past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperib_2_2023.pdf

= 1E
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}

= i
{parent=1e}
{scope}

= Solution
{parent=i}

Let $e^2=e$ with $e\ne0,1$. Then $e(1-e)=0$, and the <ideals> $eR$ and $(1-e)R$ are nonzero <rings> with identities $e$ and $1-e$. Define
$$
\Phi:R\longrightarrow eR\times(1-e)R,
\qquad
\Phi(r)=(er,(1-e)r).
$$
This is a <ring homomorphism>. Its inverse is $(x,y)\mapsto x+y$, because $ex=x$, $(1-e)y=y$, and the cross terms vanish. Hence $R\cong eR\times(1-e)R$, proving (ii).

Solved by gpt-5.6-sol high.

= ii
{parent=1e}
{scope}

= Solution
{parent=ii}

Conversely, if $R\cong R_1\times R_2$ with both factors nontrivial, the inverse image of $(1_{R_1},0)$ is an <idempotent>. It is neither zero nor one, so (i) holds. This completes the <ring product decomposition by an idempotent> equivalence.

Solved by gpt-5.6-sol high.

= Solution
{parent=1E}

For
$$
R=\{(a,b)\in\mathbb Z^2:a\equiv b\pmod2\},
$$
the parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a <ring>.

It is not an <integral> domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial <rings>. Indeed, an <idempotent> in $\mathbb Z^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=2G}

Suppose first that $X$ is connected and $f:X\to\mathbb Z$ is continuous. Its image is connected by the <continuous image of a connected space>, but $\mathbb Z$ is discrete, so its only connected nonempty subsets are singletons. Hence $f$ is constant.

Conversely, if $X=U\cup V$ is a disconnection into nonempty disjoint <open sets>, the <function> equal to zero on $U$ and one on $V$ is continuous and nonconstant. This proves the <integer-valued function criterion for connectedness>.

Now let $f:X\to\mathbb Z$ be continuous under the hypotheses on the family $\mathcal A$. Each restriction $f|_A$ is constant because $A$ is connected. If $A,B\in\mathcal A$, a point of $A\cap B$ shows that their two constants agree. Since the sets cover $X$, $f$ is constant on $X$, and the criterion proves that $X$ is connected. This is the <pairwise-intersecting connected cover> argument.

Finally, fix $y_0\in Y$. For each $x\in X$, the set
$$
A_x=(X\times\{y_0\})\cup(\{x\}\times Y)
$$
is connected: its two connected pieces meet at $(x,y_0)$. The sets $A_x$ cover $X\times Y$ and any two share $X\times\{y_0\}$. The preceding result proves that $X\times Y$ is connected, giving the <product of connected spaces> result.

Solved by gpt-5.6-sol high.

= 3A
{parent=Paper 2}
{scope}
{title2=Methods}

= Solution
{parent=3A}

The <function> is odd, so only sine coefficients occur. For $n\geq1$,
$$
b_n=\frac2\pi\int_0^\pi(x^3-\pi^2x)\sin(nx)\,dx.
$$
Integration by parts gives
$$
\int_0^\pi x\sin(nx)\,dx=-\frac{\pi(-1)^n}{n},
$$
and
$$
\int_0^\pi x^3\sin(nx)\,dx
=-\frac{\pi^3(-1)^n}{n}+\frac{6\pi(-1)^n}{n^3}.
$$
The terms of order $1/n$ cancel, leaving
$$
b_n=\frac{12(-1)^n}{n^3}.
$$
Thus
$$
x^3-\pi^2x
=12\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\sin(nx),
\qquad -\pi<x<\pi.
$$

The <Parseval identity> gives
$$
\frac1\pi\int_{-\pi}^{\pi}(x^3-\pi^2x)^2\,dx
=144\sum_{n=1}^{\infty}\frac1{n^6}.
$$
Direct integration yields
$$
\frac1\pi\int_{-\pi}^{\pi}
(x^6-2\pi^2x^4+\pi^4x^2)\,dx
=\frac{16\pi^6}{105}.
$$
Therefore
$$
\boxed{\sum_{n=1}^{\infty}\frac1{n^6}=\frac{\pi^6}{945}},
$$
as recorded in the <Fourier series of x cubed minus pi squared x>.

Solved by gpt-5.6-sol high.

= 4D
{parent=Paper 2}
{scope}
{title2=Electromagnetism}

= Solution
{parent=4D}

A capacitor consists of two conductors carrying equal and opposite charges. Its <capacitance> is
$$
C=\frac QV,
$$
where $Q$ is the magnitude of the charge on either conductor and $V$ is their potential difference.

For $a<r<b$, a coaxial Gaussian cylinder of length $\ell$ encloses charge $\lambda\ell$. <Gauss's law> gives
$$
E(r)=\frac{\lambda}{2\pi\epsilon_0r}\,e_r.
$$
Taking $V$ to mean the inner potential minus the outer potential,
$$
V=\int_a^bE(r)\,dr
=\frac{\lambda}{2\pi\epsilon_0}\log\frac ba.
$$
Since $Q=\lambda L$,
$$
\boxed{C=\frac{2\pi\epsilon_0L}{\log(b/a)}}.
$$

The field energy is
$$
\begin{aligned}
U
&=\frac{\epsilon_0}{2}
\int_a^bE(r)^2(2\pi rL)\,dr\\
&=\frac{\lambda^2L}{4\pi\epsilon_0}\log\frac ba
=\frac12(\lambda L)
\left(\frac{\lambda}{2\pi\epsilon_0}\log\frac ba\right)
=\frac12QV.
\end{aligned}
$$
These are the standard <coaxial cylindrical capacitor> formulas.

Solved by gpt-5.6-sol high.

= 5C
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics}

= a
{parent=5c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The divergence is the trace of the velocity-gradient tensor:
$$
\nabla\cdot u=A+E+I.
$$
Thus the flow is incompressible exactly when
$$
\boxed{A+E+I=0}.
$$
The constant <vector> $U_0$ is unrestricted.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The <vorticity> is
$$
\nabla\times u=(H-F,\ C-G,\ D-B).
$$
Hence the flow is irrotational exactly when
$$
\boxed{H=F,\qquad C=G,\qquad D=B},
$$
equivalently when $\Gamma$ is symmetric. Again, $U_0$ is unrestricted.

Solved by gpt-5.6-sol high.

= b
{parent=5c}
{scope}

= Solution
{parent=b}

A <streamline> $x(s)=(x(s),y(s),z(s))$ satisfies
$$
\dot x=\alpha y,
\qquad
\dot y=-\alpha x,
\qquad
\dot z=\beta,
$$
with $(x(0),y(0),z(0))=(1,0,0)$. Therefore
$$
\boxed{x(s)=\cos(\alpha s),\qquad
y(s)=-\sin(\alpha s),\qquad
z(s)=\beta s}.
$$
For $\beta\ne0$ this is a helix, with $x=\cos(\alpha z/\beta)$ and $y=-\sin(\alpha z/\beta)$; for $\beta=0$ it is the unit circle in the plane $z=0$.

Solved by gpt-5.6-sol high.

= 6H
{parent=Paper 2}
{scope}
{title2=Statistics}

= a
{parent=6h}
{scope}

= Solution
{parent=a}

Apart from constants, the log-likelihood is
$$
\ell(\theta)=-\frac1{2\sigma^2}\sum_{i=1}^n(X_i-\theta)^2.
$$
Differentiating gives $\ell'(\theta)=n(\overline X-\theta)/\sigma^2$, so
$$
\boxed{\widehat\theta_{\mathrm{MLE}}=\overline X}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=6h}
{scope}

= Solution
{parent=b}

Since
$$
\overline X\sim N\left(\theta,\frac{\sigma^2}{n}\right),
$$
an exact 95 percent confidence interval is
$$
\boxed{
\overline X\mathbin{\pm}z_{0.975}\frac{\sigma}{\sqrt n}
},
\qquad z_{0.975}\simeq1.96.
$$

Solved by gpt-5.6-sol high.

= c
{parent=6h}
{scope}

= Solution
{parent=c}

Write the posterior mean and variance as
$$
m_n=\frac{n\overline X/\sigma^2+\mu/\nu^2}
{n/\sigma^2+1/\nu^2},
\qquad
v_n=\left(\frac n{\sigma^2}+\frac1{\nu^2}\right)^{-1}.
$$
The central 95 percent posterior credible interval is
$$
\boxed{m_n\mathbin{\pm}z_{0.975}\sqrt{v_n}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=6h}
{scope}

= Solution
{parent=d}

As $n\to\infty$,
$$
m_n-\overline X
=\frac{\sigma^2(\mu-\overline X)}{n\nu^2+\sigma^2}
=O_p(n^{-1}),
$$
while
$$
\sqrt{v_n}
=\frac{\sigma}{\sqrt n}
\left(1+\frac{\sigma^2}{n\nu^2}\right)^{-1/2}
\sim\frac{\sigma}{\sqrt n}.
$$
Thus the credible interval and confidence interval have asymptotically equal centres and half-widths, and both contract around the true parameter. Their interpretations remain different: the confidence statement concerns repeated samples, whereas the credible statement concerns posterior probability. This is <normal-normal conjugacy with known observation variance>.

Solved by gpt-5.6-sol high.

= 7H
{parent=Paper 2}
{scope}
{title2=Optimisation}

= Solution
{parent=7H}

Use
$$
L(x,y,z,\lambda)
=x^2+y^4+z^6-\lambda(x+2y+3z-6).
$$
Stationarity gives
$$
2x=\lambda,
\qquad
4y^3=2\lambda,
\qquad
6z^5=3\lambda.
$$
Putting $t=\lambda/2$, this becomes
$$
x=t,\qquad y=t^{1/3},\qquad z=t^{1/5},
$$
and the constraint requires
$$
t+2t^{1/3}+3t^{1/5}=6.
$$
The left side is strictly increasing, and $t=1$ solves the equation. Hence
$$
\boxed{(x,y,z)=(1,1,1)},
\qquad
\boxed{f_{\min}=3},
\qquad
\lambda=2.
$$

The objective is convex and the constraint is affine. Its tangent-plane inequality at $(1,1,1)$ gives, for every feasible $(x,y,z)$,
$$
f(x,y,z)\geq3+(2,4,6)\cdot(x-1,y-1,z-1)=3,
$$
so the Lagrange point is globally optimal. Moreover, at the dual value $\lambda=2$, the infimum of the <Lagrangian function in constrained optimization> is attained at the same point and equals three. The primal and dual values coincide, so strong duality holds.

For the value <function>, the multiplier convention above gives the <derivative of a constrained value function>
$$
\phi'(b)=\lambda(b).
$$
At $b=6$, therefore,
$$
\boxed{\phi'(6)=2}.
$$

Solved by gpt-5.6-sol high.

= 8F
{parent=Paper 2}
{scope}
{title2=Linear Algebra}

= Solution
{parent=8F}

For an $n\times n$ <matrix> $A$, the characteristic <polynomial> is
$$
\chi_A(t)=\det(tI-A).
$$
The <Cayley-Hamilton theorem> states that $\chi_A(A)=0$.

Over $\mathbb C$, choose a <basis> in which $A$ is upper triangular, with diagonal entries $\lambda_1,\ldots,\lambda_n$. For the standard invariant flag $V_j=\langle e_1,\ldots,e_j\rangle$,
$$
(A-\lambda_jI)V_j\subseteq V_{j-1}.
$$
The factors $A-\lambda_jI$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\ldots,V_0=0$. Hence
$$
0=\prod_{j=1}^n(A-\lambda_jI)=\chi_A(A),
$$
which proves the theorem.

Direct expansion gives the commutator product rule:
$$
\begin{aligned}
[X,YZ]
&=XYZ-YZX\\
&=(XY-YX)Z+Y(XZ-ZX)\\
&=[X,Y]Z+Y[X,Z].
\end{aligned}
$$

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule gives
$$
[B,A^r]=\sum_{j=0}^{r-1}A^jCA^{r-1-j}=rA^{r-1}C.
$$
By <linearity>, for every <polynomial> $\varphi$,
$$
[B,\varphi(A)]=\varphi'(A)C.
$$

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,
$$
f'(A)C=D(f(A))=0.
$$
Assume inductively that
$$
uC^m=0,
\qquad
u=f^{(k)}(A),\quad m=2^k-1.
$$
Both $u$ and $C$ are <polynomials> in, or commute with, $A$, so $uC^m=C^mu=0$. Apply the derivation $D$ to $uC^m=0$ and multiply on the left by $C^m$:
$$
0=C^mD(u)C^m+C^muD(C^m)=C^mD(u)C^m.
$$
Since $D(u)=f^{(k+1)}(A)C$, this says
$$
f^{(k+1)}(A)C^{2m+1}
=f^{(k+1)}(A)C^{2^{k+1}-1}=0.
$$
The induction is complete. Taking $f=\chi_A$ and $k=n$ gives
$$
n!\,C^{2^n-1}=0.
$$
Thus $[B,A]$ is nilpotent, which is the <Jacobson lemma for a commuting commutator>.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

Let $Q$ act by left multiplication on the set $G/P$ of left cosets. Every orbit has size a power of $p$, while
$$
|G/P|
$$
is not divisible by $p$. Therefore at least one orbit has size one. If $gP$ is fixed, then $qgP=gP$ for every $q\in Q$, so $g^{-1}Qg\leq P$, equivalently
$$
Q\leq gPg^{-1}.
$$
This is <Sylow containment from a coset fixed point>.

The remaining <Sylow theorems> state that Sylow $p$-subgroups exist, that any two are conjugate, and that their number $n_p$ satisfies
$$
n_p\equiv1\pmod p,
\qquad
n_p\mid |G|/|P|.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

By transitivity and the <orbit-stabilizer theorem>,
$$
|G|=7|G_x|=7\cdot24=168=2^3\cdot3\cdot7.
$$
Inside $G_x\cong S_4$, there are three Sylow $2$-subgroups, each of order eight, and four Sylow $3$-subgroups, each of order three. These are also Sylow <subgroups> of $G$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Count pairs $(x,P)$ where $P$ is a Sylow $2$-subgroup fixing $x$. Each of the seven point stabilizers contains three such <subgroups>, giving 21 pairs. Conversely, a $2$-group acting on seven points has a fixed point, and a Sylow $2$-subgroup cannot fix two points because a two-point stabilizer has order four. Hence every Sylow $2$-subgroup occurs in exactly one pair, so
$$
n_2=21.
$$

Similarly, each point stabilizer contains four Sylow $3$-subgroups, giving 28 pairs. A <group> of order three acting on seven points has a fixed point, and it cannot fix two because a two-point stabilizer is a $2$-group. Thus
$$
n_3=28.
$$

The Sylow congruence and divisibility conditions give $n_7\in\{1,8\}$. Regard the faithful action as an embedding $G\leq S_7$. If $n_7=1$, its Sylow $7$-subgroup $P$ is normal, so $G\leq N_{S_7}(P)$. The stated fact gives $|N_{S_7}(P)|=7\cdot6=42$, impossible for a <subgroup> $G$ of order 168. Hence
$$
n_7=8.
$$
These are the <Sylow counts in a faithful degree-seven action with S4 point stabilizers>.

Solved by gpt-5.6-sol high.

= Solution
{parent=b}

Let $N\mathrel{\trianglelefteq}G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem gives
$$
56=7\cdot8\mid |N|.
$$
Since $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.

Every nontrivial normal <subgroup> is transitive by <normal subgroup orbits in a faithful prime-degree action>. Hence
$$
|N\cap G_x|=|N_x|=56/7=8.
$$
But $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal <subgroup> has order divisible by seven.

If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such <subgroup> would have order 84, which is divisible by seven and has just been ruled out.

Finally, if $N$ is any nontrivial normal <subgroup>, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal <subgroup>. Therefore $G$ is simple.

Solved by gpt-5.6-sol high.

= 10G
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=10G}

A <sequence> $f_n:S\to\mathbb R$ converges uniformly to $f$ if, for every $\varepsilon>0$, there is $N$ such that
$$
|f_n(x)-f(x)|<\varepsilon
$$
for every $x\in S$ and every $n\geq N$. A map $h:M\to N$ is uniformly continuous if, for every $\varepsilon>0$, there is $\delta>0$ such that
$$
d_M(x,y)<\delta\quad\Longrightarrow\quad d_N(h(x),h(y))<\varepsilon
$$
for all $x,y\in M$.

If $f\in C_0(\mathbb R^d)$, choose a ball outside which $|f|<1$. On the closed ball, continuity gives boundedness, so $f$ is bounded everywhere.

Now let $(f_n)$ be Cauchy in the uniform metric. For each $x$, $(f_n(x))$ is Cauchy in $\mathbb R$; let its <limit> be $f(x)$. Passing to the pointwise <limit> in the uniform Cauchy estimate shows that $f_n\to f$ uniformly. Hence $f$ is continuous. Given $\varepsilon>0$, choose $n$ with $\|f-f_n\|_\infty<\varepsilon/2$, and then choose $K$ so that $|f_n(x)|<\varepsilon/2$ for $\|x\|>K$. It follows that $f$ also vanishes at infinity. Thus $C_0(\mathbb R^d)$ is complete, as in <completeness of continuous functions vanishing at infinity>.

Every $f\in C_0(\mathbb R^d)$ is uniformly continuous. Given $\varepsilon>0$, choose $R$ so that $|f(x)|<\varepsilon/2$ outside the ball of radius $R$. On the compact ball of radius $R+1$, $f$ is uniformly continuous; choose the corresponding $\delta\leq1$. If two points at distance below $\delta$ are not both in that ball, then both lie outside the ball of radius $R$, and their <function> values differ by less than $\varepsilon$.

For the final <sequence>, continuity of $\varepsilon$ at zero gives, for each fixed $x$,
$$
f_n(x)=\sqrt{x^2+\varepsilon(x/n)}\longrightarrow|x|.
$$
Thus <pointwise convergence> is compulsory. <Uniform convergence> need not hold: with $\varepsilon(t)=t^2$,
$$
f_n(x)-|x|
=|x|\left(\sqrt{1+n^{-2}}-1\right),
$$
which is unbounded as a <function> of $x$ for every fixed $n$.

Under the additional bound $\varepsilon(t)\leq M|t|$, however,
$$
0\leq f_n(x)-|x|
=\frac{\varepsilon(x/n)}{\sqrt{x^2+\varepsilon(x/n)}+|x|}
\leq\frac Mn
$$
for $x\ne0$, and the difference is zero at $x=0$. Hence convergence is uniform, by the <uniform square-root perturbation under linear growth> estimate. The pointwise answer remains yes.

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 2}
{scope}
{title2=Geometry}

= Solution
{parent=11F}

The <tangent vectors> of
$$
\sigma(u,v)=(u,v,f(u,v))
$$
are
$$
\sigma_u=(1,0,f_u),
\qquad
\sigma_v=(0,1,f_v).
$$
Hence the <first fundamental form> has coefficients
$$
E=1+f_u^2,
\qquad
F=f_uf_v,
\qquad
G=1+f_v^2.
$$
Put
$$
W=\sqrt{1+f_u^2+f_v^2}.
$$
The upward <unit normal> is $N=(-f_u,-f_v,1)/W$, so the <second fundamental form> has coefficients
$$
e=\frac{f_{uu}}W,
\qquad
f_{II}=\frac{f_{uv}}W,
\qquad
g=\frac{f_{vv}}W.
$$
Thus the two forms are
$$
I=E\,du^2+2F\,du\,dv+G\,dv^2,
\qquad
II=e\,du^2+2f_{II}\,du\,dv+g\,dv^2.
$$

The <Gaussian curvature> is
$$
K=\frac{eg-f_{II}^2}{EG-F^2}.
$$
Since $EG-F^2=W^2$, the graph formula is
$$
\boxed{
K=\frac{f_{uu}f_{vv}-f_{uv}^2}
{(1+f_u^2+f_v^2)^2}
},
$$
as in <Gaussian curvature of a graph surface>.

For the final claim, fix a point of $\gamma$ and make a <rigid motion> taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,
$$
f(c(s))=0,
\qquad
\nabla f(c(s))=0.
$$
Differentiating the second identity gives
$$
\operatorname{Hess}f(c(s))\,c'(s)=0.
$$
Because $c$ is a <smooth curve>, $c'(s)\ne0$, so the Hessian is singular. Its <determinant> is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is <tangency to a plane along a curve forces zero Gaussian curvature>.

Solved by gpt-5.6-sol high.

= 12B
{parent=Paper 2}
{scope}
{title2=Complex Analysis OR Complex Methods}

= a
{parent=12b}
{scope}

= Solution
{parent=a}

Let $|f(z)|\leq K$ on $\operatorname{Re}z>0$. If $\operatorname{Re}z>c$, the circle $|w-z|=c$ lies in that half-plane. The <Cauchy derivative formula> gives
$$
|f'(z)|\leq\frac Kc.
$$
The <line segment> joining $z_1$ to $z_2$ remains in $\operatorname{Re}z>c$, so
$$
|f(z_1)-f(z_2)|
\leq\frac Kc|z_1-z_2|.
$$
Thus one may take $M=K/c$, as in the <lipschitz bound inside a bounded analytic half-plane>.

Solved by gpt-5.6-sol high.

= b
{parent=12b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The <Liouville theorem> states that every bounded <entire function> is constant.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Suppose $g(\mathbb C)$ is not dense. Then some <open disc> $D(w,r)$ is disjoint from the image, so
$$
F(z)=\frac1{g(z)-w}
$$
is entire and satisfies $|F(z)|\leq1/r$. Liouville's theorem makes $F$, and hence $g$, constant, a contradiction. Therefore every nonconstant entire $g$ has dense image, as asserted by <dense image of a nonconstant entire function>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Write $h(z)=\sum_{n=0}^{\infty}a_nz^n$. Cauchy's coefficient formula on $|z|=R$ gives
$$
a_n=\frac1{2\pi R^n}
\int_0^{2\pi}h(Re^{i\theta})e^{-in\theta}\,d\theta.
$$
Except at the two measure-zero points where $\cos\theta=0$, the hypothesis gives
$$
|h(Re^{i\theta})|
\leq R^{-1/2}|\cos\theta|^{-1/2}.
$$
Since $|\cos\theta|^{-1/2}$ is integrable,
$$
|a_n|
\leq C R^{-n-1/2}
$$
for a constant $C$ independent of $R$. Letting $R\to\infty$ shows that every $a_n=0$. Hence $h\equiv0$, in particular $h$ is constant. This proves the <entire function under a horizontal inverse-square-root bound> result.

Solved by gpt-5.6-sol high.

= 13C
{parent=Paper 2}
{scope}
{title2=Variational Principles}

= a
{parent=13c}
{scope}

= Solution
{parent=a}

Take a <variation> $y+\varepsilon\eta$, where the <differentiable function> $\eta$ obeys
$$
\eta(a)=\eta(b)=\eta'(a)=\eta'(b)=0
$$
because both $y$ and its first <derivative> have fixed endpoint values. The <first variation> of the <functional> is
$$
\delta L
=\int_a^b
\left(F_y\eta+F_{y'}\eta'+F_{y''}\eta''\right)\,dx.
$$
Applying <integration by parts> once to the second term and twice to the third gives
$$
\begin{aligned}
\delta L
={}&\int_a^b
\left[
F_y-\frac d{dx}F_{y'}
+\frac{d^2}{dx^2}F_{y''}
\right]\eta\,dx\\
&+\left[
F_{y'}\eta+F_{y''}\eta'
-\frac d{dx}(F_{y''})\eta
\right]_a^b.
\end{aligned}
$$
The endpoint conditions on $\eta$ and $\eta'$ make every boundary term zero. Since the remaining <integral> vanishes for every admissible variation, the <fundamental lemma of the calculus of variations> yields the <higher-order Euler-Lagrange equation>
$$
\boxed{
F_y-\frac d{dx}F_{y'}
+\frac{d^2}{dx^2}F_{y''}=0
}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=13c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Here
$$
F(x,y,y',y'')=\frac A2(y'')^2+\rho gy.
$$
Substitution into the <higher-order Euler-Lagrange equation> gives the fourth-order <ordinary differential equation>
$$
\boxed{Ay''''+\rho g=0}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Integrating the <ordinary differential equation> four times gives a quartic <polynomial>. It is useful to use the <linearity> of the equation and split the solution into a gravity part and a force part:
$$
y=y_0+y_F,
$$
where
$$
y_0(x)
=-\frac{\rho g}{24A}x^2(6L^2-4Lx+x^2)
$$
satisfies the clamped and torque-free <boundary conditions> with $F=0$, while
$$
y_F(x)=\frac{F}{6A}x^2(3L-x)
$$
satisfies the homogeneous equation and contributes the endpoint <force>. Direct <differentiation> verifies
$$
y(0)=y'(0)=0,\qquad y''(L)=0,\qquad -Ay'''(L)=F.
$$
Thus
$$
\boxed{
y(x)=-\frac{\rho g}{24A}x^2(6L^2-4Lx+x^2)
+\frac{F}{6A}x^2(3L-x)
},
$$
the <clamped-free beam under uniform load and endpoint force>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Evaluating the <function> at the endpoint gives the <displacement>
$$
\Delta=y(L)
=-\frac{\rho gL^4}{8A}+\frac{FL^3}{3A}.
$$
Therefore
$$
\boxed{h_0=-\frac{\rho gL^4}{8A}},
\qquad
\boxed{h=\frac{FL^3}{3A}},
\qquad
\boxed{\Delta=h_0+h}.
$$
The force-induced displacement $h$ is a <linear map> of $F$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Substitute $y=y_0+y_F$ into the energy <functional> and integrate the resulting <polynomial>:
$$
\begin{aligned}
E
&=\int_0^L
\left[\frac A2(y'')^2+\rho gy\right]\,dx\\
&=-\frac{\rho^2g^2L^5}{40A}
+\frac{F^2L^3}{6A}.
\end{aligned}
$$
The term linear in $F$ cancels. Hence
$$
\boxed{E_0=-\frac{\rho^2g^2L^5}{40A}},
\qquad
\boxed{\mathcal E=\frac{F^2L^3}{6A}},
\qquad
\boxed{E=E_0+\mathcal E}.
$$
As required, $E_0$ is independent of the <force> and $\mathcal E$ is a <quadratic function> of it.

Solved by gpt-5.6-sol high.

= v
{parent=b}
{scope}

= Solution
{parent=v}

Taking the <derivative> of the force-dependent energy gives
$$
\boxed{
\frac{d\mathcal E}{dF}
=\frac{FL^3}{3A}
=h
}.
$$
Thus the <derivative> of the additional minimized internal energy with respect to the applied <force> equals the resulting endpoint <displacement>. This is the <endpoint force derivative of clamped-free beam energy>.

Solved by gpt-5.6-sol high.

= 14A
{parent=Paper 2}
{scope}
{title2=Methods}

= a
{parent=14a}
{scope}

= Solution
{parent=a}

Apply <separation of variables> to the <Laplace equation in polar coordinates> by writing $\phi(r,\theta)=R(r)\Theta(\theta)$. Division by $R\Theta$ gives
$$
\frac{r(rR')'}R=-\frac{\Theta''}{\Theta}=n^2.
$$
The requirement that $\Theta$ be a real $2\pi$-<periodic function> restricts the separation constants to $n^2$, with angular factors $\cos(n\theta)$ and $\sin(n\theta)$. For $n\geq1$, the radial <ordinary differential equation> is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant and
$$
(rR')'=0,
$$
so $R=a_0+c_0\log r$. By <linearity>, superposition gives
$$
\boxed{
\begin{aligned}
\phi(r,\theta)
={}&a_0+c_0\log r\\
&+\sum_{n=1}^{\infty}(a_nr^n+c_nr^{-n})\cos(n\theta)\\
&+\sum_{n=1}^{\infty}(b_nr^n+d_nr^{-n})\sin(n\theta).
\end{aligned}
}
$$
This is the separated expansion of a <harmonic function> in a circular region.

Solved by gpt-5.6-sol high.

= i
{parent=a}
{scope}

= Solution
{parent=i}

Regularity at the origin excludes the <natural logarithm> and every negative radial power. Thus
$$
\boxed{c_0=0,\qquad c_n=d_n=0\quad(n\geq1)}.
$$
The coefficients $a_0,a_n,b_n$ remain arbitrary.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Regularity at infinity excludes the <natural logarithm> and every positive radial power. Thus
$$
\boxed{c_0=0,\qquad a_n=b_n=0\quad(n\geq1)}.
$$
The constant $a_0$ and the decaying coefficients $c_n,d_n$ remain arbitrary.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

An <annulus> stays away from both the origin and infinity, so every displayed radial mode is regular there. Therefore none of the coefficients is forced to vanish.

Solved by gpt-5.6-sol high.

= b
{parent=14a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Write $s=\log r$, so the two circular boundaries are $s=0$ and $s=2$. The zero angular mode must interpolate between $-1$ and $1$, giving $s-1=\log r-1$.

For the $n$th cosine mode, the radial factor has equal value $A_n$ at both boundaries. The unique <harmonic function> with those data is
$$
A_n\frac{\cosh(n(s-1))}{\cosh n}\cos(n\theta).
$$
Hence the solution of the annular <Dirichlet problem> is
$$
\boxed{
\phi(r,\theta)
=\log r-1
+\sum_{n=1}^{\infty}
A_n\frac{\cosh(n(\log r-1))}{\cosh n}\cos(n\theta)
}.
$$
At $r=1$ and $r=e^2$, the <hyperbolic cosine> quotient equals one, so the <boundary conditions> are satisfied term by term.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The mean of $f$ is zero. Its <Fourier cosine series> coefficients are
$$
\begin{aligned}
A_n
&=\frac1\pi\left[
\int_0^\pi\left(\frac\pi2-\theta\right)\cos(n\theta)\,d\theta
+\int_\pi^{2\pi}\left(\theta-\frac{3\pi}2\right)\cos(n\theta)\,d\theta
\right]\\
&=\frac{2(1-(-1)^n)}{\pi n^2}.
\end{aligned}
$$
Thus $A_n=0$ for <even> $n$ and $A_n=4/(\pi n^2)$ for <odd> $n$. Substitution into part (i) yields
$$
\boxed{
\phi(r,\theta)
=\log r-1
+\frac4\pi
\sum_{\substack{n\geq1\\ n\ {\rm odd}}}
\frac{\cosh(n(\log r-1))}{n^2\cosh n}\cos(n\theta)
}.
$$

Solved by gpt-5.6-sol high.

= 15D
{parent=Paper 2}
{scope}
{title2=Quantum Mechanics}

= a
{parent=15d}
{scope}

= Solution
{parent=a}

The <Schrodinger equation> and its complex conjugate are
$$
i\hbar\frac{\partial\psi}{\partial t}
=-\frac{\hbar^2}{2m}\nabla^2\psi+U\psi,
\qquad
-i\hbar\frac{\partial\psi^*}{\partial t}
=-\frac{\hbar^2}{2m}\nabla^2\psi^*+U\psi^*.
$$
Because the potential $U$ is real, its two contributions cancel when differentiating the <probability density> $\rho=\psi^*\psi$. Therefore
$$
\begin{aligned}
\frac{\partial\rho}{\partial t}
&=\psi^*\frac{\partial\psi}{\partial t}
+\psi\frac{\partial\psi^*}{\partial t}\\
&=\frac{i\hbar}{2m}
\left(\psi^*\nabla^2\psi-\psi\nabla^2\psi^*\right)\\
&=-\nabla\cdot
\left[
-\frac{i\hbar}{2m}
\left(\psi^*\nabla\psi-\psi\nabla\psi^*\right)
\right].
\end{aligned}
$$
The expression in square brackets is the <probability current> $J$, so
$$
\boxed{\frac{\partial\rho}{\partial t}+\nabla\cdot J=0}.
$$
This <probability continuity equation> says that probability can leave a region only through the current across its boundary.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Write the spatial factors of the stationary <wavefunction> as
$$
\psi(x,t)=e^{-iEt/\hbar}
\begin{cases}
e^{ikx}+R e^{-ikx},&x<-a,\\
C e^{kx}+D e^{-kx},&-a<x<a,\\
T e^{ikx},&x>a,
\end{cases}
$$
where
$$
k=\frac{\sqrt{2mE}}{\hbar}.
$$
Since $U_0-E=E$, the decay constant inside the <potential barrier> is also $k$. Continuity of the <wavefunction> and its first <derivative> at $x=-a$ and $x=a$ gives four linear equations. Solving them yields
$$
T=\frac{e^{-2ika}}{\cosh(2ka)}.
$$
Consequently the transmitted wave is
$$
\boxed{
\psi_{\rm tr}(x,t)
=\frac{\exp\!\left(i[k(x-2a)-Et/\hbar]\right)}
{\cosh(2ka)}
},
\qquad x>a,
$$
and its <probability density> is
$$
\boxed{
\rho_{\rm tr}(x,t)=|\psi_{\rm tr}|^2
=\operatorname{sech}^2(2ka)
}.
$$
This is <finite square barrier transmission at half barrier height>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For a plane wave $A e^{i(kx-Et/\hbar)}$, the <probability current> is
$$
J=\frac{\hbar k}{m}|A|^2.
$$
The incident amplitude is one and the transmitted amplitude is $T$, so
$$
\boxed{
\frac{J_{\rm tr}}{J_{\rm in}}
=|T|^2
=\operatorname{sech}^2(2ka)
}.
$$
This is the transmission probability for <quantum tunnelling>. In the <stationary state>, $\partial\rho/\partial t=0$, so the <continuity equation> makes the net current independent of position. The reflected current therefore supplies the remainder:
$$
\frac{|J_{\rm refl}|}{J_{\rm in}}
=1-\operatorname{sech}^2(2ka)
=\tanh^2(2ka).
$$

Solved by gpt-5.6-sol high.

= 16D
{parent=Paper 2}
{scope}
{title2=Electromagnetism}

= a
{parent=16d}
{scope}

= Solution
{parent=a}

Let $x^\mu=(ct,\mathbf x)$, let $\tau$ be <proper time>, and define the <four-velocity> and <four-momentum> by
$$
u^\mu=\frac{dx^\mu}{d\tau}=\gamma(c,\mathbf v),
\qquad
p^\mu=mu^\mu=(\mathcal E/c,\mathbf p).
$$
If $F^{\mu\nu}$ is the <electromagnetic field tensor>, the covariant <Lorentz force> law is
$$
\boxed{\frac{dp^\mu}{d\tau}=qF^{\mu\nu}u_\nu}.
$$
The temporal component and three spatial components respectively give
$$
\boxed{\frac{d\mathcal E}{dt}=q\mathbf E\cdot\mathbf v},
\qquad
\boxed{\frac{d\mathbf p}{dt}
=q(\mathbf E+\mathbf v\times\mathbf B)}.
$$
Here $\mathcal E=\gamma mc^2$ is the relativistic <energy> and $\mathbf p=\gamma m\mathbf v$ is the relativistic <momentum>. In the nonrelativistic <limit>, $\gamma\to1$, so the spatial equation becomes
$$
m\frac{d\mathbf v}{dt}
=q(\mathbf E+\mathbf v\times\mathbf B),
$$
the usual Lorentz-force law.

Solved by gpt-5.6-sol high.

= b
{parent=16d}
{scope}

= Solution
{parent=b}

The temporal component found in part (a) gives directly
$$
\boxed{
\mathcal E(t)-\mathcal E(0)
=q\int_0^t\mathbf E\cdot\mathbf v(s)\,ds
}.
$$
For a constant <electric field>, this is the work done by the field along the particle trajectory:
$$
\boxed{
\mathcal E(t)=\mathcal E(0)
+q\mathbf E\cdot[\mathbf x(t)-\mathbf x(0)]
}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=16d}
{scope}

= Solution
{parent=c}

With $\mathbf B=0$ and $\mathbf E=(0,0,E)$, the spatial <Lorentz force> equation gives
$$
p_x=p_0,\qquad p_y=0,\qquad p_z=qEt.
$$
The <relativistic energy-momentum relation> therefore yields
$$
\boxed{
\mathcal E(t)=
\sqrt{\mathcal E_0^2+c^2q^2E^2t^2}
},
\qquad
\mathcal E_0=\sqrt{m^2c^4+c^2p_0^2}.
$$
Since $\mathbf v=c^2\mathbf p/\mathcal E$,
$$
\dot z=\frac{c^2qEt}{\mathcal E(t)}.
$$
Integration from the initial <position> $z(0)=0$ gives
$$
\boxed{
z(t)=
\frac{\sqrt{\mathcal E_0^2+c^2q^2E^2t^2}-\mathcal E_0}{qE}
}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=16d}
{scope}

= Solution
{parent=d}

The $x$ component of the <velocity> is
$$
\dot x=\frac{c^2p_0}{\mathcal E(t)}.
$$
Thus
$$
x(t)=\frac{cp_0}{qE}
\operatorname{arsinh}\left(\frac{cqEt}{\mathcal E_0}\right).
$$
Solving this equation for $t$ gives
$$
\frac{cqEt}{\mathcal E_0}
=\sinh\left(\frac{qEx}{cp_0}\right).
$$
Substitution into the expression for $z(t)$ and the identity $1+\sinh^2u=\cosh^2u$ produce the trajectory
$$
\boxed{
z(x)=\frac{\mathcal E_0}{qE}
\left[
\cosh\left(\frac{qEx}{cp_0}\right)-1
\right]
}.
$$
This is <relativistic motion in a constant electric field with transverse momentum>.

Solved by gpt-5.6-sol high.

= e
{parent=16d}
{scope}

= Solution
{parent=e}

For $t\to+\infty$ with $qE>0$, an <asymptotic expansion> gives
$$
z(t)=ct-\frac{\mathcal E_0}{qE}+O(t^{-1}),
$$
so the longitudinal <velocity> tends to the speed of light. More generally, its limiting sign is the sign of $qE$.

For small $t$, the <taylor series> of the square root gives
$$
z(t)
=\frac{qEc^2}{2\mathcal E_0}t^2+O(t^4).
$$
In the nonrelativistic <limit> $\mathcal E_0\simeq mc^2$, this becomes
$$
z(t)\simeq\frac{qE}{2m}t^2
=\frac12at^2,
\qquad a=\frac{qE}{m}.
$$
The transverse motion is then uniform:
$$
\boxed{x(t)=\frac{p_0}{m}t}.
$$
Eliminating $t$ gives the nonrelativistic parabolic trajectory
$$
\boxed{
z(x)=\frac{qEm}{2p_0^2}x^2
}.
$$

Solved by gpt-5.6-sol high.

= 17B
{parent=Paper 2}
{scope}
{title2=Numerical Analysis}

= a
{parent=17b}
{scope}

= Solution
{parent=a}

Apply a one-step numerical method to the <scalar> test <ordinary differential equation>
$$
y'=\lambda y.
$$
If one step has the form
$$
y_{n+1}=R(z)y_n,
\qquad z=h\lambda,
$$
then $R$ is the <stability function> and the <linear stability domain> is
$$
\boxed{\mathcal S=\{z\in\mathbb C:|R(z)|\leq1\}}.
$$
The method is <A-stable> when
$$
\boxed{\{z:\operatorname{Re}z\leq0\}\subseteq\mathcal S},
$$
so it does not amplify any mode that the exact <differential equation> does not amplify.

Solved by gpt-5.6-sol high.

= b
{parent=17b}
{scope}

= Solution
{parent=b}

For the test equation $f(y)=\lambda y$, put $z=h\lambda$. The stage equations of this <implicit Runge-Kutta method> are
$$
\begin{pmatrix}
1-z/4&-z(1/4-a)\\
-z(1/4+a)&1-z/4
\end{pmatrix}
\begin{pmatrix}k_1\\k_2\end{pmatrix}
=\lambda y_n\begin{pmatrix}1\\1\end{pmatrix}.
$$
Substitution into the update gives the <stability function>
$$
R(z)=\frac{2+z+2a^2z^2}{2-z+2a^2z^2}.
$$
For $z=x+iy$, direct expansion gives
$$
\begin{aligned}
&|2-z+2a^2z^2|^2-|2+z+2a^2z^2|^2\\
&\hspace{35mm}=-8x(1+a^2|z|^2).
\end{aligned}
$$
This is nonnegative whenever $x\leq0$, and hence $|R(z)|\leq1$ throughout the closed left half-plane, provided the denominator has no zero there.

If $a=0$, the denominator vanishes only at $z=2$. If $a\ne0$, its zeros solve
$$
2a^2z^2-z+2=0.
$$
Real roots are positive, while a complex-conjugate pair has positive real part because the sum of the roots is $1/(2a^2)>0$. Thus there are no poles in the closed left half-plane for any real $a$. Therefore
$$
\boxed{a\in\mathbb R}
$$
is the complete set of <A-stable> parameters, as summarized by the <A-stability of a symmetric two-stage implicit Runge-Kutta family>.

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 2}
{scope}
{title2=Markov Chains}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

The statement is true. If the <Markov chain> with <transition matrix> $P$ is irreducible, then for every pair of states $i,j$ there is a path
$$
i=i_0,i_1,\ldots,i_m=j
$$
with $P_{i_{r-1}i_r}>0$ at every step. The support assumption gives $Q_{i_{r-1}i_r}>0$ for every edge of the same path. Thus every state can reach every other state under $Q$, so the second chain is also an <irreducible Markov chain>.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

The statement is true. For a state $i$, let
$$
\mathcal R_P(i)=\{n\geq1:(P^n)_{ii}>0\}
$$
be its possible return times under $P$, and define $\mathcal R_Q(i)$ similarly. Every positive $P$-path is a positive $Q$-path, so
$$
\mathcal R_P(i)\subseteq\mathcal R_Q(i).
$$
The greatest common divisor of the larger set divides that of the smaller set. Since the latter is one by the assumed <aperiodic Markov chain> property, the former is also one. Hence every state is aperiodic under $Q$.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

The statement is false. On the two-state space, take
$$
P=
\begin{pmatrix}1&0\\0&1\end{pmatrix},
\qquad
Q=
\begin{pmatrix}\tfrac12&\tfrac12\\0&1\end{pmatrix}.
$$
Every positive entry of $P$ remains positive in $Q$. Under $P$, both singleton states are closed <communicating classes>, so neither state is a <transient state>. Under $Q$, however, state one eventually moves to the absorbing state two and can never return after doing so. State one is therefore transient.

Solved by gpt-5.6-sol high.

= d
{parent=18h}
{scope}

= Solution
{parent=d}

The statement is false. Again use two states, but now take
$$
P=
\begin{pmatrix}1&0\\0&1\end{pmatrix},
\qquad
Q=
\begin{pmatrix}\tfrac12&\tfrac12\\
\tfrac12&\tfrac12
\end{pmatrix}.
$$
The support condition holds. Under $P$, the <first return time> to state one is identically one, so $\mu_1=1$. Under $Q$, the chain is irreducible with <stationary distribution> $(1/2,1/2)$. The <mean recurrence time> formula gives
$$
\eta_1=\frac1{\pi_1}=2>1=\mu_1.
$$
Thus enlarging transition support does not imply a smaller mean return time.

Solved by gpt-5.6-sol high.