operatorname{AIC}=2k-2\ell(\widehat\beta),
$$
where $k$ is the number of fitted parameters. Backward stepwise selection starts from the full model, tentatively removes each eligible term, chooses the removal producing the lowest AIC, and repeats while AIC decreases. It balances fit against model size rather than testing every coefficient at a fixed significance threshold.
Solved by gpt-5.6-sol high.
= c
{parent=13j}
{scope}
= Solution
{parent=c}
The residual deviance per residual degree of freedom is
$$
\frac{784.21}{880}\approx0.891,
$$
reasonably close to one. The first and third quartiles of the deviance residuals, $-0.591$ and $0.621$, are also fairly close to the standard-normal quartiles $\pm0.674$. There is mild evidence of underdispersion, but nothing in this summary makes the binomial value one plainly unreasonable.
Solved by gpt-5.6-sol high.
= d
{parent=13j}
{scope}
= Solution
{parent=d}
A standard moment estimator is the <Pearson dispersion estimator>
$$
\boxed{
\widehat\phi=
\frac1{n-p}\sum_{i=1}^n
\frac{(Y_i-\widehat\mu_i)^2}{V(\widehat\mu_i)}}.
$$
For Bernoulli data, $V(\mu)=\mu(1-\mu)$. The deviance divided by its residual degrees of freedom is a common alternative.
Solved by gpt-5.6-sol high.
= 14E
{parent=Paper 1}
{scope}
{title2=Further Complex Methods}
= a
{parent=14e}
{scope}
= Solution
{parent=a}
For fixed $s$, the factor $n^{-s}$ grows at most polynomially in $n$, whereas $|z|^n$ decays exponentially when $|z|<1$. The defining series for the <polylogarithm> therefore converges absolutely and locally uniformly for every complex $s$ and $|z|<1$. It is holomorphic in $z$ on the unit disc and entire in $s$ there.
Solved by gpt-5.6-sol high.
= b
{parent=14e}
{scope}
= Solution
{parent=b}
For $|z|<1$ on the negative-real portions of the <Hankel contour>, expand
$$
\frac{z}{e^{-t}-z}=\frac{ze^t}{1-ze^t}
=\sum_{n\geq1}z^ne^{nt}.
$$
Local uniform convergence permits termwise integration. The reciprocal Hankel formula gives
$$
\frac1{2\pi i}\int_{-\infty}^{(0+)}e^{nt}t^{s-1}dt
=\frac{n^{-s}}{\Gamma(1-s)},
$$
so $I(z,s)=\sum_{n\geq1}z^n/n^s=\operatorname{Li}_s(z)$ initially on the disc.
For $z\notin[1,\infty)$, choose the contour around the negative axis so that it passes between zero and every pole satisfying $e^{-t}=z$. On compact subsets of the slit $z$-plane it can be chosen uniformly, and differentiation under the integral proves holomorphy. Deforming without crossing a pole gives a single-valued analytic continuation. Near the cut, the pole $t=-\log z$ lies on opposite sides of the two admissible contours; this is why the contour must pass it consistently.
Solved by gpt-5.6-sol high.
= c
{parent=14e}
{scope}
= Solution
{parent=c}
When $s$ is a nonpositive integer, $t^{s-1}$ is meromorphic and the Hankel contour may be shrunk to a small circle about zero. The integral is then computed by the residue at zero. This gives
$$
\boxed{\operatorname{Li}_0(z)=\frac z{1-z}},
$$
$$
\boxed{\operatorname{Li}_{-1}(z)=\frac z{(1-z)^2}},
$$
$$
\boxed{\operatorname{Li}_{-2}(z)=\frac{z(1+z)}{(1-z)^3}}.
$$
These also follow by repeatedly applying $z\,d/dz$ to the geometric series.
Solved by gpt-5.6-sol high.
= d
{parent=14e}
{scope}
= Solution
{parent=d}
For real $s>0$, collapse the Hankel contour onto the negative axis and use the jump in the argument of $t^{s-1}$. The reflection formula for the <gamma function> yields
$$
\boxed{
I(z,s)=\frac z{\Gamma(s)}
\int_0^\infty\frac{t^{s-1}}{e^t-z},dt},
$$
so $K(z,s,t)=z t^{s-1}/[\Gamma(s)(e^t-z)]$. At positive integer $s$, the original Hankel expression is interpreted by analytic continuation: the pole of $\Gamma(1-s)$ cancels the corresponding zero of the contour integral.
When $z$ crosses the slit at $x>1$, the pole of the real-integral kernel at $t=\log x$ crosses the integration path. The two boundary values differ by $2\pi i$ times its residue. Since $d(e^t-z)/dt=x$ there, that residue is $(\log x)^{s-1}/\Gamma(s)$, giving the jump
$$
\boxed{\frac{2\pi i(\log x)^{s-1}}{\Gamma(s)}}.
$$
Solved by gpt-5.6-sol high.
= 15A
{parent=Paper 1}
{scope}
{title2=Cosmology}
= a
{parent=15a}
{scope}
= Solution
{parent=a}
The homogeneous part of the <continuity equation> gives $\dot{\bar n}+3H\bar n=0$. At first order,
$$
\dot{\delta n}+3H\delta n+\frac{\bar n}{a}\nabla\cdot\delta\mathbf v=0.
$$
For the <density contrast> $\delta=\delta n/\bar n$, the background equation cancels the expansion terms, leaving
$$
\boxed{\dot\delta+\frac1a\nabla\cdot\delta\mathbf v=0}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=15a}
{scope}
= Solution
{parent=b}
The homogeneous Euler equation removes the background force. Products of perturbations are second order, while differentiating the peculiar velocity inside the expanding flow produces the Hubble drag term. The first-order equation is therefore
$$
\boxed{m\bar n a(\dot{\delta\mathbf v}+H\delta\mathbf v)
=-\nabla\delta P-m\bar n\nabla\delta\Phi}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=15a}
{scope}
= Solution
{parent=c}
Take the divergence of the linear Euler equation, use $\delta P=c_s^2m\bar n\delta$, and differentiate the continuity equation. Eliminating $\nabla\cdot\delta\mathbf v$ gives
$$
\ddot\delta+2H\dot\delta
-\frac{c_s^2}{a^2}\nabla^2\delta
-\frac1{a^2}\nabla^2\delta\Phi=0.
$$
The perturbed <Poisson equation> is
$$
\nabla^2\delta\Phi=4\pi Gm\bar n a^2\delta.
$$
Thus
$$
\boxed{
\ddot\delta+2H\dot\delta
-c_s^2\frac1{a^2}(\nabla^2+k_J^2)\delta=0,
\qquad
k_J^2=\frac{4\pi Gm\bar n a^2}{c_s^2}}.
$$
Here $k_J$ is the comoving <Jeans wavenumber>.
Solved by gpt-5.6-sol high.
= d
{parent=15a}
{scope}
= Solution
{parent=d}
Neglecting the pressure and self-gravity bracket leaves
$$
\ddot\delta+2H\dot\delta=0.
$$
During <radiation domination>, $a(t)\propto t^{1/2}$ and $H=1/(2t)$, so
$$
\ddot\delta+\frac1t\dot\delta=0.
$$
Consequently
$$
\boxed{\delta(t)=A+B\log t=A+2B\log a}.
$$
Matter perturbations therefore grow only logarithmically while radiation dominates.
Solved by gpt-5.6-sol high.
= 16F
{parent=Paper 1}
{scope}
{title2=Logic and Set Theory}
= i
{parent=16f}
{scope}
= Solution
{parent=i}
The <Knaster-Tarski theorem> says that the fixed points of a monotone self-map of a complete lattice form a complete lattice. In particular,
$$
\operatorname{lfp}(f)=\bigwedge\{x:f(x)\leq x\},
\qquad
\operatorname{gfp}(f)=\bigvee\{x:x\leq f(x)\}.
$$
For the first formula, let $a$ be the displayed meet. Monotonicity gives $f(a)\leq f(x)\leq x$ for every prefixed point $x$, hence $f(a)\leq a$. Then $f(f(a))\leq f(a)$, so $f(a)$ is itself prefixed; minimality of $a$ gives $a\leq f(a)$. Thus $f(a)=a$. The dual argument gives the greatest fixed point. Applying the same construction above the join of any family of fixed points, and dually below its meet, supplies joins and meets within the fixed-point set.
A down-set contains every element below any of its members. Arbitrary unions and intersections of down-sets are down-sets. Thus the down-sets of $X$, ordered by inclusion, have joins given by unions and meets by intersections, so they form a complete lattice.
For the requested counterexample, take
$$
X=[0,1),\qquad Y=[0,1]
$$
with their usual orders. The set $X$ is a down-set in $Y$, while $Y$ is order-isomorphic to the down-set $[0,1/2]$ of $X$. They are not isomorphic because $Y$ has a greatest element and $X$ does not.
Solved by gpt-5.6-sol high.
= ii
{parent=16f}
{scope}
= Solution
{parent=ii}
Let $f:X\to A\subseteq Y$ be an order isomorphism onto a down-set, and let $g:Y\to U=X\setminus B$ be an order isomorphism onto the complement of a down-set $B$. Define increasing down-sets
$$
D_0=B,
\qquad D_{n+1}=B\cup g(f(D_n)),
\qquad D=\bigcup_{n\geq0}D_n.
$$
The order assumptions ensure inductively that each $D_n$ is a down-set. Define
$$
h(x)=
\begin{cases}
f(x),&x\in D,\\g^{-1}(x),&x\notin D.
\end{cases}
$$
The usual <Cantor-Schröder-Bernstein theorem> orbit argument shows that these two pieces partition both domain and codomain bijectively: points generated from $B$ move forward through $f$, while every remaining point lies in $U$ and moves backward through $g$. Each branch preserves order. If $x\in D$ and $x'<x$ then $x'\in D$; hence the only mixed case has $x\notin D<x'\in D$, and the initial/final-segment hypotheses put $h(x)<h(x')$. Thus $h$ is an order isomorphism and
$$
\boxed{X\cong Y}.
$$
Solved by gpt-5.6-sol high.
= 17F
{parent=Paper 1}
{scope}
{title2=Graph Theory}
= a
{parent=17f}
{scope}
= Solution
{parent=a}
A proper $k$-coloring assigns one of $k$ colors to each vertex so adjacent vertices receive different colors. The <chromatic number> $\chi(G)$ is the least such $k$. Order the vertices arbitrarily and color greedily. At most $\Delta(G)$ colors are forbidden by previously colored neighbors, so
$$
\boxed{\chi(G)\leq\Delta(G)+1}.
$$
Equality occurs for every possible maximum degree: use $K_1$ for $\Delta=0$, $K_2$ for $\Delta=1$, an odd cycle for $\Delta=2$, and $K_{\Delta+1}$ for every $\Delta\geq3$.
Solved by gpt-5.6-sol high.
= b
{parent=17f}
{scope}
= Solution
{parent=b}
A graph is $k$-connected when it has more than $k$ vertices and remains connected after deletion of fewer than $k$ vertices. In a noncomplete 3-connected graph, choose a vertex $v$ with two nonadjacent neighbors $a,b$. Since $G-\{a,b\}$ is connected, take a spanning tree rooted at $v$ and order its vertices so every vertex other than $v$ has a later tree neighbor. Color $a$ and $b$ first with the same color, then greedily color the other vertices in reverse tree order, leaving $v$ last. Every nonfinal vertex has one uncolored neighbor and therefore sees at most $\Delta-1$ colors. At $v$, the two neighbors $a,b$ share a color, so again at most $\Delta-1$ colors occur. Hence
$$
\boxed{\chi(G)\leq\Delta(G)}.
$$
This is the relevant 3-connected case of <Brooks' theorem>.
Solved by gpt-5.6-sol high.
= c
{parent=17f}
{scope}
= Solution
{parent=c}
<Euler's formula> for a connected plane graph is $|V|-|E|+|F|=2$. If the graph is triangle-free, every face has boundary length at least four, so
$$
2|E|\geq4|F|.
$$
Euler's formula then gives $|E|\leq2|V|-4$, so the average degree is less than four. There is a vertex of degree at most three. Delete it, color the remaining graph inductively with four colors, and restore it using a color absent from its at most three neighbors. Thus
$$
\boxed{\chi(G)\leq4}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=17f}
{scope}
= Solution
{parent=d}
The <edge chromatic number> $\chi'(G)$ is the least number of colors in a proper coloring of edges, where incident edges receive distinct colors. <Hall marriage theorem> says a bipartite graph has a matching saturating one side exactly when $|N(S)|\geq|S|$ for every subset $S$ of that side.
In a 4-regular bipartite graph, the $4|S|$ edges leaving $S$ all enter $N(S)$, whose vertices can receive at most $4|N(S)|$ such edges. Hence Hall's condition holds and there is a perfect matching. Remove it and repeat in the resulting 3-, 2-, and 1-regular bipartite graphs. The four perfect matchings give a four-edge-coloring, while every vertex requires four distinct colors. Therefore
$$
\boxed{\chi'(G)=4}.
$$
Solved by gpt-5.6-sol high.
= 18H
{parent=Paper 1}
{scope}
{title2=Galois Theory}
= a
{parent=18h}
{scope}
= Solution
{parent=a}
For roots $r_1,r_2,r_3$, the discriminant is
$$
\Delta(f)=\prod_{i<j}(r_i-r_j)^2,
$$
and for $f=X^3+pX+q$,
$$
\boxed{\Delta=-4p^3-27q^2}.
$$
If $f$ is irreducible, its Galois group is transitive in $S_3$: it is $A_3\cong C_3$ when $\Delta$ is a square in $K$, and $S_3$ otherwise. If $f$ has one root in $K$ and an irreducible quadratic factor, the group is $C_2$; if it splits, it is trivial. A repeated-root depressed cubic already has all roots in $K$, so adds no case.
For $X^3+3X+1$, the rational-root test proves irreducibility and
$$
\Delta=-135=-3^3\cdot5
$$
is not a rational square. Thus $G\cong S_3$. By the <Galois correspondence>[fundamental theorem of Galois theory], the complete subfield list is
$$
\mathbb Q,quad L,quad
\mathbb Q(r_1),\mathbb Q(r_2),\mathbb Q(r_3),quad
\mathbb Q(\sqrt{-135})=\mathbb Q(\sqrt{-15}),
$$
where the three cubic fields fix the three order-two subgroups and the quadratic field fixes $A_3$.
Solved by gpt-5.6-sol high.
= b
{parent=18h}
{scope}
= Solution
{parent=b}
If $L/K$ is Galois, form the norm polynomial
$$
F(X)=\prod_{\sigma\in\operatorname{Gal}(L/K)}\sigma(f)(X).
$$
The Galois group permutes the factors, so $F\in K[X]$. Since the identity factor is $f$, taking $g=F/f\in L[X]$ proves the claim.
If $L/K$ is merely finite separable, place it in a finite Galois closure $M/K$ and use the same product over $\operatorname{Gal}(M/K)$. It gives $F\in K[X]$ divisible by $f$ in $M[X]$. Polynomial division of $F$ by $f$ uses only coefficients in $L$, and the remainder is zero, so the quotient $g=F/f$ actually lies in $L[X]$. It is nonzero and satisfies $fg=F\in K[X]$.
Solved by gpt-5.6-sol high.
= 19H
{parent=Paper 1}
{scope}
{title2=Representation Theory}
= Solution
{parent=19H}
<Maschke's theorem> says that every invariant subspace of a finite-dimensional complex representation of a finite group has an invariant complement. Average any projection over the group to make it equivariant. Induction on dimension therefore decomposes every representation into irreducibles.
The <character of a representation> is $\chi_V(g)=\operatorname{tr}(g|_V)$. For a $G$-set $X$, the permutation representation $\mathbb CX$ has basis $X$, and its character is the number of fixed points:
$$
\chi_{\mathbb CX}(g)=|X^g|.
$$
For the regular action,
$$
\chi_{\mathbb CG}(1)=|G|,
\qquad \chi_{\mathbb CG}(g)=0\quad(g\ne1).
$$
The multiplicity of an irreducible $V_i$ is the character inner product
$$
\langle\chi_{\rm reg},\chi_i\rangle
=\chi_i(1)=\dim V_i,
$$
so
$$
\boxed{\mathbb CG\cong\bigoplus_i(\dim V_i)V_i}.
$$
If $V\cong\bigoplus_i n_iV_i$, <Schur lemma> gives
$$
\boxed{\dim\operatorname{Hom}_G(V,V)=\sum_i n_i^2}.
$$
For $V=\mathbb CG$, this is $\sum_i(\dim V_i)^2=|G|$.
Now suppose $\chi(g)=0$ off the identity. The multiplicity of $V_i$ in $V$ is
$$
\langle\chi,\chi_i\rangle
=\frac{\dim V}{|G|}\dim V_i.
$$
Thus $\dim V/|G|$ is a nonnegative integer and $V$ is that many copies of the regular representation. Finally,
$$
\chi_{W\otimes\mathbb CG}(g)
=\chi_W(g)\chi_{\rm reg}(g)
=(\dim W)\chi_{\rm reg}(g),
$$
so
$$
\boxed{W\otimes\mathbb CG\cong(\dim W)\mathbb CG}.
$$
Solved by gpt-5.6-sol high.
= 20H
{parent=Paper 1}
{scope}
{title2=Number Fields}
= a
{parent=20h}
{scope}
= Solution
{parent=a}
Write $K=\mathbb Q(\alpha)$ with minimal polynomial of degree $n$. An embedding is determined by the image of $\alpha$, and each of the $n$ distinct complex roots gives one embedding, so there are exactly $n$.
For an integral basis $\omega_1,\ldots,\omega_n$, the field discriminant is
$$
d_K=\det(\sigma_i(\omega_j))^2.
$$
Order the embeddings with all real embeddings first and each nonreal embedding adjacent to its conjugate. Replacing each conjugate pair of rows by their real and imaginary parts extracts one factor $2i$ from the determinant. The remaining determinant is real and nonzero. Squaring shows
$$
\boxed{\operatorname{sgn}(d_K)=(-1)^s},
$$
where $s$ is the number of conjugate pairs.
Solved by gpt-5.6-sol high.
= b
{parent=20h}
{scope}
= Solution
{parent=b}
The polynomial $X^3+2X^2+1$ is irreducible by the rational-root test. The power basis $1,\theta,\theta^2$ has discriminant
$$
\operatorname{disc}(X^3+2X^2+1)=-59.
$$
For the order $\mathbb Z[\theta]\subseteq\mathcal O_L$,
$$
\operatorname{disc}(\mathbb Z[\theta])
=[\mathcal O_L:\mathbb Z[\theta]]^2d_L.
$$
Since $59$ is squarefree, the index can only be one. Hence
$$
\boxed{\mathcal O_L=\mathbb Z[\theta]}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=20h}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
If the embedding with unit modulus is real, it sends $\alpha$ to $\pm1$; irreducibility of the minimal polynomial then forces $\alpha=\pm1$. Otherwise $\overline{\sigma_j(\alpha)}=1/\sigma_j(\alpha)$ is another conjugate. Hence $\alpha^{-1}$ is conjugate to $\alpha$, so their monic minimal polynomials coincide. The minimal polynomial is therefore reciprocal up to its constant sign, and its constant term has absolute value one. Taking the product of all conjugates gives
$$
\boxed{|N_{K/\mathbb Q}(\alpha)|=1}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
If $\alpha\in\mathcal O_K$, its norm is an integer. The preceding result gives norm $\pm1$. The characteristic polynomial of multiplication by $\alpha$ then expresses $\alpha^{-1}$ as an integral polynomial in $\alpha$, so $\alpha^{-1}\in\mathcal O_K$. Therefore
$$
\boxed{\alpha\text{ is a unit of }\mathcal O_K}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=c}
{scope}
= Solution
{parent=iii}
Take $K=\mathbb Q(i)$ and
$$
\alpha=\frac{3+4i}{5}.
$$
Its two embeddings give $(3\pm4i)/5$, both of modulus one. But $alpha\notin\mathbb Z[i]=\mathcal O_K$, since its real and imaginary parts are not integers. This is the required example.
Solved by gpt-5.6-sol high.
= 21I
{parent=Paper 1}
{scope}
{title2=Algebraic Topology}
= Solution
{parent=21I}
Chain maps $f,g:C_*\to C_*'$ are <chain homotopy>[chain homotopic] when maps $h_n:C_n\to C'_{n+1}$ satisfy
$$
f-g=\partial h+h\partial.
$$
For a cycle $z$, $f(z)-g(z)=\partial h(z)$ is a boundary, so $f_*=g_*$ on homology.
Fix a vertex $e_0$ of $\Delta^n$ and define the cone operator by adjoining $e_0$ to an oriented simplex, with zero when it is already present. The simplicial boundary formula gives
$$
\partial h+h\partial=\operatorname{id}
$$
on the reduced chain complex. Thus its identity is null-homotopic and
$$
\widetilde H_k(\Delta^n)=0
$$
for every $k$; equivalently, $H_0(\Delta^n)\cong\mathbb Z$ and higher homology vanishes.
For the 2-skeleton $K$ of $\Delta^6$,
$$
\dim C_0=7,
\quad\dim C_1=21,
\quad\dim C_2=35.
$$
It is connected and simply connected, so $H_0\cong\mathbb Z$ and $H_1=0$. Euler characteristic then gives
$$
21=1+\operatorname{rank}H_2,
\qquad
\boxed{H_2(K)\cong\mathbb Z^{20}}.
$$
For $\sigma=(0123)(456)$, no vertex or edge is fixed. The only invariant 2-simplex is $[456]$, on which the 3-cycle preserves orientation, so the chain traces are $0,0,1$. The <Lefschetz number>[Lefschetz trace formula] gives
$$
1=\operatorname{tr}(f_*|H_0)+\operatorname{tr}(f_*|H_2),
$$
therefore
$$
\boxed{\operatorname{tr}(f_*|H_2(K;\mathbb Q))=0}.
$$
Solved by gpt-5.6-sol high.
= 22G
{parent=Paper 1}
{scope}
{title2=Linear Analysis}
= a
{parent=22g}
{scope}
= Solution
{parent=a}
Let $m_N=\inf_{n\geq N}x_n$ and $M_N=\sup_{n\geq N}x_n$. Positivity and normalization give
$$
m_N\leq\phi((x_n)_{n\geq N})\leq M_N.
$$
Shift invariance makes the middle term equal to $\phi(x)$. Letting $N\to\infty$ yields
$$
\boxed{\liminf x_n\leq\phi(x)\leq\limsup x_n}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=22g}
{scope}
= Solution
{parent=b}
No. Shift invariance makes all $\phi(e_n)$ equal. Positivity gives
$$
0\leq N\phi(e_1)=\phi(e_1+\cdots+e_N)\leq\phi(\mathbf1)=1,
$$
so $\phi(e_n)=0$. If an $\ell^1$ sequence represented $\phi$, then $y_n=\phi(e_n)=0$ for every $n$, which would give $\phi(\mathbf1)=0$, a contradiction.
Solved by gpt-5.6-sol high.
= c
{parent=22g}
{scope}
= Solution
{parent=c}
Take $x_n=(-1)^n$. Its shift is $-x$, so linearity and shift invariance give
$$
\phi(x)=\phi(-x)=-\phi(x).
$$
Thus every such functional assigns the common value
$$
\boxed{\phi(x)=0},
$$
although the sequence does not converge.
Solved by gpt-5.6-sol high.
= d
{parent=22g}
{scope}
= Solution
{parent=d}
Let $S$ be the left shift and define
$$
A_px=\frac1p\sum_{k=1}^pS^kx.
$$
Shift invariance gives $\phi(A_px)=\phi(x)$. The hypothesis says $A_px$ converges uniformly in $n$, hence in the $\ell^\infty$ norm, to the constant sequence $y\mathbf1$. By continuity,
$$
\boxed{\phi(x)=\lim_{p\to\infty}\phi(A_px)
=\phi(y\mathbf1)=y}.
$$
Solved by gpt-5.6-sol high.
= 23G
{parent=Paper 1}
{scope}
{title2=Analysis of Functions}
= Solution
{parent=23G}
The <Lebesgue differentiation theorem> states that if $f\in L^1_{\mathrm{loc}}(\mathbb R^n)$, then
$$
\lim_{r\downarrow0}\frac1{\lambda(B_r(x))}\int_{B_r(x)}f(y)\,d\lambda(y)=f(x)
$$
for <almost everywhere>[Lebesgue almost every] $x$. The <Radon-Nikodym theorem> states that if two sigma-finite measures satisfy $\nu\ll\mu$, then there is a nonnegative measurable function $h=d\nu/d\mu$, unique $\mu$-almost everywhere, such that $\nu(E)=\int_Eh\,d\mu$.
For any $t\in[0,1]$, take $z=0$ and let $B$ be a planar sector of angle $2\pi t$. Every disc centred at the origin meets this sector in the proportion
$$
\frac{\lambda(B\cap B_r(0))}{\lambda(B_r(0))}=t,
$$
so its <measure density> at $0$ is $t$.
Apply the differentiation theorem to the <indicator function> $1_A$. Its averages over balls are exactly $\rho_{\lambda,A}(x)$, so the <Lebesgue density theorem> gives
$$
\rho_{\lambda,A}(x)=1_A(x)\in\{0,1\}
$$
almost everywhere. If $\lambda(A)=0$, every numerator vanishes, so the density is zero wherever its denominator is nonzero. Conversely, if the density vanishes almost everywhere, the displayed identity gives $1_A=0$ almost everywhere, hence $\lambda(A)=0$.
Finally suppose $\nu$ and $\lambda$ are <mutually absolutely continuous measures>. Write $d\nu=h\,d\lambda$. Then $0<h(x)<\infty$ almost everywhere by the <positive Radon-Nikodym derivative> result. At almost every $x$, the differentiation theorem applied to both $h$ and $h1_A$ gives
$$
\rho_{\nu,A}(x)
=\lim_{r\downarrow0}
\frac{\int_{B_r(x)}h1_A\,d\lambda}{\int_{B_r(x)}h\,d\lambda}
=\frac{h(x)1_A(x)}{h(x)}=1_A(x).
$$
Thus the $\nu$-density also exists and belongs to $\{0,1\}$ at $\lambda$-almost every point.
Solved by gpt-5.6-sol high.
= 24F
{parent=Paper 1}
{scope}
{title2=Riemann Surfaces}
= a
{parent=24f}
{scope}
= Solution
{parent=a}
The <uniformization theorem> says that every simply connected <Riemann surfaces>[Riemann surface] is conformally equivalent to exactly one of the <Riemann sphere>, the <complex plane>, and the <unit disc>. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a <Möbius transformation> would produce a nonconstant bounded entire function, contradicting the <Liouville theorem>. It is therefore conformally equivalent to the disc, which is the <Riemann mapping theorem>.
Solved by gpt-5.6-sol high.
= b
{parent=24f}
{scope}
= Solution
{parent=b}
Use the principal branch $\operatorname{Log}z=\log|z|+i\arg z$, with $-\pi<\arg z<\pi$. Then
$$
\boxed{f(z)=i\operatorname{Log}z=-\arg z+i\log|z|}
$$
is a <conformal equivalence> from the slit plane to the vertical strip. Its imaginary part tends to $-\infty$ as $z\to0$ and to $+\infty$ as $z\to\infty$.
Solved by gpt-5.6-sol high.
= c
{parent=24f}
{scope}
= Solution
{parent=c}
The map $h(z)=e^{-iz}$ is a conformal equivalence from $E$ to the plane slit along the nonpositive real axis. Conjugating $\phi$ by $h$ gives an automorphism of the slit plane whose two prime ends at $0$ and $\infty$ are fixed. Taking the principal square root identifies the slit plane with the right half-plane; the automorphisms of a half-plane that fix $0$ and $\infty$ are precisely $w\mapsto aw$ with $a>0$. Undoing the square root changes only the positive scale, so
$$
e^{-i\phi(z)}=a e^{-iz}
$$
for some $a>0$. Since $E$ is connected and both sides use the same branch,
$$
\boxed{\phi(z)=z+i\log a}.
$$
Thus $\phi$ is translation by a purely imaginary number.
Solved by gpt-5.6-sol high.
= d
{parent=24f}
{scope}
= Solution
{parent=d}
If $f_0(z)=i\operatorname{Log}z$ is the projection from part (b), then any other Mercator projection $f$ gives an automorphism $f\circ f_0^{-1}$ of $E$ preserving its two ends. Part (c) therefore yields
$$
\boxed{f(z)=i\operatorname{Log}z+ic\qquad(c\in\mathbb R)}.
$$
Every map in this family plainly has the required limiting behaviour.
Solved by gpt-5.6-sol high.
= 25H
{parent=Paper 1}
{scope}
{title2=Algebraic Geometry}
= Solution
{parent=25H}
For an irreducible <algebraic variety> $X$ and $p\in X$, the local ring $\mathcal O_{X,p}$ consists of germs of <rational function>[rational functions] regular on a neighbourhood of $p$. If $X\subseteq\mathbb A^n$ has ideal $I(X)$, its <Zariski tangent space> is
$$
T_pX=\{v\in k^n:df_p(v)=0\text{ for every }f\in I(X)\}.
$$
The given variety is the <blowup of the affine plane at the origin>. On the chart $W\ne0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.
If $(X,Y)\ne(0,0)$, the equation forces
$$
[W:Z]=[X:Y].
$$
Consequently $\pi$ restricts to an isomorphism away from the origin, and is therefore <birational variety>[birational]. Above the origin, however,
$$
\pi^{-1}(0,0)=\{(0,0)\}\times\mathbb P^1,
$$
so $\pi$ is not injective and cannot be an <isomorphism of algebraic varieties>. Birationality gives
$$
\boxed{k(V)=k(\mathbb A^2)=k(X,Y)}.
$$
For any morphism $\varphi:V\to V'$ with $V'$ affine, its restriction to the exceptional curve $E\cong\mathbb P^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V'$ therefore have constant pullbacks. Since $E$ has more than one point, $\varphi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.
Over $\mathbb C$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex <projective variety> is Euclidean compact. Since <compact space>[compactness] is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.
Solved by gpt-5.6-sol high.
= 26I
{parent=Paper 1}
{scope}
{title2=Differential Geometry}
= a
{parent=26i}
{scope}
= Solution
{parent=a}
The <Gauss map> sends $p$ to the chosen <unit normal> $N(p)$. Differentiating $\langle N,N\rangle=1$ shows that $dN_p$ maps $T_pS$ into itself. The <shape operator> $S_p=-dN_p$ is self-adjoint because its associated <second fundamental form> is symmetric. Its <eigenvalue>[eigenvalues] $k_1,k_2$ are the <principal curvature>[principal curvatures]. A point is <umbilical point>[umbilical] when $k_1=k_2$, and the surface is <minimal surface>[minimal] when its <mean curvature> $H=(k_1+k_2)/2$ vanishes everywhere.
In an orthonormal principal basis, $dN_p$ has diagonal matrix $\operatorname{diag}(-k_1,-k_2)$. It is conformal exactly when
$$
\langle dN_pu,dN_pv\rangle=\lambda(p)\langle u,v\rangle,
\qquad \lambda(p)>0,
$$
which is equivalent to $k_1^2=k_2^2\ne0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.
Solved by gpt-5.6-sol high.
= b
{parent=26i}
{scope}
= Solution
{parent=b}
A minimal surface can have a planar point, where $k_1=k_2=0$. There $dN=0$, which fails the stipulated conformality condition $\lambda>0$. Thus minimality need not make the Gauss map conformal everywhere; a plane is the simplest counterexample.
Conversely, conformality gives $|k_1|=|k_2|\ne0$. It permits either $k_1=-k_2$, which is minimal, or $k_1=k_2\ne0$, which is umbilical and nonminimal. On a round <unit sphere>[sphere], the Gauss map is conformal while both principal curvatures are equal and nonzero. Thus conformality alone does not imply minimality.
Solved by gpt-5.6-sol high.
= c
{parent=26i}
{scope}
= Solution
{parent=c}
No. If the image lies in a great circle, there is a fixed unit vector $a$ with $\langle N,a\rangle=0$, so $a$ is tangent to $S$ everywhere. Since the ambient derivative of this constant vector is zero, the <shape operator> satisfies $S(a)=0$. One principal curvature is therefore zero. Minimality makes their sum zero, so both vanish and $S=0$. The Gauss map is then locally constant, and connectedness makes it constant on all of $S$, contradicting its image being a great circle.
Solved by gpt-5.6-sol high.
= 27G
{parent=Paper 1}
{scope}
{title2=Probability and Measure}
= a
{parent=27g}
{scope}
= Solution
{parent=a}
Let $(X_n)$ be <independent random variables> and let
$$
\mathcal T=\bigcap_{n\geq1}\sigma(X_n,X_{n+1},\ldots)
$$
be their <tail sigma-algebra>. The <Kolmogorov zero-one law> states that every $A\in\mathcal T$ has probability zero or one.
Indeed, $A$ is independent of $\sigma(X_1,\ldots,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma(X_1,X_2,\ldots)$, and <independence extended from generating pi-systems> shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,
$$
\mathbb P(A)=\mathbb P(A\cap A)=\mathbb P(A)^2,
$$
so $\mathbb P(A)\in\{0,1\}$.
Solved by gpt-5.6-sol high.
= b
{parent=27g}
{scope}
= Solution
{parent=b}
On a probability space carrying an infinite <IID random variables>[IID] sequence $(X_n)$ with law $m$, map $\omega$ to $(X_1(\omega),X_2(\omega),\ldots)$. Its pushforward measure $\mu$ satisfies
$$
\mu\!\left(\prod_{n\geq1}A_n\right)=\prod_{n\geq1}m(A_n)
$$
for every cylinder set. Cylinder sets form a <pi-system> generating $\sigma(\mathcal C)$, so the <sigma-finite uniqueness theorem for measures> proves uniqueness. This is the countable <product measure> $m^{\otimes\mathbb N}$.
For a cylinder $A=A_1\times\cdots\times A_N\times\mathbb R\times\cdots$,
$$
\mu(\theta^{-1}A)=m(\mathbb R)m(A_1)\cdots m(A_N)=\mu(A).
$$
The same generating-class argument extends this equality to every measurable set, so $\theta$ is a <measure-preserving transformation>.
If $A=\theta^{-1}A$, then $A=\theta^{-n}A$ for every $n$, so membership in $A$ is independent of the first $n$ coordinates. Thus $A$ belongs to the tail sigma-algebra. The zero-one law gives $\mu(A)\in\{0,1\}$, proving that the shift is <ergodic measure-preserving transformation>[ergodic].
Solved by gpt-5.6-sol high.
= 28J
{parent=Paper 1}
{scope}
{title2=Applied Probability}
= a
{parent=28j}
{scope}
= Solution
{parent=a}
A <Q-matrix> satisfies $q_{ij}\geq0$ for $i\ne j$ and $q_{ii}=-\sum_{j\ne i}q_{ij}$; on a countable state space the exit rate $-q_{ii}$ is required to be finite. For a finite-state <continuous-time Markov chain> with <transition semigroup of a continuous-time Markov chain>[transition semigroup] $P(t)$,
$$
Q=P'(0)=\lim_{t\downarrow0}\frac{P(t)-I}{t},
\qquad P(t)=e^{tQ}.
$$
The backward and forward <Kolmogorov equations> are respectively
$$
P'(t)=QP(t),\qquad P'(t)=P(t)Q.
$$
Solved by gpt-5.6-sol high.
= b
{parent=28j}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Put $s=\lambda+\mu$. Solving either Kolmogorov equation gives
$$
P(t)=\frac1s
\begin{pmatrix}
\lambda+\mu e^{-st}&\mu(1-e^{-st})\\
\lambda(1-e^{-st})&\mu+\lambda e^{-st}
\end{pmatrix}.
$$
These entries have the correct initial values and their rows sum to one.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Direct multiplication gives $Q^2=-sQ$, hence for $n\geq1$,
$$
Q^n=(-s)^{n-1}Q.
$$
Therefore the <matrix exponential> is
$$
e^{tQ}=I+\sum_{n\geq1}\frac{t^nQ^n}{n!}
=I+\frac{1-e^{-st}}sQ,
$$
which expands to exactly the transition matrix found in part (i).
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
Writing $\pi=(\pi_1,\pi_2)$, the equations $\pi Q=0$ and $\pi_1+\pi_2=1$ give
$$
\boxed{\pi=\left(\frac\lambda{s},\frac\mu{s}\right)}.
$$
This is the <invariant distribution of a continuous-time Markov chain>[invariant distribution]. Since $e^{-st}\to0$, both rows of $P(t)$ converge to $\pi$, directly verifying convergence to equilibrium.
Solved by gpt-5.6-sol high.
= iv
{parent=b}
{scope}
= Solution
{parent=iv}
The <Markov property> and conditioning on the intermediate state give the bridge probability
$$
\mathbb P(X_t=2\mid X_0=1,X_{3t}=1)
=\frac{p_{12}(t)p_{21}(2t)}{p_{11}(3t)}.
$$
Substitution from part (i) yields
$$
\boxed{
\frac{\lambda\mu(1-e^{-st})(1-e^{-2st})}
{s(\lambda+\mu e^{-3st})}}
\qquad(s=\lambda+\mu).
$$
Solved by gpt-5.6-sol high.
= 29K
{parent=Paper 1}
{scope}
{title2=Principles of Statistics}
= a
{parent=29k}
{scope}
= Solution
{parent=a}
Continuous differentiability gives the first-order expansion
$$
\Phi(\widehat\theta_n)-\Phi(\theta_0)
=\nabla\Phi(\theta_0)^T(\widehat\theta_n-\theta_0)
+o(\|\widehat\theta_n-\theta_0\|).
$$
The assumed convergence in distribution implies $\sqrt n(\widehat\theta_n-\theta_0)=O_p(1)$, so after multiplication by $\sqrt n$ the remainder is $o_p(1)$. The <Slutsky theorem> gives the multivariate <delta method>
$$
\boxed{\sqrt n\bigl(\Phi(\widehat\theta_n)-\Phi(\theta_0)\bigr)
\xrightarrow d\nabla\Phi(\theta_0)^TZ}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=29k}
{scope}
= Solution
{parent=b}
For observations $X_1,\ldots,X_n$, the <log-likelihood> is, up to an additive constant,
$$
\ell(\sigma)=-n\log\sigma-\frac1{2\sigma^2}\sum_iX_i^2.
$$
Its unique maximum is
$$
\widehat\sigma_{\rm MLE}=\sqrt{\frac1n\sum_iX_i^2}.
$$
Now $\mathbb E[X_i^2]=\sigma^2$ and, using $\mathbb E[Z^4]=3$, $\operatorname{Var}(X_i^2)=2\sigma^4$. The <central limit theorem> followed by the <delta method> for $g(x)=\sqrt x$ gives
$$
\boxed{\sqrt n(\widehat\sigma_{\rm MLE}-\sigma)
\xrightarrow dN(0,\sigma^2/2)}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=29k}
{scope}
= Solution
{parent=c}
The one-observation <score function> is
$$
S_\sigma(X)=-\frac1\sigma+\frac{X^2}{\sigma^3}.
$$
Its variance, or equivalently minus the expected second derivative of the log-likelihood, is the <Fisher information matrix>[Fisher information]
$$
\boxed{I(\sigma)=\frac2{\sigma^2}}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=29k}
{scope}
= Solution
{parent=d}
Since $\rho=\sigma^2$ and the reparametrization is one-to-one, invariance of the <maximum-likelihood estimator> gives
$$
\boxed{\widehat\rho_{\rm MLE}=\widehat\sigma_{\rm MLE}^{,2}
=\frac1n\sum_{i=1}^nX_i^2}.
$$
Solved by gpt-5.6-sol high.
= 30K
{parent=Paper 1}
{scope}
{title2=Stochastic Financial Models}
= a
{parent=30k}
{scope}
= Solution
{parent=a}
Let $V(n,x)$ be the minimum expected remaining cost after time $n$, given $X_n=x$. The terminal condition is $V(N,x)=x^2$, and the <Bellman equation> is
$$
V(n-1,x)=\inf_{u\in\mathbb R}
\left\{u^2+\mathbb E[V(n,x+u+\xi_n)]\right\}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=30k}
{scope}
= Solution
{parent=b}
Set $V(n,x)=A_n+B_nx+C_nx^2$. The terminal values are $(A_N,B_N,C_N)=(0,0,1)$. Since the noise is centred, backward induction keeps $B_n=0$. Completing the square in the Bellman equation gives
$$
u^*=-\frac{C_n}{1+C_n}x,qquad
C_{n-1}=\frac{C_n}{1+C_n},qquad
A_{n-1}=A_n+C_n\sigma^2.
$$
Hence
$$
\boxed{C_n=\frac1{N-n+1},\quad B_n=0,\quad
A_n=\sigma^2\sum_{j=n+1}^{N}\frac1{N-j+1}}
$$
(with the empty sum equal to zero). This proves the asserted quadratic form of the <value function>.
Solved by gpt-5.6-sol high.
= c
{parent=30k}
{scope}
= Solution
{parent=c}
At time $n$, the minimizing feedback from part (b) is
$$
u_n^*=-\frac{X_{n-1}}{N-n+2}.
$$
Starting with $u_1^*=-X_0/(N+1)$ and repeatedly substituting
$X_j=X_{j-1}+u_j^*+\xi_j$ gives by induction
$$
X_{n-1}=\frac{N-n+2}{N+1}X_0
+\sum_{j=1}^{n-1}\frac{N-n+2}{N-j+1}\xi_j.
$$
Therefore
$$
\boxed{u_n^*=-\frac{X_0}{N+1}-\frac{\xi_1}{N}
-\frac{\xi_2}{N-1}-\cdots-\frac{\xi_{n-1}}{N-n+2}},
$$
as required.
Solved by gpt-5.6-sol high.
= 31J
{parent=Paper 1}
{scope}
{title2=Mathematics of Machine Learning}
= a
{parent=31j}
{scope}
= Solution
{parent=a}
For a binary function class $\mathcal F$, the <shattering coefficient> is
$$
s(\mathcal F,n)=\max_{x_1,\ldots,x_n}
\left|\{(f(x_1),\ldots,f(x_n)):f\in\mathcal F\}\right|.
$$
Its <VC dimension> is the largest $n$ for which $s(\mathcal F,n)=2^n$, with value infinity if there is no largest such $n$. The <Sauer-Shelah lemma> states that, when $\operatorname{VC}(\mathcal F)=d<\infty$,
$$
s(\mathcal F,n)\leq\sum_{j=0}^{d}\binom nj.
$$
Solved by gpt-5.6-sol high.
= b
{parent=31j}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Choose $2m$ ordered points. Every subset of them has at most $m$ maximal consecutive runs, and those runs can be covered by at most $m$ closed intervals. Thus the points are shattered. On $2m+1$ ordered points, however, the alternating labeling
$$
1,0,1,0,\ldots,0,1
$$
has $m+1$ separated positive runs and cannot be realized by a union of $m$ intervals. Therefore
$$
\boxed{\operatorname{VC}(\mathcal F_1)=2m}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
These are indicators of affine half-spaces in $\mathbb R^p$. Adjoining a constant coordinate embeds their defining functions in the $(p+1)$-dimensional vector space
$$
\{x\mapsto\mu+x^T\beta\}.
$$
The <VC dimension of a vector space> bound therefore gives
$$
\boxed{\operatorname{VC}(\mathcal F_2)\leq p+1}.
$$
In fact equality holds, but only the upper bound is required.
Solved by gpt-5.6-sol high.
= c
{parent=31j}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
Fix sample points $(x_i,z_i)$, $1\leq i\leq n$. A labeling by $h(x,z)=f(x)g(z)$ is the coordinatewise product of one labeling that $\mathcal F$ realizes on $x_1,\ldots,x_n$ and one that $\mathcal G$ realizes on $z_1,\ldots,z_n$. There are at most the product of the two numbers of choices, so
$$
\boxed{s(\mathcal H,n)\leq s(\mathcal F,n)s(\mathcal G,n)}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
For fixed $x_1,\ldots,x_n$, the vector
$$
(h(x_1),\ldots,h(x_n))
$$
is again the coordinatewise product of a labeling realized by $\mathcal F$ and one realized by $\mathcal G$, now on the same sample. Hence there are at most
$$
s(\mathcal F,n)s(\mathcal G,n)
$$
possible products. Maximizing over samples proves the claimed inequality.
Solved by gpt-5.6-sol high.
= d
{parent=31j}
{scope}
= i
{parent=d}
{scope}
= Solution
{parent=i}
For one real coordinate, a union of $m$ intervals labels a fixed $n$-point sample by choosing at most $2m$ endpoints among the $n+1$ gaps, so the number of labelings is at most $(n+1)^{2m}$. Membership in a Cartesian product is the product of the $p$ coordinate-membership indicators. Applying part (c) repeatedly gives
$$
\boxed{s(\mathcal F_3,n)\leq\bigl((n+1)^{2m}\bigr)^p
=(n+1)^{2mp}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=d}
{scope}
= Solution
{parent=ii}
A convex polygon with $m$ sides is an intersection of $m$ affine half-planes. Affine half-plane indicators in $\mathbb R^2$ have VC dimension at most $3$, so the <Sauer-Shelah growth bound> gives at most $(n+1)^3$ labelings on $n$ points. An intersection corresponds to the coordinatewise product of its $m$ half-plane indicators. Repeated use of part (c) therefore gives
$$
\boxed{s(\mathcal F_4,n)\leq\bigl((n+1)^3\bigr)^m
=(n+1)^{3m}}.
$$
Solved by gpt-5.6-sol high.
= 32B
{parent=Paper 1}
{scope}
{title2=Dynamical Systems}
= a
{parent=32b}
{scope}
= Solution
{parent=a}
For the unperturbed <Hamiltonian> system, choose a conserved energy $H(x,y)$. Along the perturbed system,
$$
\dot H=\varepsilon\bigl(H_xp_1+H_yp_2\bigr).
$$
The <energy balance method> integrates this expression around each unperturbed periodic orbit. A zero of the resulting $O(\varepsilon)$ energy change is a necessary leading-order condition for a nearby periodic orbit. A simple zero at which the drift changes from positive inside to negative outside predicts a stable orbit; the opposite change predicts an unstable one.
Solved by gpt-5.6-sol high.
= b
{parent=32b}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
At $\varepsilon=0$,
$$
H(x,y)=2x^2+\frac12y^2
$$
is conserved, and its orbit of amplitude $a>0$ may be written
$$
x=a\cos\theta,\qquad y=-2a\sin\theta,\qquad \theta=2t.
$$
The energy change over one period, to first order in $\varepsilon$, is
$$
\begin{aligned}
\Delta H
&=\varepsilon\int_0^\pi
\left[k(1-2x^2)y^2-(1-3x^2)y^4\right]dt+O(\varepsilon^2)\\
&=\varepsilon\pi a^2\left[2k-(k+6)a^2+3a^4\right]+O(\varepsilon^2)\\
&=3\varepsilon\pi a^2(a^2-2)(a^2-k/3)+O(\varepsilon^2).
\end{aligned}
$$
Here the stated trigonometric integrals, together with
$\int_0^{2\pi}\cos^2\theta\sin^4\theta\,d\theta=\pi/8$, were used. The two candidate leading-order amplitudes are therefore
$$
\boxed{a=\sqrt2,\qquad a=\sqrt{k/3}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
For $0<k<6$, the roots satisfy $\sqrt{k/3}<\sqrt2$. The averaged drift is positive below the first root, negative between the roots, and positive above the second. Hence the orbit of amplitude $\sqrt{k/3}$ is stable and the orbit of amplitude $\sqrt2$ is unstable.
For $k>6$, their order reverses. The same sign calculation shows that the orbit of amplitude $\sqrt2$ is stable and that of amplitude $\sqrt{k/3}$ is unstable. Thus in either case the smaller-amplitude candidate is stable and the larger one unstable.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
At $k=6$, the two zeros coalesce and
$$
\Delta H=3\varepsilon\pi a^2(a^2-2)^2+O(\varepsilon^2).
$$
The first-order drift has a double zero and does not change sign there. The energy balance calculation is therefore degenerate: it neither proves persistence of a periodic orbit nor determines its stability near $k=6$. Higher-order terms are needed.
Solved by gpt-5.6-sol high.
= 33E
{parent=Paper 1}
{scope}
{title2=Integrable Systems}
= a
{parent=33e}
{scope}
= Solution
{parent=a}
Since $L^T=L$ and $B^T=-B$,
$$
[B,L]^T=(BL-LB)^T=L(-B)-(-B)L=[B,L],
$$
so the commutator is symmetric. If $Lv=\lambda v$ with $\lVert v\rVert=1$, symmetry permits a differentiable orthonormal eigenbasis locally, and
$$
\dot\lambda=v^T\dot Lv
=v^T(BL-LB)v
=\lambda v^TBv-\lambda v^TBv=0.
$$
Thus a <Lax pair> evolution is <isospectral Lax equation>[isospectral].
For the harmonic oscillator,
$$
H(q,p)=\frac12(p^2+\omega^2q^2),\qquad
\dot q=p,\qquad \dot p=-\omega^2q.
$$
With the stated $L$, take
$$
B=\begin{pmatrix}0&-\omega/2\\ \omega/2&0\end{pmatrix}.
$$
Direct calculation gives
$$
[B,L]=
\begin{pmatrix}-\omega^2q&\omega p\\
\omega p&\omega^2q\end{pmatrix}
=\dot L.
$$
The eigenvalues of $L$ are
$$
\lambda_\pm=\pm\sqrt{p^2+\omega^2q^2}.
$$
Their constancy proves conservation of the oscillator energy $H=\lambda_+^2/2$.
Solved by gpt-5.6-sol high.
= b
{parent=33e}
{scope}
= Solution
{parent=b}
Differentiating the first auxiliary equation in $t$, the second in $x$, and equating $\psi_{xt}$ and $\psi_{tx}$ leaves precisely
$$
q_t+q_{xxx}=0,
$$
so the <Airy equation> is their <compatibility condition for an overdetermined linear system>[compatibility condition].
Multiplication of $\psi_x-ik\psi=q$ by $e^{-ikx}$ gives
$$
(e^{-ikx}\psi)_x=e^{-ikx}q.
$$
Hence
$$
\psi_+(x,t,k)=e^{ikx}\int_{-\infty}^{x}e^{-iky}q(y,t)\,dy.
$$
For $\operatorname{Im}k\geq0$, the kernel $e^{ik(x-y)}$ is bounded when $y\leq x$; dominated differentiation makes $\psi_+$ analytic for $\operatorname{Im}k>0$. Rapid decrease gives
$$
\lim_{x\to+\infty}e^{-ikx}\psi_+(x,t,k)=\widehat q(k,t).
$$
For real $k$, multiply the time equation by $e^{-ikx}$ and let $x\to+\infty$. The rapidly decreasing terms on its right vanish, leaving
$$
\widehat q_t-ik^3\widehat q=0.
$$
Therefore
$$
\widehat q(k,t)=e^{ik^3t}\widehat q(k,0),
$$
and the <Fourier inversion theorem> gives
$$
\boxed{q(x,t)=\frac1{2\pi}\int_{\mathbb R}
e^{i(kx+k^3t)}\widehat q(k,0)\,dk}.
$$
Solved by gpt-5.6-sol high.
= 34A
{parent=Paper 1}
{scope}
{title2=Principles of Quantum Mechanics}
= a
{parent=34a}
{scope}
= Solution
{parent=a}
From the <number operator> commutator $[N,A]=-A$,
$$
N(A|n\rangle)=(n-1)A|n\rangle,
$$
so $A|n\rangle$ is proportional to $|n-1\rangle$. Its squared norm is
$$
\lVert A|n\rangle\rVert^2=\langle n|A^\dagger A|n\rangle=n.
$$
With the conventional phase choice,
$$
\boxed{A|n\rangle=\sqrt n\,|n-1\rangle}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=34a}
{scope}
= Solution
{parent=b}
Orthogonality of the energy eigenstates gives
$$
\langle z|z\rangle=e^{-|z|^2}\sum_{n=0}^{\infty}\frac{|z|^{2n}}{n!}=1.
$$
Using the <creation and annihilation operators>[lowering-operator] action from part (a) and shifting the summation index,
$$
A|z\rangle=e^{-|z|^2/2}\sum_{n\geq1}
\frac{z^n\sqrt n}{\sqrt{n!}}|n-1\rangle
=z|z\rangle.
$$
Thus this <coherent state> is a normalized eigenstate of $A$ with eigenvalue $z$.
Solved by gpt-5.6-sol high.
= c
{parent=34a}
{scope}
= Solution
{parent=c}
The number probabilities are
$$
\mathbb P(N=n)=e^{-|z|^2}\frac{|z|^{2n}}{n!},
$$
a <Poisson distribution> of parameter $|z|^2$. Hence
$$
\boxed{\langle N\rangle=|z|^2,\qquad
(\Delta N)^2=|z|^2,\qquad \Delta N=|z|}.
$$
The relative uncertainty is $1/|z|=1/\sqrt{\langle N\rangle}$, which tends to zero as the mean occupation tends to infinity.
Solved by gpt-5.6-sol high.
= d
{parent=34a}
{scope}
= Solution
{parent=d}
The <quantum harmonic oscillator> Hamiltonian is
$H=\hbar\omega(N+\tfrac12)$. Therefore
$$
e^{-iHt/\hbar}|z\rangle
=e^{-i\omega t/2}|ze^{-i\omega t}\rangle,
$$
so the evolved state remains an $A$-eigenstate, with eigenvalue
$\boxed{ze^{-i\omega t}}$.
The coherent-state overlap gives
$$
\left|\langle z|ze^{-i\omega t}\rangle\right|^2
=\exp\!\left[-2|z|^2(1-\cos\omega t)\right]
=\exp\!\left[-4|z|^2\sin^2\frac{\omega t}{2}\right].
$$
For the classical period $T=2\pi/\omega$, this probability equals one whenever $t=kT$.
Solved by gpt-5.6-sol high.
= 35D
{parent=Paper 1}
{scope}
{title2=Applications of Quantum Mechanics}
= a
{parent=35d}
{scope}
= Solution
{parent=a}
For incidence from the left, normalize the asymptotic wavefunction by
$$
\psi(x)\sim e^{ikx}+r e^{-ikx}\quad(x\to-\infty),
\qquad
\psi(x)\sim t e^{ikx}\quad(x\to+\infty).
$$
For incidence from the right, define $r',t'$ analogously with the directions reversed. Conservation of the <probability current> gives
$$
|r|^2+|t|^2=1,\qquad |r'|^2+|t'|^2=1.
$$
Applying the same conserved sesquilinear current to the two scattering solutions gives
$$
t'^*r+r'^*t=0.
$$
These are exactly the column orthonormality relations $S^\dagger S=I$ for the stated <one-dimensional S-matrix>, so $S$ is unitary.
Solved by gpt-5.6-sol high.
= b
{parent=35d}
{scope}
= Solution
{parent=b}
The <parity> of a <wavefunction> is its <eigenvalue> under <spatial reflection>: $\psi(-x)=+\psi(x)$ for <even function>[even] parity and $\psi(-x)=-\psi(x)$ for <odd function>[odd] parity. If $V(-x)=V(x)$, the <Hamiltonian operator>[Hamiltonian] commutes with reflection, so its <energy eigenspaces> admit a parity <eigenbasis>. In particular, every <nondegenerate energy eigenvalue>[nondegenerate] bound-state <eigenstate>[energy eigenfunction] has definite parity; within a <state degeneracy>[degenerate] scattering eigenspace one must choose the even and odd <linear combinations>.
Solved by gpt-5.6-sol high.
= c
{parent=35d}
{scope}
= Solution
{parent=c}
Put
$$
\kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.
$$
An even solution is proportional to $\cosh(\kappa x)$ inside the <potential barrier>. Normalize the right exterior region as
$$
\psi(x)=e^{-ikx}+(r+t)e^{ikx}\qquad(x>a/2).
$$
Continuity of $\psi$ and $\psi'$ at $x=a/2$ gives, with
$\gamma=\kappa\tanh(\kappa a/2)$,
$$
\frac{ik\bigl((r+t)e^{ika/2}-e^{-ika/2}\bigr)}
{(r+t)e^{ika/2}+e^{-ika/2}}=\gamma.
$$
Solving,
$$
\boxed{r+t=e^{-ika}\frac{ik+\gamma}{ik-\gamma}},
\qquad
\gamma=\kappa\tanh(\kappa a/2).
$$
Its modulus is one, as expected for the even-parity eigenvalue of the unitary S-matrix.
Solved by gpt-5.6-sol high.
= 36A
{parent=Paper 1}
{scope}
{title2=Statistical Physics}
= a
{parent=36a}
{scope}
= Solution
{parent=a}
A <grand canonical ensemble> describes a system that exchanges both energy and particles with a reservoir, at fixed $T,V,\mu$. If microstate $n$ has occupation $N_n$ and one-particle energy $E_n$, then
$$
\boxed{\mathcal Z=\sum_{\{N_n\}}
\exp\!\left[-\beta\sum_nN_n(E_n-\mu)\right]},
\qquad \beta=(k_BT)^{-1},
$$
with the allowed occupations chosen for the particle statistics.
Solved by gpt-5.6-sol high.
= b
{parent=36a}
{scope}
= Solution
{parent=b}
The <grand potential> is
$$
\boxed{\Phi=E-TS-\mu\langle N\rangle
=-k_BT\log\mathcal Z}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=36a}
{scope}
= Solution
{parent=c}
Extensivity makes $\Phi(T,V,\mu)$ proportional to $V$. Comparing its differential
$d\Phi=-S\,dT-p\,dV-\langle N\rangle d\mu$
with Euler scaling gives
$$
\boxed{\Phi=-pV}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=36a}
{scope}
= i
{parent=d}
{scope}
= Solution
{parent=i}
The one-particle translational partition function is
$$
q_1=\frac{V}{h^3}\int_{\mathbb R^3}
e^{-\beta p^2/(2m)}\,d^3p
=\frac{V}{\lambda_T^3},
\qquad
\lambda_T=\frac{h}{\sqrt{2\pi mk_BT}}.
$$
For a classical <ideal gas>,
$$
\boxed{\mathcal Z=\sum_{N=0}^\infty
\frac{(e^{\beta\mu}q_1)^N}{N!}
=\exp\!\left(\frac{Ve^{\beta\mu}}{\lambda_T^3}\right)},
$$
and hence
$$
\boxed{\Phi=-\frac{k_BTVe^{\beta\mu}}{\lambda_T^3}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=d}
{scope}
= Solution
{parent=ii}
Differentiating $\log\mathcal Z$ with respect to $\beta\mu$ gives
$$
\boxed{\langle N\rangle=\frac{Ve^{\beta\mu}}{\lambda_T^3}},
\qquad
(\Delta N)^2=\frac{\partial^2\log\mathcal Z}
{\partial(\beta\mu)^2}=\langle N\rangle.
$$
Thus
$$
\frac{\Delta N}{\langle N\rangle}
=\frac1{\sqrt{\langle N\rangle}},
$$
which vanishes in the thermodynamic limit.
Solved by gpt-5.6-sol high.
= iii
{parent=d}
{scope}
= Solution
{parent=iii}
Combining $\Phi=-pV$ with part (i) and the expression for $\langle N\rangle$ gives the <ideal gas>[ideal-gas equation of state]
$$
\boxed{pV=\langle N\rangle k_BT}.
$$
Solved by gpt-5.6-sol high.
= e
{parent=36a}
{scope}
= Solution
{parent=e}
For relativistic one-particle energy
$\epsilon(p)=\sqrt{p^2c^2+m^2c^4}$, replace $q_1$ by
$$
q_1^{\rm rel}=\frac{V}{h^3}\int_{\mathbb R^3}
e^{-\beta\epsilon(p)}\,d^3p.
$$
Classical particle counting still gives
$$
\mathcal Z=\exp(e^{\beta\mu}q_1^{\rm rel}),\qquad
\langle N\rangle=e^{\beta\mu}q_1^{\rm rel},
\qquad
(\Delta N)^2=\langle N\rangle.
$$
Since $q_1^{\rm rel}$ is proportional to $V$, $\Phi=-k_BT\langle N\rangle=-pV$, and therefore
$$
\boxed{pV=\langle N\rangle k_BT,\qquad
\frac{\Delta N}{\langle N\rangle}
=\frac1{\sqrt{\langle N\rangle}}}.
$$
The classical ideal-gas equation of state is independent of the dispersion relation.
Solved by gpt-5.6-sol high.
= 37B
{parent=Paper 1}
{scope}
{title2=Electrodynamics}
= a
{parent=37b}
{scope}
= Solution
{parent=a}
Dot the vacuum <Ampere-Maxwell equation> with $\mathbf E$ and <Faraday's law> with $\mathbf B/\mu_0$. The vector identity
$\nabla\cdot(\mathbf E\times\mathbf B)
=\mathbf B\cdot\nabla\times\mathbf E-\mathbf E\cdot\nabla\times\mathbf B$
gives the <Poynting theorem>
$$
\partial_t\left(\frac{\epsilon_0E^2}{2}
+\frac{B^2}{2\mu_0}\right)
+\nabla\cdot\left(\frac1{\mu_0}\mathbf E\times\mathbf B\right)=0.
$$
Thus
$$
\boxed{\mathbf N=\frac1{\mu_0}\mathbf E\times\mathbf B}
$$
is the <Poynting vector>, the electromagnetic energy-flux density.
Solved by gpt-5.6-sol high.
= b
{parent=37b}
{scope}
= Solution
{parent=b}
Differentiate $\mathbf g=\epsilon_0\mathbf E\times\mathbf B$, use the vacuum <Maxwell equations>, and use $\nabla\cdot\mathbf E=\nabla\cdot\mathbf B=0$. Componentwise vector identities then give
$$
\partial_tg_j+\partial_i\sigma_{ij}=0,
$$
where
$$
\boxed{\sigma_{ij}
=\epsilon_0\left(\frac12E^2\delta_{ij}-E_iE_j\right)
+\frac1{\mu_0}\left(\frac12B^2\delta_{ij}-B_iB_j\right)}.
$$
This is the negative of the conventional <Maxwell stress tensor>; with the sign used here, its divergence is the outward momentum flux.
Solved by gpt-5.6-sol high.
= c
{parent=37b}
{scope}
= Solution
{parent=c}
With $x^0=ct$, so that $\partial_0=c^{-1}\partial_t$, the $\nu=0$ component of
$\partial_\mu T^{\mu\nu}=0$ is
$$
c^{-1}\partial_t\epsilon+\partial_i(N_i/c)=0,
$$
and the $\nu=j$ component is
$$
\partial_tg_j+\partial_i\sigma_{ij}=0.
$$
These are exactly the energy and momentum conservation laws from parts (a) and (b), now combined into conservation of the electromagnetic <stress-energy tensor>.
Solved by gpt-5.6-sol high.
= d
{parent=37b}
{scope}
= Solution
{parent=d}
Let $J_k=\int\epsilon_{k\ell j}x_\ell g_j\,d^3x$. Local momentum conservation and integration by parts, with the boundary term vanishing for a localized field, give
$$
\dot J_k
=-\int\epsilon_{k\ell j}x_\ell\partial_i\sigma_{ij}\,d^3x
=\int\epsilon_{kij}\sigma_{ij}\,d^3x=0,
$$
because $\sigma_{ij}$ is symmetric. Hence the total electromagnetic angular momentum is independent of time.
Solved by gpt-5.6-sol high.
= e
{parent=37b}
{scope}
= Solution
{parent=e}
For $T^{\mu\nu}$ to be symmetric in every inertial frame, its mixed components must satisfy
$$
T^{0i}=cg_i=c\epsilon_0(\mathbf E\times\mathbf B)_i
=T^{i0}=\frac{N_i}{c}
=\frac{(\mathbf E\times\mathbf B)_i}{\mu_0c}.
$$
For arbitrary fields this requires
$$
\boxed{\mu_0\epsilon_0=\frac1{c^2}}.
$$
Solved by gpt-5.6-sol high.
= 38D
{parent=Paper 1}
{scope}
{title2=General Relativity}
= a
{parent=38d}
{scope}
= Solution
{parent=a}
For $k=-1$, the <Friedmann-Lemaitre-Robertson-Walker metric> is
$$
ds^2=-dt^2+a^2(t)\left[d\chi^2+\sinh^2\chi
\,(d\theta^2+\sin^2\theta\,d\phi^2)\right].
$$
The empty first <Friedmann equation>, with $\Lambda=0$, gives
$$
\frac{\dot a^2-1}{a^2}=0.
$$
Choosing the expanding branch and shifting the origin of proper time yields
$$
\boxed{a(t)=t,\qquad t>0}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=38d}
{scope}
= Solution
{parent=b}
For an FLRW spacetime,
$$
R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}
=12\left[\left(\frac{\ddot a}{a}\right)^2
+\left(\frac{\dot a^2+k}{a^2}\right)^2\right].
$$
With $a=t$ and $k=-1$, both terms vanish, so
$$
\boxed{R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}=0}.
$$
Indeed the vacuum equations give a vanishing Ricci tensor, and conformal flatness of an FLRW metric then makes the whole Riemann tensor vanish. The apparent singularity at $t=0$ is therefore only a coordinate singularity.
Solved by gpt-5.6-sol high.
= c
{parent=38d}
{scope}
= Solution
{parent=c}
Introduce
$$
T=t\cosh\chi,\qquad R=t\sinh\chi.
$$
Then
$$
-dT^2+dR^2+R^2d\Omega^2
=-dt^2+t^2(d\chi^2+\sinh^2\chi\,d\Omega^2),
$$
which is the <Minkowski metric>. Since $t>0$,
$$
T^2-R^2=t^2>0,\qquad T>0,
$$
so these coordinates cover exactly the interior of the future light cone of its vertex $p$. The $t=\text{constant}$ slices are future hyperboloids. Their spatial isometries are precisely the proper orthochronous <Lorentz transformation>[Lorentz transformations] of Minkowski spacetime that fix $p$.
Solved by gpt-5.6-sol high.
= 39C
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics II}
= a
{parent=39c}
{scope}
= Solution
{parent=a}
Steadiness, axial and rotational symmetry, and the absence of any imposed radial or azimuthal motion make the velocity independent of $z$ and $\theta$ and directed along the cylinders. Incompressibility is then automatic, so
$$
\boxed{\mathbf u=u(r)\mathbf e_z}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=39c}
{scope}
= Solution
{parent=b}
The radial momentum equation makes $p$ independent of $r$. The axial <Navier-Stokes equation> reduces to
$$
\boxed{0=-\frac{dp}{dz}
+\mu\frac1r\frac d{dr}\left(r\frac{du}{dr}\right)+\rho g}.
$$
The no-slip boundary conditions are
$$
\boxed{u(\alpha R)=V,\qquad u(R)=0}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=39c}
{scope}
= Solution
{parent=c}
The reduced axial equation shows that $dp/dz$ is constant, because its other terms depend only on $r$. Both open ends are at the same atmospheric pressure. Since a linear pressure cannot take the same value at the two ends unless its slope vanishes,
$$
\boxed{\frac{dp}{dz}=0}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=39c}
{scope}
= Solution
{parent=d}
Put
$$
G=\frac{\rho g}{4\mu},\qquad
A=\frac{V-GR^2(1-\alpha^2)}{\log\alpha}.
$$
Integrating the equation in part (b) and imposing both no-slip conditions gives
$$
\boxed{u(r)=G(R^2-r^2)+A\log\frac rR}.
$$
For $V=0$, gravity drives downward annular Poiseuille flow, positive in the interior and zero at both walls. Positive $V$ raises the inner-wall endpoint to $V$ and strengthens the downward flow; sufficiently negative $V$ creates a region of upward flow near the inner cylinder. In each case the graph is the displayed quadratic-plus-logarithmic profile.
Solved by gpt-5.6-sol high.
= e
{parent=39c}
{scope}
= Solution
{parent=e}
The shear stress is $\tau_{rz}=\mu u'(r)$, where
$$
u'(r)=-\frac{\rho g}{2\mu}r+\frac Ar.
$$
Thus the axial forces exerted by the fluid per unit length on the inner and outer cylinders are
$$
\boxed{F_{\rm in}=2\pi\alpha R\mu u'(\alpha R)
=2\pi\mu A-\pi\rho g\alpha^2R^2},
$$
$$
\boxed{F_{\rm out}=-2\pi R\mu u'(R)
=\pi\rho gR^2-2\pi\mu A}.
$$
Their sum is
$$
F_{\rm in}+F_{\rm out}=\pi\rho gR^2(1-\alpha^2),
$$
the weight per unit length of the fluid in the annulus. This is the expected overall vertical force balance.
Solved by gpt-5.6-sol high.
= 40C
{parent=Paper 1}
{scope}
{title2=Waves}
= a
{parent=40c}
{scope}
= Solution
{parent=a}
Write $\rho=\rho_0+\widetilde\rho$, $p=p_0+\widetilde p$, and let $\widetilde{\mathbf u}$ be small. Linearized mass and momentum conservation are
$$
\partial_t\widetilde\rho+\rho_0\nabla\cdot\widetilde{\mathbf u}=0,
\qquad
\rho_0\partial_t\widetilde{\mathbf u}=-\nabla\widetilde p.
$$
The linearized equation of state is
$\widetilde p=c_0^2\widetilde\rho$. Differentiate the first equation in time and eliminate the velocity with the second to obtain the <wave equation>
$$
\boxed{\partial_t^2\widetilde p-c_0^2\nabla^2\widetilde p=0}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=40c}
{scope}
= Solution
{parent=b}
With the convention $\widetilde p=\operatorname{Re}(\widehat p e^{i\omega t})$, a wave travelling in the positive $x$-direction has
$$
\boxed{\widehat p(x)=P e^{-ikx},\qquad k=\omega/c_0}.
$$
The linearized momentum equation gives
$$
\boxed{\widetilde u=\frac{\widetilde p}{\rho_0c_0}}
$$
for this right-moving wave; the sign is reversed for a left-moving wave.
Solved by gpt-5.6-sol high.
= c
{parent=40c}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
In $x<0$, write
$$
\widehat p=e^{-ikx}+Re^{ikx}.
$$
In the cavity, the rigid condition $\widehat u(L)=0$ gives
$\widehat p=C\cos k(x-L)$. Continuity of fluid velocity at the membrane and the membrane equation
$$
-\omega^2m\,\widehat\xi
=\widehat p(0^-)-\widehat p(0^+),
\qquad \widehat u(0)=i\omega\widehat\xi,
$$
give, after eliminating $C$ and $\widehat\xi$,
$$
\boxed{R=\frac{\cos\beta+(\alpha-i)\sin\beta}
{\cos\beta+(\alpha+i)\sin\beta}},
\qquad
\alpha=\frac{\omega m}{\rho_0c_0},\quad
\beta=\frac{\omega L}{c_0}.
$$
The numerator is the complex conjugate of the denominator, hence $|R|=1$. No mean acoustic energy can pass through the rigid end, so all incident energy is ultimately reflected.
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
The membrane is stationary when the incident and reflected velocities cancel, so $R=1$. The formula gives this while allowing a nonzero cavity pressure when
$$
\boxed{\sin\beta=0,\qquad \beta=n\pi}.
$$
These are standing-wave resonances of the closed cavity: both the membrane at $x=0$ and the rigid wall at $x=L$ are velocity nodes, while the pressure field is nontrivial.
Solved by gpt-5.6-sol high.
= iii
{parent=c}
{scope}
= Solution
{parent=iii}
Take the heavy-membrane limit
$$
\boxed{\alpha=\frac{\omega m}{\rho_0c_0}\to\infty}
$$
away from the resonances $\sin\beta=0$. Then $R\to1$, the membrane velocity tends to zero, and the cavity amplitude obtained from the matching equations is $O(\alpha^{-1})$ relative to the pressure in $x<0$. The membrane therefore acts as an almost rigid reflecting wall and scarcely excites the cavity.
Solved by gpt-5.6-sol high.
= 41C
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}
= a
{parent=41c}
{scope}
= Solution
{parent=a}
Since $H$ is diagonalizable, write $H=V\Lambda V^{-1}$. Then
$$
z^{(k)}=H^kz^{(0)}=V\Lambda^kV^{-1}z^{(0)}.
$$
If the <spectral radius> $\rho(H)<1$, every eigenvalue satisfies $|\lambda_i|<1$, so $\Lambda^k\to0$ and hence $z^{(k)}\to0$ for every initial vector. Conversely, taking $z^{(0)}$ to be an eigenvector of any eigenvalue $\lambda$ shows that convergence for every initial vector forces $\lambda^k\to0$, hence $|\lambda|<1$. Therefore
$$
\boxed{z^{(k)}\to0\text{ for all }z^{(0)}
\quad\Longleftrightarrow\quad \rho(H)<1}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=41c}
{scope}
= Solution
{parent=b}
Because $A$ is <Hermitian positive-definite matrix>[symmetric positive definite],
$$
\nabla f(x)=Ax-b,\qquad \nabla^2f(x)=A\succ0.
$$
Thus the only stationary point is $x_*=A^{-1}b$, and
$$
f(x)=f(x_*)+\frac12(x-x_*)^TA(x-x_*).
$$
The final term is positive unless $x=x_*$, proving that $A^{-1}b$ is the unique minimizer.
Solved by gpt-5.6-sol high.
= c
{parent=41c}
{scope}
= Solution
{parent=c}
The <gradient descent> iteration is
$$
\boxed{x^{(k+1)}=(I-\alpha A)x^{(k)}+\alpha b}.
$$
For the error $e^{(k)}=x^{(k)}-A^{-1}b$,
$$
e^{(k+1)}=(I-\alpha A)e^{(k)}.
$$
Part (a) shows that convergence for every initial vector is equivalent to
$|1-\alpha\lambda_i|<1$ for every eigenvalue of $A$. Since all eigenvalues are positive,
$$
\boxed{0<\alpha<\frac2{\lambda_{\max}(A)}}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=41c}
{scope}
= Solution
{parent=d}
Since $r^{(k)}=b-Ax^{(k)}=-\nabla f(x^{(k)})$, exact line search minimizes
$f(x^{(k)}+\alpha r^{(k)})$. Differentiating with respect to $\alpha$ gives
$$
-(r^{(k)})^Tr^{(k)}
+\alpha(r^{(k)})^TAr^{(k)}=0,
$$
so
$$
\boxed{\alpha^{(k)}
=\frac{(r^{(k)})^Tr^{(k)}}
{(r^{(k)})^TAr^{(k)}}}.
$$
Moreover,
$$
r^{(k+1)}=r^{(k)}-\alpha^{(k)}Ar^{(k)},
$$
and substitution of the displayed step size yields
$$
\boxed{(r^{(k)})^Tr^{(k+1)}=0}.
$$
Thus consecutive residuals are orthogonal.
Solved by gpt-5.6-sol high.