past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperib_1_2021.pdf

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

= Solution
{parent=1E}

The map
$$
\ell_v:V\to\mathbb R,
\qquad
\ell_v(x)=\langle v,x\rangle
$$
is a nonzero <linear functional>: otherwise $v$ would lie in the <radical of a bilinear form>, contradicting nondegeneracy. Its kernel is $v^\perp$, so the <rank-nullity theorem> gives $\dim v^\perp=n-1$. Antisymmetry gives $\langle v,v\rangle=-\langle v,v\rangle=0$, hence $v\in v^\perp$.

Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^\perp$, it is also orthogonal to $v$, and therefore to
$$
W\oplus\mathbb Rv=v^\perp.
$$
For a <nondegenerate bilinear form>, $(v^\perp)^\perp=\mathbb Rv$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap\mathbb Rv=\{0\}$, proving that the restriction to $W$ is nondegenerate.

The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional <symplectic vector space> has even dimension.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2F}

The condition $Df_p\ne0$ says that $0$ is a <regular value> of $f$. The <regular level set theorem> therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in $\mathbb R^3$, hence a <smooth surface>.

For
$$
F(x,y,z)=x^2+y^2-\cosh(z^2),
$$
the <gradient> is
$$
\nabla F=(2x,2y,-2z\sinh(z^2)).
$$
On $F^{-1}(0)$, $x^2+y^2=\cosh(z^2)\geq1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne0$ there, and the given set is a smooth surface.

Not every smooth surface in $\mathbb R^3$ is a global zero set. Every set $f^{-1}(0)$ is <closed set>[closed] because $f$ is continuous, whereas the open unit disk
$$
\{(x,y,0):x^2+y^2<1\}
$$
is a smooth surface but is not closed in $\mathbb R^3$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3B}

The integrand has an order-two <pole> at $0$ and a simple pole at $2$. Write it near zero as $h(z)/z^2$, where
$$
h(z)=\frac{z^2+e^z}{z-2}.
$$
Its <residue> at zero is
$$
h'(0)=\left.\frac{(2z+e^z)(z-2)-(z^2+e^z)}{(z-2)^2}\right|_{z=0}
=-\frac34.
$$
At $z=2$ the residue is
$$
\frac{2^2+e^2}{2^2}=1+\frac{e^2}{4}.
$$
The <residue theorem> now gives
$$
\boxed{
g(x)=\begin{cases}
-\dfrac{3\pi i}{2},&0<x<2,\\[4pt]
\dfrac{\pi i}{2}(1+e^2),&x>2.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4D}

Introduce a <Lagrange multiplier> $\lambda$ for the normalization and vary
$$
J[u]=\int_D\bigl(|\nabla u|^2-\lambda u^2\bigr)\,dx\,dy.
$$
For a smooth variation $u+\varepsilon\eta$ with $\eta=0$ on $\partial D$, <integration by parts> gives
$$
\left.\frac d{d\varepsilon}J[u+\varepsilon\eta]\right|_{\varepsilon=0}
=2\int_D(\nabla u\mathbin{\cdot}\nabla\eta-\lambda u\eta)
=-2\int_D(\nabla^2u+\lambda u)\eta.
$$
The <fundamental lemma of the calculus of variations> therefore yields the <Euler-Lagrange equation>
$$
\boxed{\nabla^2u+\lambda u=0}.
$$

Multiplying by $u$ and integrating, while using
$$
\nabla\mathbin{\cdot}(u\nabla u)=|\nabla u|^2+u\nabla^2u,
$$
the <divergence theorem> and $u=0$ on the boundary give
$$
0=I[u]+\int_Du\nabla^2u=I[u]-\lambda\int_Du^2.
$$
The normalization is one, so the multiplier equals the stationary value:
$$
\boxed{\lambda=I[u]}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5B}

Apply symmetric <Gaussian elimination> without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric <Schur complement>. If $d_k\leq0$, stop and report that $A$ is not <positive-definite matrix>[positive definite]. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an <LDL decomposition>
$$
A=LDL^T,
$$
where $L$ is unit lower triangular and $D=\operatorname{diag}(d_1,\ldots,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,
$$
x^TAx=(L^Tx)^TD(L^Tx)>0
$$
because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square gives
$$
\begin{pmatrix}s\\y\end{pmatrix}^{T}
\begin{pmatrix}a&b^T\\b&C\end{pmatrix}
\begin{pmatrix}s\\y\end{pmatrix}
=a\left(s+\frac{b^Ty}{a}\right)^2
+y^T\left(C-\frac{bb^T}{a}\right)y.
$$
Choosing $s=-b^Ty/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also <Sylvester's criterion>.

At step $k$, updating the remaining matrix costs $O((n-k)^2)$ arithmetic operations. Hence the total is
$$
\sum_{k=1}^nO((n-k)^2)=O(n^3),
$$
which proves the existence of the required algorithm.

Solved by gpt-5.6-sol high.

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

= a
{parent=6h}
{scope}

= Solution
{parent=a}

A <sufficient statistic> $T$ is one for which the conditional distribution of the full sample $(X_1,\ldots,X_n)$ given $T$ does not depend on $p$. Take
$$
T=\sum_{i=1}^nX_i.
$$
For a binary sample $x$ with $\sum_i x_i=t$, its <likelihood function> is
$$
p^t(1-p)^{n-t},
$$
which depends on the data only through $t$. By the <Fisher-Neyman factorization theorem>, $T$ is sufficient. Equivalently, conditionally on $T=t$, the sample is uniform over the $\binom nt$ binary vectors containing $t$ ones, independently of $p$.

Solved by gpt-5.6-sol high.

= b
{parent=6h}
{scope}

= Solution
{parent=b}

The <Rao-Blackwell theorem> states that if $T$ is sufficient and $U$ is an estimator with finite variance, then
$$
U^*=\mathbb E[U\mid T]
$$
has the same expectation as $U$ and no larger variance; it preserves unbiasedness. The <tower property of conditional expectation> gives
$$
\mathbb E U^*=\mathbb E U.
$$
The <law of total variance> gives
$$
\operatorname{Var}(U)
=\operatorname{Var}(\mathbb E[U\mid T])
+\mathbb E[\operatorname{Var}(U\mid T)]
\geq\operatorname{Var}(U^*).
$$
Sufficiency ensures that $U^*$ is a statistic whose definition does not depend on the unknown parameter. The inequality is strict exactly when the conditional variance is positive with positive probability.

Solved by gpt-5.6-sol high.

= c
{parent=6h}
{scope}

= Solution
{parent=c}

The estimator $U=X_1X_2$ is <unbiased estimator>[unbiased] for $p^2$ because the <independent random variables> $X_1,X_2$ satisfy $\mathbb E[X_1X_2]=p^2$. Given $T=t$, all placements of the $t$ successes are equally likely, so
$$
\mathbb E[X_1X_2\mid T=t]
=\frac{\binom{n-2}{t-2}}{\binom nt}
=\frac{t(t-1)}{n(n-1)}.
$$
Thus the Rao-Blackwellized estimator is
$$
\boxed{\widehat{p^2}=\frac{T(T-1)}{n(n-1)}}.
$$
It is unbiased by the tower property. Since $n\geq3$ and $p\in(0,1)$, the event $T=2$ has positive probability, and conditionally on it $X_1X_2$ takes both zero and one with positive probability. Hence $\mathbb E[\operatorname{Var}(X_1X_2\mid T)]>0$, so the new estimator has strictly smaller variance.

Solved by gpt-5.6-sol high.

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

= a
{parent=7h}
{scope}

= Solution
{parent=a}

For $0\leq t\leq1$ and $x,y\in\mathbb R^d$, <convex function>[convexity] gives
$$
f_i(tx+(1-t)y)\leq tf_i(x)+(1-t)f_i(y).
$$
Taking the maximum over $i$ and then bounding each term by the corresponding endpoint maxima yields
$$
\max_i f_i(tx+(1-t)y)
\leq t\max_i f_i(x)+(1-t)\max_i f_i(y).
$$
Thus the finite <pointwise maximum of convex functions> is convex. Summing the original inequalities over $i$ proves that $\sum_i f_i$ is convex as well.

Solved by gpt-5.6-sol high.

= b
{parent=7h}
{scope}

= Solution
{parent=b}

Because $x\mapsto c^Tx$ is a <linear map>,
$$
g(tx+(1-t)y)
=f(tc^Tx+(1-t)c^Ty)
\leq tf(c^Tx)+(1-t)f(c^Ty).
$$
Hence composition of a convex function with an affine map is convex, and $g$ is convex.

Solved by gpt-5.6-sol high.

= c
{parent=7h}
{scope}

= Solution
{parent=c}

The <softplus function>
$$
h(s)=\log(1+e^s)
$$
is convex because
$$
h''(s)=\frac{e^s}{(1+e^s)^2}\geq0.
$$
Part (b) therefore shows that every $\beta\mapsto h(a_i^T\beta)$ is convex. The <absolute value function> is convex, so $\beta\mapsto|\beta_j|$ is convex for every coordinate $j$. Finally, part (a) says that a finite sum of convex functions is convex. Therefore
$$
Q(\beta)=\sum_{i=1}^n\log(1+e^{a_i^T\beta})+\sum_{j=1}^d|\beta_j|
$$
is convex. The second sum is the <L1 norm> regularizer.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 1}
{scope}
{title2=Linear Algebra  }

= a
{parent=8e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <nilpotent Jordan block> $J_d$ shifts each standard basis vector one place toward the first coordinate. Therefore
$$
(J_d^n)_{ij}
=\begin{cases}
1,&j-i=n,\\
0,&\text{otherwise}.
\end{cases}
$$
Thus $J_d^0=I$, for $1\leq n<d$ the matrix has ones precisely on its $n$th superdiagonal, and
$$
\boxed{J_d^n=0\quad\text{for }n\geq d}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Since $\lambda I$ commutes with $J_d$, the <binomial theorem for commuting matrices> and part (i) give
$$
\boxed{(\lambda I+J_d)^n
=\sum_{k=0}^{\min(n,d-1)}
\binom nk\lambda^{\,n-k}J_d^k}.
$$
Equivalently, its $k$th superdiagonal is constant with value $\binom nk\lambda^{n-k}$ for $0\leq k<d$, with the convention that this value is zero when $k>n$.

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

If $\phi v=\mu v$ for a nonzero <eigenvector> $v$, then
$$
0=\phi^nv=\mu^nv.
$$
Hence every <eigenvalue> satisfies $\mu=0$. Since a complex endomorphism has an eigenvalue, zero is the only possible eigenvalue.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The <Jordan normal form> can contain only blocks $J_r(0)$ with eigenvalue zero. Moreover,
$$
J_r(0)^n=0
\quad\Longleftrightarrow\quad r\leq n.
$$
Thus the possible blocks are precisely the nilpotent Jordan blocks of sizes $1\leq r\leq n$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Assume $\phi^2=0$. Then
$$
\operatorname{im}\phi\subseteq\ker\phi.
$$
Set $W_1=\operatorname{im}\phi$, choose a <direct-sum complement> $U$ such that $\ker\phi=U\oplus W_1$, and choose a complement $W_2$ such that $V=\ker\phi\oplus W_2$.

The restriction $\phi|_{W_2}:W_2\to W_1$ is injective because $W_2\cap\ker\phi=0$, and it is surjective because every image $\phi(v)$ equals $\phi(w_2)$ after decomposing $v=k+w_2$. It is therefore a <linear isomorphism>[isomorphism], so $\dim W_2=\dim W_1$ and $\phi(W_2)=W_1$. Also $\phi(U)=\phi(W_1)=0$. Hence
$$
\boxed{V=U\oplus W_1\oplus W_2}
$$
has all the required properties.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=9G}

Suppose first that every <ideal> of $R$ is finitely generated. For an ascending chain
$$
I_1\subseteq I_2\subseteq\cdots,
$$
the union $I=\bigcup_nI_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\geq N$. Thus $R$ satisfies the <ascending chain condition> and is a <Noetherian ring>.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\ldots,a_n$, choose
$$
a_{n+1}\in I\setminus(a_1,\ldots,a_n).
$$
This creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\varphi:R\to S$ is surjective and $J\lhd S$, then $\varphi^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\ldots,a_m$, then $J$ is generated by $\varphi(a_1),\ldots,\varphi(a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The <Hilbert basis theorem> states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I\lhd R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\ldots,f_s\in I$, and let $N=\max_i\deg f_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha\in\mathbb C$, the ring $\mathbb Z[\alpha]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb Z[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb Z[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb Z[X]$, and quotients preserve Noetherianity.

For a <unique factorization domain> that is not Noetherian, take
$$
k[x_1,x_2,\ldots].
$$
Every polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. But
$$
(x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots
$$
is a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb R,\mathbb R)$ is not Noetherian. For $n\geq1$, let
$$
I_n=\{f:f(x)=0\text{ for all }|x|\leq1/n\}.
$$
These are ideals and $I_n\subsetneq I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.

Solved by gpt-5.6-sol high.

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

= i
{parent=10f}
{scope}

= Solution
{parent=i}

Assume inverse images of <open set>[open sets] are open. If $x_n\to a$ and $V$ is any open neighbourhood of $f(a)$, then $f^{-1}(V)$ is an open neighbourhood of $a$. By the definition of <convergence in a metric space>, $x_n\in f^{-1}(V)$ eventually, and hence $f(x_n)\in V$ eventually. Thus $f(x_n)\to f(a)$.

Solved by gpt-5.6-sol high.

= ii
{parent=10f}
{scope}

= Solution
{parent=ii}

Assume the sequential condition, and let $A\subseteq Y$ be open. If $f^{-1}(A)$ were not open, some $a\in f^{-1}(A)$ would have no ball contained in $f^{-1}(A)$. For each $n$, choose
$$
x_n\notin f^{-1}(A),
\qquad
d_X(x_n,a)<\frac1n.
$$
Then $x_n\to a$, so $f(x_n)\to f(a)\in A$. Because $A$ is open, this forces $f(x_n)\in A$ eventually, a contradiction. Therefore $f^{-1}(A)$ is open. This proves the <sequential characterization of continuity in metric spaces>.

Solved by gpt-5.6-sol high.

= a
{parent=10f}
{scope}

= Solution
{parent=a}

This is always true: it is the <Heine-Cantor theorem>. If uniform continuity failed, there would be an $\varepsilon>0$ and sequences $x_n,y_n\in X$ such that
$$
d_X(x_n,y_n)<\frac1n,
\qquad
d_Y(f(x_n),f(y_n))\geq\varepsilon.
$$
By <sequential compactness of a compact metric space>, some subsequence $x_{n_k}$ converges to $x\in X$. The triangle inequality gives $y_{n_k}\to x$. Continuity then makes both image subsequences converge to $f(x)$, contradicting their separation by $\varepsilon$.

Solved by gpt-5.6-sol high.

= b
{parent=10f}
{scope}

= Solution
{parent=b}

This may be false because continuity only forces the <continuous image of a compact space> $f(X)$ to be compact, not the whole codomain. For example, let $X=\{0\}$, let $Y=\mathbb R$, and set $f(0)=0$. The domain is compact and $f$ is continuous, but $\mathbb R$ is not compact.

Solved by gpt-5.6-sol high.

= c
{parent=10f}
{scope}

= Solution
{parent=c}

This is always true. The continuous image $f(X)$ of a <connected space> is connected. Its <closure> is also connected: if the closure were separated into disjoint nonempty relatively open sets, connectedness would put $f(X)$ inside one of them, preventing its closure from meeting the other. Since $f(X)$ is dense in $Y$, its closure is $Y$, so $Y$ is connected.

Solved by gpt-5.6-sol high.

= d
{parent=10f}
{scope}

= Solution
{parent=d}

This is always true for metric spaces. Let $x_n\to x$. If $f(x_n)$ did not converge to $f(x)$, some subsequence would remain at least $\varepsilon>0$ from $f(x)$. Compactness of $Y$ gives a further subsequence
$$
f(x_{n_k})\to y.
$$
Then $(x_{n_k},f(x_{n_k}))\to(x,y)$. The graph is closed, so $(x,y)$ belongs to it and $y=f(x)$, contradicting the $\varepsilon$ separation. Thus $f(x_n)\to f(x)$ for every convergent sequence, and the equivalence proved above makes $f$ continuous. This is the <closed-graph criterion with compact codomain>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=11F}

For an oriented <smooth surface> $S$, the <Gauss map> sends $p$ to the chosen unit <normal vector> $N(p)\in S^2$. Since $|N|^2=1$, differentiation shows that $DN_p(X)$ is perpendicular to $N(p)$ and hence lies in $T_pS$.

In a local parametrization $\phi(u,v)$ with $n=N\circ\phi$, differentiating
$$
n\mathbin{\cdot}\phi_u=n\mathbin{\cdot}\phi_v=0
$$
gives
$$
n_u\mathbin{\cdot}\phi_v=-n\mathbin{\cdot}\phi_{uv}
=n_v\mathbin{\cdot}\phi_u.
$$
Thus the bilinear form $(X,Y)\mapsto DN_p(X)\mathbin{\cdot}Y$ is symmetric, so $DN_p$ is <self-adjoint operator>[self-adjoint]. The <Gaussian curvature> is
$$
\kappa=\det(DN_p).
$$
Writing the coefficients of the <first fundamental form> as
$$
E=\phi_u^2,\qquad F=\phi_u\mathbin{\cdot}\phi_v,\qquad G=\phi_v^2
$$
and those of the <second fundamental form> as
$$
e=n\mathbin{\cdot}\phi_{uu},\qquad
f=n\mathbin{\cdot}\phi_{uv},\qquad
g=n\mathbin{\cdot}\phi_{vv},
$$
one obtains
$$
\boxed{\kappa=\frac{eg-f^2}{EG-F^2}}.
$$

At an <umbilical point>[umbilic point], the self-adjoint map $DN_p$ has a repeated eigenvalue, so it is a scalar map. If every point is umbilic, there is a function $\lambda$ with
$$
n_u=\lambda\phi_u,\qquad n_v=\lambda\phi_v.
$$
Equality of mixed partial derivatives gives
$$
\lambda_v\phi_u=\lambda_u\phi_v.
$$
The two tangent vectors are linearly independent, hence $\lambda_u=\lambda_v=0$. Since $\mathbb R^2$ is connected, $\lambda$ is constant.

If $\lambda=0$, then $n$ is constant and
$$
\partial_u(n\mathbin{\cdot}\phi)
=\partial_v(n\mathbin{\cdot}\phi)=0,
$$
so $S$ lies in a plane. If $\lambda\ne0$, then
$$
\partial_u(n-\lambda\phi)
=\partial_v(n-\lambda\phi)=0.
$$
Thus $n-\lambda\phi=c$ is constant, and
$$
\left|\phi+\frac c\lambda\right|
=\frac{|n|}{|\lambda|}
=\frac1{|\lambda|}.
$$
Therefore $S$ lies in a sphere of radius $1/|\lambda|$. This proves that the surface is part of a plane or part of a sphere.

Solved by gpt-5.6-sol high.

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

= a
{parent=12g}
{scope}

= Solution
{parent=a}

The <Laurent theorem> says that if $f$ is analytic on an annulus
$$
r<|z-a|<R,
$$
then it has a unique <Laurent series>
$$
f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n
$$
converging locally uniformly on that annulus, where
$$
c_n=\frac1{2\pi i}\oint_C\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta
$$
for any positively oriented circle $C$ in the annulus around $a$.

An <isolated singularity> at $a$ is a point at which $f$ is not analytic although it is analytic on some punctured neighbourhood. It is removable when every $c_n$ with $n<0$ vanishes; it is a pole of order $m$ when $c_{-m}\ne0$ and $c_n=0$ for $n<-m$; and it is essential when infinitely many negative-index coefficients are nonzero.

For $0<|z|<1$,
$$
\frac1{z(z-1)}
=-\frac1z\frac1{1-z}
=-\sum_{n=0}^{\infty}z^{n-1}.
$$
For $|z|>1$,
$$
\frac1{z(z-1)}
=\frac1{z^2}\frac1{1-z^{-1}}
=\sum_{n=0}^{\infty}z^{-n-2}.
$$
The coefficients are unique after the annulus is fixed; these expansions differ because they represent the function on different annuli. At zero the first expansion has principal part $-z^{-1}$, so zero is a simple pole with residue $-1$.

Solved by gpt-5.6-sol high.

= b
{parent=12g}
{scope}

= Solution
{parent=b}

Put $g=1/f$. The hypothesis $|f(z)|\to\infty$ gives $g(z)\to0$ as $z\to a$, so defining $g(a)=0$ makes $g$ continuous on $U$ and analytic there by the stated assumption. Its zero at $a$ has some finite order $m\geq1$, and hence
$$
g(z)=(z-a)^m q(z),
\qquad q(a)\ne0.
$$
Therefore
$$
f(z)=(z-a)^{-m}\frac1{q(z)}
$$
has a pole of order $m$: its Laurent series has $c_{-m}\ne0$ and $c_n=0$ for $n<-m$.

Now let $f$ be entire and tend to infinity at infinity. The function
$$
h(z)=f(1/z)
$$
tends to infinity as $z\to0$, so the preceding argument says that $h$ has a pole at zero. If the <Taylor series> of $f$ is $f(w)=\sum_{n\geq0}a_nw^n$, then
$$
h(z)=\sum_{n\geq0}a_nz^{-n}.
$$
A pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a <polynomial>.

Solved by gpt-5.6-sol high.

= c
{parent=12g}
{scope}

= Solution
{parent=c}

For
$$
g(z)=\frac{e^z-1}{z\log(1+z)},
$$
both $e^z-1$ and $\log(1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its <residue> is
$$
\operatorname{Res}(g,0)
=\lim_{z\to0}\frac{e^z-1}{\log(1+z)}
=\boxed{1}.
$$

For $h(z)=\sin z\sin(1/z)$, multiplication of the two Laurent series shows that all powers are even:
$$
h(z)=
\sum_{p,q\geq0}
\frac{(-1)^{p+q}}{(2p+1)!(2q+1)!}\,
z^{2(p-q)}.
$$
There are infinitely many negative powers, so zero is an <essential singularity>. There is no $z^{-1}$ term, and therefore
$$
\boxed{\operatorname{Res}(h,0)=0}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=13c}
{scope}

= Solution
{parent=a}

With $\xi=x+ct$ and $\eta=x-ct$,
$$
\partial_x=\partial_\xi+\partial_\eta,
\qquad
\partial_t=c\partial_\xi-c\partial_\eta,
$$
so the <wave equation> becomes
$$
u_{tt}-c^2u_{xx}=-4c^2u_{\xi\eta}=0.
$$
Thus $u=F(\xi)+G(\eta)$. At $t=0$ the initial data give
$$
F(x)+G(x)=\phi(x),
\qquad
cF'(x)-cG'(x)=\psi(x).
$$
Solving for $F'$ and $G'$ and integrating gives <D'Alembert formula>[d'Alembert's formula]>
$$
\boxed{
u(x,t)=\frac{\phi(x+ct)+\phi(x-ct)}2
+\frac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds }.
$$

Solved by gpt-5.6-sol high.

= b
{parent=13c}
{scope}

= Solution
{parent=b}

Extend the forcing oddly across the boundary:
$$
f_{\rm odd}(y,s)=
\begin{cases}
f(y,s),&y\geq0,\\
-f(-y,s),&y<0.
\end{cases}
$$
The initial displacement $\sin x$ is already odd, so its homogeneous evolution on the line is $\sin x\cos(ct)$. Applying <Duhamel principle>[Duhamel's principle] to the odd extension gives
$$
\boxed{
u(x,t)=\sin x\cos(ct)
+\frac1{2c}\int_0^t
\int_{x-c(t-s)}^{x+c(t-s)}
f_{\rm odd}(y,s)\,dy\,ds }.
$$
The odd-reflection method makes $u(0,t)=0$. At $t=0$ the double integral vanishes together with its first time derivative, so the prescribed initial displacement and velocity are also satisfied.

Solved by gpt-5.6-sol high.

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

= i
{parent=14c}
{scope}

= Solution
{parent=i}

For a normalized wavefunction obeying the infinite-wall <Dirichlet boundary condition>, integration by parts gives the energy expectation
$$
\langle H\rangle
=\int_0^a\left(\frac{\hbar^2}{2m}|\psi'(x)|^2
+U(x)|\psi(x)|^2\right)\,dx\geq0.
$$
Thus every <energy eigenvalue> is nonnegative.

For $0<E<U_0$, the <time-independent Schrodinger equation> and the wall conditions give
$$
\psi(x)=
\begin{cases}
A\sin(kx),&0\leq x\leq a/2,\\
B\sinh(l(a-x)),&a/2\leq x\leq a,
\end{cases}
$$
where $k=\sqrt{2mE}/\hbar$ and $l=\sqrt{2m(U_0-E)}/\hbar$. Continuity of $\psi$ and $\psi'$ at the finite potential step gives
$$
A\sin(ka/2)=B\sinh(la/2),
$$
$$
Ak\cos(ka/2)=-Bl\cosh(la/2).
$$
Dividing and rearranging yields the <bound-state quantization condition>
$$
\boxed{\frac1k\tan\frac{ka}{2}
=-\frac1l\tanh\frac{la}{2}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=14c}
{scope}

= Solution
{parent=ii}

Before the change, the normalized <ground state> of the infinite square well is
$$
\psi_0(x)=\sqrt{\frac2a}\sin\frac{\pi x}{a}.
$$
For an allowed post-quench energy $E\in(0,U_0)$ satisfying part (i), define the unnormalized eigenfunction
$$
\chi_E(x)=
\begin{cases}
\sin(kx),&0\leq x\leq a/2,\\[2pt]
\dfrac{\sin(ka/2)}{\sinh(la/2)}
\sinh(l(a-x)),&a/2\leq x\leq a.
\end{cases}
$$
Its normalization factor is
$$
N_E^{-2}=
\int_0^{a/2}\sin^2(kx)\,dx
+\frac{\sin^2(ka/2)}{\sinh^2(la/2)}
\int_{a/2}^{a}\sinh^2(l(a-x))\,dx.
$$
The <Born rule> therefore gives
$$
\boxed{
\operatorname{prob}(E)
=\frac{2N_E^2}{a}
\left|
\int_0^{a/2}\sin\frac{\pi x}{a}\sin(kx)\,dx
+\frac{\sin(ka/2)}{\sinh(la/2)}
\int_{a/2}^{a}\sin\frac{\pi x}{a}\sinh(l(a-x))\,dx
\right|^2 }.
$$
For a value of $E$ that is not an eigenvalue, this probability is zero. The sudden change leaves the wavefunction fixed, while the energy eigenbasis changes.

Solved by gpt-5.6-sol high.

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

= a
{parent=15d}
{scope}

= Solution
{parent=a}

Since the <magnetic field> is $B=\nabla\times A$, <Stokes theorem> gives, for any oriented surface $S$ bounded by $C$,
$$
\Phi=\int_SB\mathbin{\cdot}dS
=\int_S(\nabla\times A)\mathbin{\cdot}dS
=\boxed{\oint_CA\mathbin{\cdot}dx}.
$$
Under a <gauge transformation> $A\mapsto A+\nabla\chi$, the integral changes by
$$
\oint_C\nabla\chi\mathbin{\cdot}dx=0
$$
because $C$ is closed. The flux expression is therefore gauge independent.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

The magnetostatic <Ampere-Maxwell equation> is $\nabla\times B=\mu_0J$. Substituting $B=\nabla\times A$ and using the <Coulomb gauge> $\nabla\mathbin{\cdot}A=0$ gives
$$
\mu_0J=\nabla\times(\nabla\times A)
=\nabla(\nabla\mathbin{\cdot}A)-\nabla^2A
=-\nabla^2A.
$$
The free-space <Green function of the Laplacian> therefore gives
$$
A(x)=\frac{\mu_0}{4\pi}
\int_{\mathbb R^3}\frac{J(x')}{|x-x'|}\,d^3x'.
$$
For the thin wire current stated in the question this becomes
$$
\boxed{
A(x)=\frac{\mu_0I}{4\pi}\oint_C\frac{dx'}{|x-x'|}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

Substituting the wire potential from part (b) into the flux formula from part (a) gives <Neumann's mutual-inductance formula>
$$
L_{12}
=\frac{\Phi_{12}}{I_2}
=\frac{\mu_0}{4\pi}
\oint_{C_1}\oint_{C_2}
\frac{dx_1\mathbin{\cdot}dx_2}{|x_1-x_2|}.
$$
Interchanging the two curves proves $L_{12}=L_{21}$.

Parametrize the coaxial circles by
$$
x_1=(a\cos\phi,a\sin\phi,0),
\qquad
x_2=(b\cos\psi,b\sin\psi,c).
$$
Then, with $\theta=\phi-\psi$ and $R=\sqrt{a^2+b^2+c^2}$,
$$
dx_1\mathbin{\cdot}dx_2
=ab\cos\theta\,d\phi\,d\psi,
\qquad
|x_1-x_2|=R\sqrt{1-q\cos\theta},
$$
where $q=2ab/R^2$. One angular integration contributes $2\pi$, so
$$
L_{12}
=\frac{\mu_0ab}{2R}
\int_0^{2\pi}\frac{\cos\theta\,d\theta}
{\sqrt{1-q\cos\theta}}.
$$
Since $ab=qR^2/2$, this is
$$
\boxed{
L_{12}=\frac{\mu_0R}{4}f(q)
=\frac{\mu_0}{4}\sqrt{a^2+b^2+c^2}\,f(q)}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=16a}
{scope}

= Solution
{parent=a}

For a <velocity potential>, $u=\nabla\phi$. Writing $\theta=x-t$ gives
$$
\boxed{u=(\varepsilon y\cos\theta,\varepsilon\sin\theta)}.
$$
Its <divergence> is
$$
\boxed{\nabla\mathbin{\cdot}u
=-\varepsilon y\sin\theta}.
$$
Thus this potential flow is generally compressible.

Solved by gpt-5.6-sol high.

= b
{parent=16a}
{scope}

= Solution
{parent=b}

Both $\sin(x-t)$ and $\cos(x-t)$ have zero average over one period, so $\langle u\rangle=0$ at every fixed point.

The <material derivative> gives the particle acceleration
$$
a=\frac{\partial u}{\partial t}+(u\mathbin{\cdot}\nabla)u.
$$
Direct differentiation yields
$$
a_x=\varepsilon y\sin\theta
+\varepsilon^2(1-y^2)\sin\theta\cos\theta,
$$
$$
a_y=-\varepsilon\cos\theta
+\varepsilon^2y\cos^2\theta.
$$
Therefore the <Eulerian time average> at fixed $(x,y)$ is
$$
\boxed{\langle a\rangle=(0,\varepsilon^2y/2)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=16a}
{scope}

= Solution
{parent=c}

The dyed particle satisfies the <Lagrangian trajectory> equations
$$
\dot x=\varepsilon y\cos(x-t),
\qquad
\dot y=\varepsilon\sin(x-t),
\qquad
x(0)=y(0)=0.
$$
For the proposed approximation, $x=O(\varepsilon^2)$ and
$$
\dot y=-\varepsilon\sin t
=\varepsilon\sin(x-t)+O(\varepsilon^3).
$$
Also
$$
\dot x
=\varepsilon^2(\cos^2t-\cos t)
=\varepsilon y\cos(x-t)+O(\varepsilon^3).
$$
The initial conditions hold, verifying
$$
x=\varepsilon^2\left(\frac14\sin2t+\frac t2-\sin t\right),
\qquad
y=\varepsilon(\cos t-1)
$$
through order $\varepsilon^2$.

Over one period, the periodic terms return to their initial values while the secular term changes $x$ by $\varepsilon^2\pi$. Hence the dyed particle has <Stokes drift>
$$
\boxed{\overline v_{\rm particle}
=\left(\frac{\varepsilon^2}{2},0\right)}
$$
to this order, despite the zero Eulerian mean velocity.

Solved by gpt-5.6-sol high.

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

= i
{parent=17b}
{scope}

= Solution
{parent=i}

The <local truncation error> is the error made by one numerical step started from the exact solution:
$$
\tau_{n+1}
=y(t_{n+1})-y(t_n)
-h\phi(t_n,y(t_n),h).
$$
A one-step method has local order $p+1$ when $\tau_{n+1}=O(h^{p+1})$ uniformly for $t_n$ in each fixed bounded time interval.

Solved by gpt-5.6-sol high.

= ii
{parent=17b}
{scope}

= Solution
{parent=ii}

Let $e_n=y^n-y(t_n)$. Subtracting the exact one-step relation from the numerical method and applying the stated <Lipschitz continuity> gives
$$
\|e_{n+1}\|
\leq(1+hL)\|e_n\|+\|\tau_{n+1}\|.
$$
If $\|\tau_{n+1}\|\leq Ch^{p+1}$, iteration yields the <discrete Gronwall inequality>
$$
\|e_n\|
\leq(1+hL)^n\|e_0\|
+Ch^{p+1}\sum_{j=0}^{n-1}(1+hL)^j.
$$
For $nh\leq t^*$,
$$
(1+hL)^n\leq e^{nhL}\leq e^{t^*L}
$$
and
$$
h^{p+1}\sum_{j=0}^{n-1}(1+hL)^j
\leq\frac{e^{t^*L}-1}{L}h^p.
$$
Consequently
$$
\boxed{
\max_{0\leq n\leq\lfloor t^*/h\rfloor}
\|y^n-y(nh)\|
\leq e^{t^*L}\|e_0\|+O(h^p)}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=17b}
{scope}

= Solution
{parent=iii}

Here
$$
\phi(u,h)=\frac14\left[
f(u)+3f\left(u+\frac{2h}{3}f(u)\right)\right].
$$
For $0<h\leq1$, the Lipschitz bound on $f$ gives
$$
|\phi(u,h)-\phi(v,h)|
\leq\left(K+\frac{K^2}{2}\right)|u-v|,
$$
so part (ii) applies with a constant independent of sufficiently small $h$.

The exact solution has the <Taylor series>[Taylor expansion]>
$$
y(t+h)=y(t)+hf(y(t))
+\frac{h^2}{2}f'(y(t))f(y(t))+O(h^3).
$$
Meanwhile,
$$
f\left(y+\frac{2h}{3}f(y)\right)
=f(y)+\frac{2h}{3}f'(y)f(y)+O(h^2),
$$
so one numerical step from $y$ is
$$
y+h\phi(y,h)
=y+hf(y)+\frac{h^2}{2}f'(y)f(y)+O(h^3).
$$
The local error is therefore $O(h^3)$. Taking $p=2$ in part (ii) proves the required second-order global-error bound. This method is a two-stage <Runge-Kutta method>.

Solved by gpt-5.6-sol high.

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

= a
{parent=18h}
{scope}

= Solution
{parent=a}

The <moment-generating function> of $W_i\sim\operatorname{Exp}(1)$ is
$$
M_{W_i}(t)=\frac1{1-t},
\qquad t<1.
$$
Independence makes the moment-generating function of the sum equal the product:
$$
M_{\sum_iW_i}(t)=(1-t)^{-n}.
$$
This is the moment-generating function of the <gamma distribution> $\Gamma(n,1)$, and moment-generating functions determine distributions in a neighbourhood of zero. Hence
$$
\boxed{\sum_{i=1}^nW_i\sim\Gamma(n,1)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

For $y\geq0$ and $X\sim U(0,1)$,
$$
\Pr(-\log X\leq y)
=\Pr(X\geq e^{-y})
=1-e^{-y}.
$$
The distribution function is zero for $y<0$, so this is exactly the distribution function of $\operatorname{Exp}(1)$. Therefore
$$
\boxed{-\log X\sim\operatorname{Exp}(1)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

The <Neyman-Pearson lemma> states that, for testing one simple hypothesis with density $f_0$ against another with density $f_1$, a size-$\alpha$ test that rejects for the largest values of the <likelihood ratio> $f_1/f_0$ is most powerful among all tests of size at most $\alpha$, with boundary randomization if needed.

Solved by gpt-5.6-sol high.

= d
{parent=18h}
{scope}

= Solution
{parent=d}

The normalized density is
$$
f_\theta(x)=(\theta+1)x^\theta\mathbf1_{(0,1)}(x).
$$
For $\theta=1$ against $\theta=0$, the sample likelihood ratio is
$$
\frac{L(1)}{L(0)}
=2^n\prod_{i=1}^nX_i
=2^ne^{-S},
\qquad
S=-\sum_{i=1}^n\log X_i.
$$
It is strictly decreasing in $S$. Under $H_0$, parts (a) and (b) give
$$
S\sim\Gamma(n,1).
$$
If $q_\alpha$ is the lower $\alpha$-quantile of this gamma distribution, the <Neyman-Pearson lemma> gives the most powerful size-$\alpha$ critical region
$$
\boxed{S\leq q_\alpha}.
$$

For any fixed $\theta>0$,
$$
\frac{L(\theta)}{L(0)}
=(\theta+1)^ne^{-\theta S}
$$
is again strictly decreasing in the same statistic $S$. Thus the same critical region is most powerful against every $\theta>0$ and is consequently a <uniformly most powerful test> of $H_0:\theta=0$ against $H_1:\theta>0$.

Solved by gpt-5.6-sol high.

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

= a
{parent=19h}
{scope}

= Solution
{parent=a}

A random time $T$ is a <stopping time> when the event $\{T\leq n\}$ is determined by $X_0,\ldots,X_n$. The <strong Markov property> says that, conditionally on $T<\infty$ and $X_T=x$, the process $(X_{T+k})_{k\geq0}$ is a fresh Markov chain started at $x$, independent of the history before $T$.

Use the state space $\{1,2,4\}\times\mathbb Z_{\geq0}$, with $(2,0)$ absorbing. Observe the chain only when it is at square 2. From $(2,k)$, the change $Y$ in wealth by the next return to square 2 has distribution
$$
\Pr(Y=-1)=\frac12,\qquad
\Pr(Y=1)=\frac18,\qquad
\Pr(Y=2)=\frac38.
$$
Indeed, heads lands on square 3 and returns to square 2 after losing £1. After tails reaches square 4, the remaining two or three moves give the other cases.

For $r=2/3$,
$$
\mathbb E[r^Y]
=\frac12r^{-1}+\frac18r+\frac38r^2=1.
$$
Thus $r^{M_j}$ is a <martingale> for the embedded wealth random walk $M_j$. Stopping when it first reaches $m-1$ or a large upper level and then letting that level tend to infinity gives
$$
\Pr_{(2,m)}(\text{ever hit }(2,m-1))=r=\boxed{\frac23}.
$$
The upper-bound contribution vanishes because $0<r<1$; equivalently, this is the smaller probability solution of the first-step equation.

Solved by gpt-5.6-sol high.

= b
{parent=19h}
{scope}

= Solution
{parent=b}

By part (a) and repeated use of the <strong Markov property>, the probability of descending from $(2,k)$ to $(2,0)$ is
$$
\left(\frac23\right)^k.
$$
Starting from square 1 with £$m$, heads moves to $(2,m)$, whereas tails lands on square 3, loses £1, and moves to $(2,m-1)$. Therefore the loss probability is
$$
\frac12\left(\frac23\right)^m
+\frac12\left(\frac23\right)^{m-1}
=\boxed{\frac56\left(\frac23\right)^{m-1}}.
$$

Solved by gpt-5.6-sol high.