past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperib_1_2023.pdf

= 1F
{parent=Paper 1}
{scope}
{title2=Linear Algebra}

= Solution
{parent=1F}

Choose <bases> of $V$ and $W$, of dimensions $v$ and $w$. A <linear map> is uniquely determined by the arbitrary images of the $v$ <basis> <vectors>, each with $w$ coordinates, so
$$
\dim L(V,W)=vw.
$$

Put $a=\dim A$ and $b=\dim B$, and extend <bases> of $A,B$ to <bases> of $V,W$. The condition $\phi(A)\subseteq B$ forces the $(w-b)a$ <matrix> entries from $A$ to a complement of $B$ to vanish and imposes no other restriction. Hence
$$
\dim X=vw-a(w-b).
$$

For the last part let $r=\dim(S\cap T)=s+t-v$. Choose
$$
V=(S\cap T)\oplus S_0\oplus T_0.
$$
A map in $Y$ sends $S\cap T$ into itself, $S_0$ into $S$, and $T_0$ into $T$, with these choices independent. Therefore
$$
\dim Y=r^2+s(s-r)+t(t-r),
\qquad r=s+t-v.
$$

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 1}
{scope}
{title2=Geometry}

= Solution
{parent=2F}

A <topological surface> is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of $\mathbb R^2$.

The antipodal action on $S^2$ is free. The <surface quotient by a free finite action> applies: Around each point choose a small <open disc> disjoint from its antipodal image; the quotient map restricts to a homeomorphism from that disc onto an open neighbourhood in the quotient. Compactness gives Hausdorffness and second countability descends from the sphere. Thus the quotient is the <real projective plane>, in particular a topological surface.

For the second quotient write a point away from the poles as
$$
(\sqrt{1-z^2}\cos\phi,\sqrt{1-z^2}\sin\phi,z).
$$
The map
$$
q(\phi,z)=(2\phi,z)
$$
in these cylindrical coordinates extends continuously over the poles and identifies exactly $\phi$ with $\phi+\pi$. It therefore induces a continuous bijection $S^2/\!\approx\,\to S^2$. The domain is compact and the sphere Hausdorff, so this bijection is a homeomorphism.

Solved by gpt-5.6-sol high.

= 3B
{parent=Paper 1}
{scope}
{title2=Complex Analysis OR Complex Methods}

= a
{parent=3b}
{scope}

= Solution
{parent=a}

The <Laurent series> is
$$
e^{1/z}=\sum_{k=0}^{\infty}\frac{1}{k!z^k},
\qquad 0<|z|<\infty.
$$
Thus zero is an essential singularity.

Solved by gpt-5.6-sol high.

= b
{parent=3b}
{scope}

= Solution
{parent=b}

On $|z|\geq\rho$, put $w=1/z$, so $|w|\leq1/\rho$. The Taylor <polynomials>
$$
f_n(z)=\sum_{k=0}^n\frac{w^k}{k!}
$$
converge uniformly there to $e^w$. Moreover
$$
|e^w|=e^{\Re w}\geq e^{-1/\rho}.
$$
For all sufficiently large $n$, the uniform error is below half this lower bound, so $|f_n(z)|>0$ whenever $|z|\geq\rho$. Every zero therefore lies in $|z|<\rho$.

Solved by gpt-5.6-sol high.

= 4C
{parent=Paper 1}
{scope}
{title2=Variational Principles}

= Solution
{parent=4C}

For a regular constraint $g=0$, an interior constrained extremum satisfies
$$
\nabla f=\lambda\nabla g,
$$
together with the constraint; boundary cases must also be checked.

Here maximize $f=xyz$ subject to $x^2+y^2+z^2=L^2$. The <Lagrange multiplier> equations give
$$
yz=2\lambda x,\qquad xz=2\lambda y,\qquad xy=2\lambda z.
$$
At a positive maximizer these imply $x=y=z=L/\sqrt3$. Boundary points have zero volume, so the maximum is
$$
\boxed{\frac{L^3}{3\sqrt3}}.
$$

Solved by gpt-5.6-sol high.

= 5B
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= a
{parent=5b}
{scope}

= Solution
{parent=a}

For any $h$,
$$
\|A(c^*+h)-y\|_2^2
=\|Ac^*-y\|_2^2+2h^TA^T(Ac^*-y)+\|Ah\|_2^2.
$$
If the <normal equation> $A^TAc^*=A^Ty$ holds, the last two terms reduce to $\|Ah\|^2\geq0$, so $c^*$ minimizes. Conversely, at a minimizer the directional <derivative> in every $h$ vanishes, forcing $A^T(Ac^*-y)=0$.

Solved by gpt-5.6-sol high.

= b
{parent=5b}
{scope}

= Solution
{parent=b}

If $A$ has full column rank, then for every nonzero $h$,
$$
h^TA^TAh=\|Ah\|^2>0.
$$
Thus $A^TA$ is positive definite and invertible, and the normal equation has the unique solution
$$
c^*=(A^TA)^{-1}A^Ty.
$$

Solved by gpt-5.6-sol high.

= 6H
{parent=Paper 1}
{scope}
{title2=Statistics}

= a
{parent=6h}
{scope}

= Solution
{parent=a}

For a null parameter space $\Theta_0\subseteq\Theta$, the generalized likelihood ratio is
$$
\Lambda=\frac{\sup_{\theta\in\Theta_0}L(\theta)}
{\sup_{\theta\in\Theta}L(\theta)}.
$$
The <Wilks theorem> states, under standard regularity conditions and under the null, that
$$
-2\log\Lambda\Rightarrow\chi^2_{\dim\Theta-\dim\Theta_0}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=6h}
{scope}

= Solution
{parent=b}

The null hypothesis is that mortality is independent of carbolic-acid use, equivalently that the two mortality probabilities are equal. The two-sided alternative is that they differ.

Use the <Chi-squared test of independence>, equivalently the likelihood-ratio statistic, with Pearson's statistic
$$
X^2=\sum_{i,j}\frac{(O_{ij}-E_{ij})^2}{E_{ij}},
$$
where $E_{ij}$ are the fitted counts under independence using the observed row and column totals. Under the null, $X^2$ is asymptotically $\chi^2_1$. A size-$\alpha$ test rejects when
$$
X^2>\chi^2_{1,1-\alpha}.
$$

Solved by gpt-5.6-sol high.

= 7H
{parent=Paper 1}
{scope}
{title2=Optimisation}

= Solution
{parent=7H}

The <minimum-cost flow> problem chooses, for each directed edge $(i,j)\in E$, a flow $x_{ij}$ and solves
$$
\min_x\sum_{(i,j)\in E}C_{ij}x_{ij}
$$
subject to
$$
\sum_{j:(j,i)\in E}x_{ji}-\sum_{j:(i,j)\in E}x_{ij}=b_i,
\qquad
M_{ij}\leq x_{ij}\leq\overline M_{ij}.
$$
Feasibility requires $\sum_i b_i=0$.

Set $y_{ij}=x_{ij}-M_{ij}$. Then
$$
0\leq y_{ij}\leq\overline M_{ij}-M_{ij},
$$
and its balance <vector> is
$$
b_i'=b_i+\sum_jM_{ij}-\sum_jM_{ji}.
$$
The objective becomes $\sum C_{ij}y_{ij}$ plus the constant $\sum C_{ij}M_{ij}$. Translation by $M$ is a bijection between feasible flows and preserves their ordering by cost, so the transformed zero-lower-bound problem is equivalent.

Solved by gpt-5.6-sol high.

= 8F
{parent=Paper 1}
{scope}
{title2=Linear Algebra}

= a
{parent=8f}
{scope}

= Solution
{parent=a}

The statement is true. By <Jordan normal form>, the characteristic <polynomial> records the total size of the blocks for each <eigenvalue>, while the minimal <polynomial> records the largest block for each <eigenvalue>. In dimension three this determines every block partition: for algebraic multiplicity three the possibilities $3$, $2+1$, and $1+1+1$ have largest block sizes three, two, and one; multiplicities one and two are equally immediate. Thus the two <matrices> have the same Jordan form and are conjugate. This is the <characteristic and minimal polynomials determine similarity in dimension three> phenomenon.

Solved by gpt-5.6-sol high.

= b
{parent=8f}
{scope}

= Solution
{parent=b}

The statement is false. At <eigenvalue> $2$, algebraic multiplicity two and minimal-polynomial exponent two force one block $J_2(2)$. At <eigenvalue> $1$, the blocks have total size five and largest size two, so there are only two possible partitions,
$$
2+2+1\qquad\hbox{or}\qquad2+1+1+1.
$$
Hence there are exactly two conjugacy classes satisfying the given <polynomial> data, represented by
$$
J_2(2)\oplus J_2(1)\oplus J_2(1)\oplus J_1(1)
$$
and
$$
J_2(2)\oplus J_2(1)\oplus J_1(1)\oplus J_1(1)\oplus J_1(1).
$$
There cannot be three mutually non-conjugate examples.

Solved by gpt-5.6-sol high.

= c
{parent=8f}
{scope}

= Solution
{parent=c}

The statement is true. Since $\alpha$ is invertible, every <eigenvalue> $\lambda$ is nonzero. If $\alpha$ had a nontrivial Jordan block $\lambda I+N$, then for the stated positive integer $N_0$,
$$
(\lambda I+N)^{N_0}
=\lambda^{N_0}I+N_0\lambda^{N_0-1}N+\cdots
$$
would still have a nonzero nilpotent part and could not be diagonalizable. Therefore all Jordan blocks of $\alpha$ have size one. This also proves <diagonalizability inherited from an invertible power>.

Solved by gpt-5.6-sol high.

= d
{parent=8f}
{scope}

= Solution
{parent=d}

The statement is false. The real <matrix>
$$
A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$
has the two distinct complex <eigenvalues> $i$ and $-i$, so it is diagonalizable over $\mathbb C$. It has no real <eigenvalue> and hence no real eigenbasis, so it is not diagonalizable over $\mathbb R$.

Solved by gpt-5.6-sol high.

= e
{parent=8f}
{scope}

= Solution
{parent=e}

The statement is true. Suppose $A,B$ are real and $AP=PB$ for an invertible complex <matrix> $P=X+iY$, with $X,Y$ real. Taking real and imaginary parts gives
$$
AX=XB,\qquad AY=YB.
$$
The <polynomial> $q(t)=\det(X+tY)$ is not identically zero, since $q(i)=\det P\ne0$. Choose a real $t$ outside its finite zero set. Then $Q=X+tY$ is real and invertible, and $AQ=QB$. Thus $A$ and $B$ are conjugate over $\mathbb R$, as stated by <complex similarity of real matrices implies real similarity>.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 1}
{scope}
{title2=Groups, Rings and Modules}

= i
{parent=9e}
{scope}

= Solution
{parent=i}

Assume (i). For $a,b\in R$, write the <ideal> $(a,b)=(d)$. Then $d=ra+sb$ for some $r,s\in R$. Since $a,b\in(d)$, the element $d$ divides both; and every common divisor of $a,b$ divides the displayed linear combination $d$. Thus $d$ is a greatest common divisor and (ii) holds.

Solved by gpt-5.6-sol high.

= ii
{parent=9e}
{scope}

= Solution
{parent=ii}

Assume (ii), and let an $R$-submodule $M\subseteq F$ be generated by $x_1,\ldots,x_m$. Choose a common nonzero denominator $d$ and write $x_j=a_j/d$ with $a_j\in R$. Repeated application of (ii) shows that
$$
(a_1,\ldots,a_m)=(g)
$$
for one element $g$: the gcd at each step is a linear combination of the elements processed so far. Therefore
$$
M=R\frac gd,
$$
so (iii) holds. This is the finite-ideal property defining a <Bézout domain>.

Solved by gpt-5.6-sol high.

= iii
{parent=9e}
{scope}

= Solution
{parent=iii}

Assume (iii). Because $R$ is a <Noetherian ring>, every <ideal> $I\subseteq R$ is finitely generated. Regard $I$ as an $R$-submodule of $F$. By (iii), $I=Rx$ for some $x\in F$; since $x\in I\subseteq R$ unless $I=0$, this is a principal <ideal> of $R$. Thus every <ideal> is principal and (i) holds.

Together with the previous two implications, this proves (i), (ii), and (iii) equivalent.

Solved by gpt-5.6-sol high.

= iv
{parent=9e}
{scope}

= Solution
{parent=iv}

If (i) holds, the <submodule theorem for free modules over a principal ideal domain> says that any submodule $N\subseteq R^n$ is free. Its rank is at most $n$, since tensoring the inclusion with $F$ embeds $N\otimes_R F$ into $F^n$. Hence $N$ has at most $n$ generators, proving (iv).

Conversely, apply (iv) with $n=1$. Every <ideal> is an $R$-submodule of $R$ and therefore has one generator, which is (i). Thus all four conditions are equivalent.

Solved by gpt-5.6-sol high.

= Solution
{parent=9E}

Suppose the additive <group> of the <integral> domain $R$ is isomorphic to $\mathbb Z^n$. Every <ideal> $I$ is then an additive <subgroup> of a finitely generated free abelian <group>, so it has finitely many additive generators $x_1,\ldots,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb Z$ in $R$. Thus every <ideal> is finitely generated, proving that $R$ is Noetherian by <noetherianity from finite additive rank>.

For an example that fails (i)--(iv), take
$$
R=\mathbb Z[\sqrt{-5}].
$$
It is an <integral> domain and has additive <group> $\mathbb Z^2$. The <ideal>
$$
I=(2,1+\sqrt{-5})
$$
has index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is $\mathbb F_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have <determinant> and absolute index
$$
a^2+5b^2=2,
$$
which has no integer solution. Hence $I$ is not principal, as detailed in <nonprincipal ideal in the integers adjoined a square root of minus five>. Condition (i), and therefore all four equivalent conditions, fails.

Solved by gpt-5.6-sol high.

= 10G
{parent=Paper 1}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=10G}

A <Cauchy sequence> $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\varepsilon>0$ there is $N$ such that $d(x_m,x_n)<\varepsilon$ whenever $m,n\geq N$. A <complete metric space> is one in which every Cauchy <sequence> converges to a point of the space.

Every Cauchy <sequence> is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\geq N$, and put
$$
R=1+\max_{1\leq j<N}d(x_j,x_N).
$$
Then every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing <sequence> of nonempty closed sets with $\operatorname{diam}F_n\to0$. Choose $x_n\in F_n$. Given $\varepsilon>0$, choose $N$ with $\operatorname{diam}F_N<\varepsilon$. For $m,n\geq N$, both points lie in $F_N$, so $d(x_m,x_n)<\varepsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore
$$
x\in\bigcap_{n=1}^{\infty}F_n.
$$

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. Define
$$
F_n=\overline{\{x_m:m\geq n\}}.
$$
These sets are nonempty, closed, and decreasing. The Cauchy property implies $\operatorname{diam}F_n\to0$; taking a closure does not change the diameter. Choose $x\in\bigcap_nF_n$. Since $x_n,x\in F_n$,
$$
d(x_n,x)\leq\operatorname{diam}F_n\longrightarrow0.
$$
Thus every Cauchy <sequence> converges and $M$ is complete. This proves the <Cantor intersection theorem> characterization.

The <contraction mapping theorem> states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda\in\Lambda$, the map $T_\lambda(x)=T(\lambda,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^*(\lambda)$. This defines the required unique <function>. Fix $\lambda_0$. The fixed-point identities and the triangle inequality give
$$
\begin{aligned}
d(x^*(\lambda),x^*(\lambda_0))
&\leq d(T(\lambda,x^*(\lambda)),T(\lambda,x^*(\lambda_0)))\\
&\quad+d(T(\lambda,x^*(\lambda_0)),T(\lambda_0,x^*(\lambda_0)))\\
&\leq k\,d(x^*(\lambda),x^*(\lambda_0))
+d(T(\lambda,x^*(\lambda_0)),T(\lambda_0,x^*(\lambda_0))).
\end{aligned}
$$
Consequently
$$
d(x^*(\lambda),x^*(\lambda_0))
\leq\frac{d(T(\lambda,x^*(\lambda_0)),T(\lambda_0,x^*(\lambda_0)))}{1-k}.
$$
The numerator tends to zero as $\lambda\to\lambda_0$ by the assumed continuity for the fixed point $x^*(\lambda_0)$. Hence $x^*$ is continuous, an instance of <continuous dependence of the fixed point of a uniform contraction>.

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 1}
{scope}
{title2=Geometry}

= Solution
{parent=11F}

A subset $S\subset\mathbb R^3$ is a smooth surface if every point has a neighbourhood in $S$ parametrized by a map $X:U\to S$, where $U\subset\mathbb R^2$ is open, $X$ is a homeomorphism onto that neighbourhood, $X$ is smooth, and $DX$ has rank two everywhere. These are the <embedded surface parametrization> conditions.

For the given set use local angular intervals in the parametrization
$$
X(\theta,t)=(\phi(t)\cos\theta,\phi(t)\sin\theta,t).
$$
It is locally one-to-one and has <tangent vectors>
$$
X_\theta=(-\phi\sin\theta,\phi\cos\theta,0),
\qquad
X_t=(\phi'\cos\theta,\phi'\sin\theta,1).
$$
Their cross product has magnitude
$$
|X_\theta\times X_t|=\phi(t)\sqrt{1+\phi'(t)^2}>0,
$$
so the <derivative> has rank two. The angular charts cover $\Sigma$, proving that it is a smooth surface.

The area between heights $a_0$ and $b_0$ is therefore
$$
2\pi\int_{a_0}^{b_0}\phi(t)\sqrt{1+\phi'(t)^2}\,dt.
$$
By hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,
$$
\phi(t)\sqrt{1+\phi'(t)^2}=r,
$$
and squaring gives
$$
r^2=\phi(t)^2+\phi(t)^2\phi'(t)^2.
$$

If $0<\phi(t)<r$, then $\phi'$ never vanishes, so its sign $\sigma\in\{1,-1\}$ is constant. The last equation gives
$$
\phi'=\sigma\frac{\sqrt{r^2-\phi^2}}{\phi},
$$
and hence
$$
\frac d{dt}\sqrt{r^2-\phi(t)^2}=-\sigma.
$$
Thus $\sqrt{r^2-\phi(t)^2}=-\sigma(t-t_0)$ for a constant $t_0$, and
$$
(t-t_0)^2+\phi(t)^2=r^2.
$$
The graph lies on a circle of radius $r$, exactly as described by <constant strip-area density of a surface of revolution>.

Solved by gpt-5.6-sol high.

= 12G
{parent=Paper 1}
{scope}
{title2=Complex Analysis P OR Complex Methods}

= a
{parent=12g}
{scope}

= Solution
{parent=a}

Termwise <differentiation> inside the disc gives
$$
f'(z)=\sum_{n=1}^{\infty}(1-z)^{n-1}=\frac1z.
$$
Consequently
$$
\frac d{dz}\bigl(ze^{-f(z)})
=e^{-f(z)}(1-zf'(z))=0.
$$
Since $f(1)=0$, the constant is one, so $e^{f(z)}=z$. Thus $f$ is an analytic branch of the logarithm on $D(1,1)$ with the required value.

Given $a\in D$, write $a=|a|e^{i\theta_a}$ with $0<\theta_a<2\pi$. On $|z/a-1|<1$, define
$$
\ell_a(z)=f(z/a)+\log|a|+i\theta_a.
$$
Then $e^{\ell_a(z)}=z$ and $\operatorname{Im}\ell_a(a)=\theta_a$. After shrinking the neighbourhood of $a$, continuity keeps its imaginary part in $(0,2\pi)$.

Solved by gpt-5.6-sol high.

= b
{parent=12g}
{scope}

= Solution
{parent=b}

Near each $a\in D$, the local branch from part (a), chosen to have value $\log|a|+i\theta(a)$ at $a$, agrees with
$$
L(z)=\log|z|+i\theta(z).
$$
Indeed, two logarithms of the same nonzero number differ by $2\pi i$ times an integer, and the integer is locally constant; it is zero at $a$. Hence $L$ is locally analytic and therefore analytic on $D$. This is the <analytic logarithm on the positive-axis slit plane>.

For $-1<\alpha<1$, an analytic branch is
$$
z^\alpha=\exp(\alpha L(z)).
$$

Solved by gpt-5.6-sol high.

= c
{parent=12g}
{scope}

= Solution
{parent=c}

The <residue theorem> states that if a meromorphic <function> has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour <integral> is $2\pi i$ times the sum of the enclosed residues.

Apply it to
$$
F(z)=\frac{e^{\alpha L(z)}}{(1+z)^2}
$$
on a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha<1$ and $\alpha>-1$, respectively. On the upper bank the numerator tends to $x^\alpha$, while on the lower bank it tends to $e^{2\pi i\alpha}x^\alpha$ and the direction is reversed. Therefore the limiting contour <integral> is
$$
(1-e^{2\pi i\alpha})I.
$$

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,
$$
\operatorname{Res}_{z=-1}F
=\left.\frac d{dz}e^{\alpha L(z)}\right|_{z=-1}
=-\alpha e^{i\pi\alpha}.
$$
The residue theorem now gives
$$
(1-e^{2\pi i\alpha})I=-2\pi i\alpha e^{i\pi\alpha}.
$$
For $\alpha\ne0$, division and
$$
1-e^{2\pi i\alpha}=-2ie^{i\pi\alpha}\sin(\pi\alpha)
$$
yield
$$
\boxed{I=\frac{\pi\alpha}{\sin(\pi\alpha)}}.
$$
At $\alpha=0$, the <integral> is one, agreeing with the continuous <limit>. This is the <positive-axis keyhole beta integral>.

Solved by gpt-5.6-sol high.

= 13A
{parent=Paper 1}
{scope}
{title2=Methods}

= a
{parent=13a}
{scope}

= Solution
{parent=a}

The Lagrange identity for this <Sturm-Liouville theory> operator is
$$
y_0L(y;\lambda_0)-yL(y_0;\lambda_0)
=\frac d{dx}\left[p(y_0y'-yy_0')\right].
$$
Integrating from zero to one, the boundary term vanishes because both $y$ and $y_0$ vanish at both endpoints. Since $L(y_0;\lambda_0)=0$ and $L(y;\lambda_0)=f$, it follows that
$$
\int_0^1y_0f\,dx=0.
$$
This is the <solvability condition at a Sturm-Liouville eigenvalue>.

Solved by gpt-5.6-sol high.

= b
{parent=13a}
{scope}

= Solution
{parent=b}

Because the parameter enters through $\lambda wy$,
$$
L(y;\lambda_0)
=L(y;\lambda)-(\lambda-\lambda_0)wy
=y^{m+1}-\varepsilon^m\mu wy.
$$
Use this as $f$ in part (a). The necessary orthogonality condition is
$$
0=\int_0^1y_0\left(y^{m+1}-\varepsilon^m\mu wy\right)dx.
$$
With $y=\varepsilon y_0+\varepsilon^2y_1$ and $\mu=O(1)$,
$$
y^{m+1}=\varepsilon^{m+1}y_0^{m+1}+O(\varepsilon^{m+2}),
\qquad
wy=\varepsilon wy_0+O(\varepsilon^2).
$$
Consequently
$$
0=\varepsilon^{m+1}
\left[
\int_0^1y_0^{m+2}\,dx
-\mu\int_0^1wy_0^2\,dx
\right]
+O(\varepsilon^{m+2}).
$$
The normalization makes the second <integral> one. Divide by $\varepsilon^{m+1}$ to obtain
$$
\boxed{\mu=\int_0^1y_0^{m+2}\,dx+O(\varepsilon)}.
$$
This is the <leading nonlinear eigenvalue shift in a Sturm-Liouville problem>.

Solved by gpt-5.6-sol high.

= 14D
{parent=Paper 1}
{scope}
{title2=Quantum Mechanics}

= a
{parent=14d}
{scope}

= Solution
{parent=a}

For a normalized state, write
$$
A'=\widehat A-\langle\widehat A\rangle_\psi,
\qquad
B'=\widehat B-\langle\widehat B\rangle_\psi.
$$
Then $\Delta_\psi A=\|A'\psi\|$ and $\Delta_\psi B=\|B'\psi\|$. The Schwarz inequality gives
$$
(\Delta_\psi A)(\Delta_\psi B)
\geq |(A'\psi,B'\psi)|
\geq |\operatorname{Im}(A'\psi,B'\psi)|.
$$
Since $A'$ and $B'$ are Hermitian and their commutator equals $[\widehat A,\widehat B]$,
$$
2i\operatorname{Im}(A'\psi,B'\psi)
=(\psi,[\widehat A,\widehat B]\psi).
$$
Therefore
$$
(\Delta_\psi A)(\Delta_\psi B)
\geq\frac12\left|(\psi,[\widehat A,\widehat B]\psi)\right|,
$$
which is the <Robertson uncertainty principle>.

Solved by gpt-5.6-sol high.

= b
{parent=14d}
{scope}

= Solution
{parent=b}

Because $\widehat X-i\lambda\widehat Y$ is the adjoint of $\widehat X+i\lambda\widehat Y$,
$$
f(\lambda)=\|(\widehat X+i\lambda\widehat Y)\psi\|^2\geq0.
$$
Expanding gives the real quadratic
$$
f(\lambda)
=\langle\widehat X^2\rangle
+i\lambda\langle[\widehat X,\widehat Y]\rangle
+\lambda^2\langle\widehat Y^2\rangle.
$$
Its discriminant must be nonpositive. Since the expectation of a commutator of Hermitian operators is purely imaginary,
$$
4\langle\widehat X^2\rangle\langle\widehat Y^2\rangle
\geq\left|\langle[\widehat X,\widehat Y]\rangle\right|^2.
$$
Taking $\widehat X=\widehat A-\langle\widehat A\rangle$ and $\widehat Y=\widehat B-\langle\widehat B\rangle$ and then taking square roots gives the stated uncertainty relation. This is the <quadratic-norm proof of the Heisenberg uncertainty relation>.

Solved by gpt-5.6-sol high.

= c
{parent=14d}
{scope}

= Solution
{parent=c}

For a <differentiable> wave <function>,
$$
[\widehat x,\widehat p]\psi
=-i\hbar x\psi'+i\hbar(x\psi)'
=i\hbar\psi.
$$
Thus $[\widehat x,\widehat p]=i\hbar I$. Substitution into part (a) yields
$$
\boxed{\Delta_\psi x\,\Delta_\psi p\geq\frac\hbar2}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=14d}
{scope}

= Solution
{parent=d}

For $\psi(x)=Ce^{-\alpha x^2}$,
$$
\psi''=(4\alpha^2x^2-2\alpha)\psi.
$$
The stationary <Schrodinger equation> becomes
$$
\left[
\frac{\hbar^2\alpha}{m}
+\left(k-\frac{2\hbar^2\alpha^2}{m}\right)x^2
\right]\psi=E\psi.
$$
Hence, choosing the positive root and a positive normalization constant,
$$
\alpha=\frac1\hbar\sqrt{\frac{mk}{2}},
\qquad
E=\hbar\sqrt{\frac{k}{2m}},
\qquad
C=\left(\frac{2\alpha}{\pi}\right)^{1/4}.
$$
The state is even, so $\langle x\rangle=\langle p\rangle=0$. The Gaussian <integrals> give
$$
(\Delta x)^2=\frac1{4\alpha},
\qquad
(\Delta p)^2=\hbar^2\alpha,
$$
and therefore $\Delta x\Delta p=\hbar/2$. These values are collected in <gaussian eigenstate for a quadratic potential>.

Finally, equality in the derivation of the uncertainty relation requires the centred <vectors> to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,
$$
(\widehat p-p_0)\psi=is(\widehat x-x_0)\psi.
$$
In position space this says
$$
\psi'=\left(\frac{ip_0}{\hbar}-\frac{s(x-x_0)}{\hbar}\right)\psi,
$$
whose normalizable solutions are
$$
\psi(x)=C_0\exp\left(\frac{ip_0x}{\hbar}-\frac{s(x-x_0)^2}{2\hbar}\right).
$$
Thus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the <equality case of the Heisenberg uncertainty relation>.

Solved by gpt-5.6-sol high.

= 15D
{parent=Paper 1}
{scope}
{title2=Electromagnetism}

= Solution
{parent=15D}

In SI units the <Maxwell equations> in free space are
$$
\nabla\cdot E=\frac{\rho}{\epsilon_0},
\qquad
\nabla\cdot B=0,
$$
$$
\nabla\times E=-\frac{\partial B}{\partial t},
\qquad
\nabla\times B=\mu_0J+\mu_0\epsilon_0\frac{\partial E}{\partial t}.
$$

Solved by gpt-5.6-sol high.

= a
{parent=15d}
{scope}

= Solution
{parent=a}

Take the divergence of the <Ampère-Maxwell equation>. Since the divergence of a curl is zero, <Gauss's law> gives
$$
0=\mu_0\nabla\cdot J
+\mu_0\epsilon_0\frac{\partial}{\partial t}(\nabla\cdot E)
=\mu_0\left(\nabla\cdot J+\frac{\partial\rho}{\partial t}\right).
$$
Thus
$$
\frac{\partial\rho}{\partial t}+\nabla\cdot J=0,
$$
which is <charge conservation from Maxwell equations>. Integrating over a fixed volume and using the divergence theorem yields
$$
\frac{dQ}{dt}=-\int_{\partial V}J\cdot n\,dS.
$$
Charge in $V$ is conserved provided no current crosses $\partial V$. For total charge in all space, the corresponding assumption is sufficient decay of $J$ so that the flux at infinity vanishes.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

In empty space the fields are divergence-free. Taking the curl of <Faraday's law> and using
$$
\nabla\times(\nabla\times E)=\nabla(\nabla\cdot E)-\nabla^2E
$$
gives
$$
\nabla^2E-\mu_0\epsilon_0\frac{\partial^2E}{\partial t^2}=0.
$$
The same calculation, starting from the <Ampère-Maxwell equation>, gives the corresponding equation for $B$. Thus every component satisfies the <electromagnetic wave equation> with speed
$$
c=\frac1{\sqrt{\mu_0\epsilon_0}}
\simeq3.0\times10^8\ \mathrm{m\,s^{-1}}.
$$
Its agreement with the measured speed of light identifies light as an <electromagnetic wave>.

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

Dot the <Ampère-Maxwell equation> with $E$, dot <Faraday's law> with $B/\mu_0$, and use the stated <vector> identity. The result is the local balance law
$$
\frac{\partial}{\partial t}
\left[
\frac12\left(\epsilon_0E^2+\frac{B^2}{\mu_0}\right)
\right]
+\nabla\cdot S=-J\cdot E,
\qquad
S=\frac1{\mu_0}E\times B.
$$
Here $S$ is the <Poynting vector>. Integration over $V$ gives the <Poynting theorem>
$$
\frac{dU}{dt}
=-\int_{\partial V}S\cdot n\,dS
-\int_VJ\cdot E\,d^3x.
$$
The field energy decreases through outward electromagnetic energy flux and through work done on charges.

Solved by gpt-5.6-sol high.

= d
{parent=15d}
{scope}

= Solution
{parent=d}

The source-free equations imply $k\cdot E_0=0$, $\omega=c|k|$, and
$$
B=\frac{k\times E_0}{\omega}\sin(k\cdot x-\omega t).
$$
Therefore
$$
S=\frac{E_0\times(k\times E_0)}{\mu_0\omega}
\sin^2(k\cdot x-\omega t)
=\epsilon_0c|E_0|^2\sin^2(k\cdot x-\omega t)\,\widehat k.
$$
It is parallel to $k$, and its period average is
$$
\langle S\rangle
=\frac12\epsilon_0c|E_0|^2\widehat k.
$$
The average electric and magnetic energy densities are equal:
$$
\langle u_E\rangle=\langle u_B\rangle
=\frac14\epsilon_0|E_0|^2.
$$
Hence $\langle u\rangle=\epsilon_0|E_0|^2/2$ and $\langle S\rangle=c\langle u\rangle\widehat k$, as summarized by <energy density and flux of a plane electromagnetic wave>.

Solved by gpt-5.6-sol high.

= e
{parent=15d}
{scope}

= Solution
{parent=e}

For isotropic radiation, the average power is spread over a sphere. At $r=0.10\ \mathrm m$, the <intensity of an isotropic radiator> is
$$
\langle|S|\rangle=\frac{P}{4\pi r^2}
=\frac1{4\pi(0.10)^2}
\simeq7.96\ \mathrm{W\,m^{-2}}.
$$
Using the result of part (d),
$$
|E_0|=\sqrt{\frac{2\langle|S|\rangle}{\epsilon_0c}}
\simeq77\ \mathrm{V\,m^{-1}}.
$$

Solved by gpt-5.6-sol high.

= 16C
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics}

= a
{parent=16c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Across the two vertical faces of a control rectangle, the mass fluxes are
$$
\rho u(y,t)\,\delta y
$$
with opposite signs, since $u$ is independent of $x$. There is no flux across the horizontal faces because the $y$-velocity is zero. The net mass flux is therefore zero, equivalently
$$
\nabla\cdot u=\frac{\partial u}{\partial x}=0.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The <pressure> force in the $x$-direction on a rectangle of dimensions $\delta x\times\delta y$ is
$$
-\frac{\partial p}{\partial x}\,\delta x\delta y+o(\delta x\delta y).
$$
The shear stress is $\tau_{xy}=\mu u_y$, so the net viscous force is
$$
\mu u_{yy}\,\delta x\delta y+o(\delta x\delta y).
$$
There is no convective <acceleration>. Dividing the <momentum> balance by the area and taking the <limit> gives
$$
\boxed{\rho u_t=-\frac{dp}{dx}+\mu u_{yy}}.
$$
The $y$-momentum balance gives $p_y=0$, so the <pressure> depends only on $x$.

Solved by gpt-5.6-sol high.

= b
{parent=16c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For steady flow the <momentum> equation is
$$
\mu u''=G.
$$
Applying no slip, $u(0)=0$ and $u(h)=U$, gives the <Couette-Poiseuille flow in a thin gap>
$$
\boxed{u(y)=\frac{Uy}{h}+\frac{G}{2\mu}y(y-h)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The fluid shear stress is
$$
\tau_{xy}(y)=\mu u'(y)
=\frac{\mu U}{h}+G\left(y-\frac h2\right).
$$
The condition $\tau_{xy}(0)=0$ therefore gives
$$
\boxed{G_0=\frac{2\mu U}{h^2}}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

At $G=G_0$, the linear terms cancel and
$$
u(y)=U\frac{y^2}{h^2}.
$$
The volume flux per unit width is
$$
Q=\int_0^h u(y)\,dy
=\boxed{\frac{Uh}{3}}.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

At the upper wall,
$$
\mu u'(h)=\frac{2\mu U}{h}.
$$
This is the shear exerted by the plate on the fluid in the positive $x$-direction. By action and reaction, the flow exerts on the top plate the stress
$$
\boxed{-\frac{2\mu U}{h}e_x}.
$$

The same result follows from a force balance on a rectangle of length $\delta x$ and full height $h$. The <pressure> force is $-G_0h\,\delta x\,e_x$, the bottom shear vanishes, and equilibrium requires the top plate to exert $G_0h\,\delta x\,e_x$ on the fluid. Thus the fluid exerts $-G_0h\,e_x=-2\mu Ue_x/h$ on the plate. These results form the <Couette-Poiseuille flow with a stress-free stationary wall>.

Solved by gpt-5.6-sol high.

= 17B
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= a
{parent=17b}
{scope}

= Solution
{parent=a}

The <function> is not Lipschitz on $\mathbb R$. For $\varepsilon>0$,
$$
\frac{|f(\varepsilon)-f(-\varepsilon)|}{|\varepsilon-(-\varepsilon)|}
=\frac2{2\varepsilon}=\frac1\varepsilon,
$$
which is unbounded as $\varepsilon\downarrow0$.

Solved by gpt-5.6-sol high.

= b
{parent=17b}
{scope}

= Solution
{parent=b}

The <function>
$$
y(t)=\begin{cases}
y_0-t,&0\leq t\leq y_0,\\
0,&y_0\leq t\leq T
\end{cases}
$$
is continuous, satisfies the equation away from $\widetilde t=y_0$, and has unequal one-sided <derivatives> there.

It is unique in the stated class. While a solution is positive its <derivative> is $-1$, so continuity forces it to follow $y_0-t$ until it first reaches zero at $t=y_0$. It cannot subsequently make a positive excursion: on any connected interval $(a,b)$ where $y>0$ and $y(a)=0$, integration of $y'=-1$ gives $y(t)=-(t-a)<0$. A negative excursion similarly contradicts $y'=1$. Hence the solution remains zero. This is the <sign-decay differential equation>.

Solved by gpt-5.6-sol high.

= c
{parent=17b}
{scope}

= Solution
{parent=c}

Yes. The Euler recurrence is
$$
y_{n+1}=y_n-h\operatorname{sign}(y_n).
$$
Write $y_0=mh+r$, where $m=\lfloor y_0/h\rfloor$ and $0\leq r<h$. For $n\leq m$, the numerical values agree with the exact linear descent. If $r=0$, the method reaches zero and stays there. If $0<r<h$, then
$$
y_m=r,\qquad y_{m+1}=r-h,
$$
and the recurrence thereafter alternates between $r$ and $r-h$. The exact solution is then zero, while both numerical values have magnitude at most $h$. Thus, uniformly for $0\leq n\leq N$,
$$
|y_n-y(nh)|\leq h=O(h).
$$
This is the <explicit Euler method for the sign-decay equation>.

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 1}
{scope}
{title2=Statistics}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

Independence gives
$$
f_\mu(x_1,x_2)
=\frac1{2\pi}
\exp\left[-\frac12\left((x_1-\mu)^2+(x_2-\mu)^2\right)\right].
$$

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

Expanding the exponent gives
$$
f_\mu(x_1,x_2)
=\frac1{2\pi}e^{-(x_1^2+x_2^2)/2}
e^{\mu(x_1+x_2)-\mu^2}.
$$
The <Fisher-Neyman factorization theorem> therefore shows that $T=X_1+X_2$ is sufficient.

It is also minimal sufficient. For two sample points $x,y\in\mathbb R^2$,
$$
\frac{f_\mu(x)}{f_\mu(y)}
=C(x,y)\exp\left(\mu[T(x)-T(y)]\right),
$$
where $C$ is independent of $\mu$. This ratio is independent of $\mu$ exactly when $T(x)=T(y)$. The <likelihood-ratio criterion for minimal sufficiency> applies, proving the claim and the general <normal sample sum with known variance> result in this case.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Writing $X_1=\mu+Z$ with $Z\sim N(0,1)$ gives
$$
\mathbb E S=\mathbb E X_1^2-1=\mu^2,
$$
so $S$ is unbiased. Its mean square error is therefore its variance. Since
$$
\mathbb E X_1^4
=\mathbb E(\mu+Z)^4
=\mu^4+6\mu^2+3,
$$
we obtain
$$
\operatorname{MSE}(S)
=\operatorname{var}(X_1^2)
=\mu^4+6\mu^2+3-(\mu^2+1)^2
=\boxed{4\mu^2+2}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=18h}
{scope}

= Solution
{parent=d}

The pair $(X_1,T)$ is jointly normal, with
$$
\mathbb EX_1=\mu,\quad \mathbb ET=2\mu,\quad
\operatorname{var}(X_1)=1,\quad
\operatorname{var}(T)=2,\quad
\operatorname{cov}(X_1,T)=1.
$$
The <conditional distribution of a bivariate normal variable> is therefore
$$
X_1\mid T=t\sim N\left(\frac t2,\frac12\right).
$$
Equivalently, its conditional density is
$$
f_{X_1\mid T=t}(x)=\frac1{\sqrt\pi}
\exp\left[-\left(x-\frac t2\right)^2\right].
$$

Solved by gpt-5.6-sol high.

= e
{parent=18h}
{scope}

= Solution
{parent=e}

The <Rao-Blackwell theorem> suggests conditioning $S$ on $T$. Part (d) gives
$$
\widetilde S
=\mathbb E[X_1^2-1\mid T]
=\frac12+\frac{T^2}{4}-1
=\frac{T^2}{4}-\frac12.
$$
It remains unbiased. Since $T\sim N(2\mu,2)$ and a normal variable with mean $m$ and variance $v$ satisfies $\operatorname{var}(T^2)=2v^2+4m^2v$,
$$
\operatorname{MSE}(\widetilde S)
=\frac1{16}\operatorname{var}(T^2)
=\frac1{16}(8+32\mu^2)
=\boxed{\frac12+2\mu^2}.
$$
This is strictly below $2+4\mu^2$ for every $\mu$. The calculation is recorded as the <Rao-Blackwell estimator of a squared normal mean>.

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 1}
{scope}
{title2=Markov Chains}

= a
{parent=19h}
{scope}

= Solution
{parent=a}

From any vertex, following the parent edge repeatedly reaches the root with positive probability $p^k$. From the root, any prescribed binary string $(b_1,\ldots,b_k)$ can be reached by following its bits, with positive probability $p_{b_1}\cdots p_{b_k}$. Thus every state communicates with every other state and the chain is irreducible.

At the root there is a self-loop of probability $p>0$, so the root has period one. All states of an irreducible chain have the same period; hence the chain is aperiodic.

Solved by gpt-5.6-sol high.

= b
{parent=19h}
{scope}

= Solution
{parent=b}

Put
$$
q=p_0+p_1=1-p.
$$
The length process $L_n=\ell(X_n)$ is a nearest-neighbour chain on the nonnegative integers: away from zero it moves up with probability $q$ and down with probability $p$, while at zero it moves up with probability $q$ and stays put with probability $p$.

Returns of the original chain to the root are exactly returns of $L_n$ to zero, so their recurrence classifications agree. The <reflected biased random walk on the nonnegative integers> is transient when $q>p$, null recurrent when $q=p$, and positive recurrent when $q<p$. Since $p=1-q$, the conditions are respectively
$$
\begin{array}{c|c}
\text{classification}&\text{condition}\\ \hline
\text{transient}&p_0+p_1>\tfrac12,\\
\text{null recurrent}&p_0+p_1=\tfrac12,\\
\text{positive recurrent}&p_0+p_1<\tfrac12.
\end{array}
$$
Irreducibility transfers the classification from the root to every state.

Solved by gpt-5.6-sol high.

= c
{parent=19h}
{scope}

= Solution
{parent=c}

In the positive-recurrent case, let $q=p_0+p_1<p$. Detailed balance for the length chain requires
$$
\pi_kq=\pi_{k+1}p,
$$
so, with $r=q/p<1$,
$$
\pi_k=\pi_0r^k.
$$
Normalization gives $\pi_0=1-r$. The self-loop at zero makes the length chain aperiodic, so the convergence theorem for irreducible positive-recurrent aperiodic Markov chains gives
$$
\boxed{
\lim_{n\to\infty}\mathbb P(\ell(X_n)=k\mid X_0=\varnothing)
=\left(1-\frac{p_0+p_1}{p}\right)
\left(\frac{p_0+p_1}{p}\right)^k
}.
$$

Solved by gpt-5.6-sol high.