= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperia_1_2023.pdf

= 1A
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=1a}
{scope}

= Solution
{parent=a}

For the stated angular-frequency <Fourier transform> convention, the <convolution theorem> gives
$$
\widetilde{f*g}(k)=\widetilde f(k)\widetilde g(k),
\qquad
(f*g)(x)=\int_{-\infty}^{\infty}f(\xi)g(x-\xi)\,d\xi.
$$
The <Fourier inversion theorem> and the assumed absolute integrability therefore imply
$$
\boxed{h(x)=\int_{-\infty}^{\infty}f(\xi)g(x-\xi)\,d\xi.}
$$

Solved by gpt-5.6-sol high.

Since $|-i|=1$ and the principal argument is $-\pi/2$,
$$
\operatorname{Log}(-i)=-\frac{\pi i}{2}.
$$
All logarithms are $-\pi i/2+2\pi i n$, $n\in\mathbb Z$.

Solved by gpt-5.6-sol high.

= b
{parent=1a}
{scope}

= Solution
{parent=b}

The <Fourier transform of a derivative> gives
$$
\widetilde{p'}(k)=ik\widetilde p(k).
$$
Since $p'=g$,
$$
\boxed{\widetilde p(k)=\frac{\widetilde g(k)}{ik}},
\qquad k\neq0.
$$
An arbitrary additive constant in $p$ affects only the zero-frequency <Dirac delta function> when transforms are interpreted as distributions.

Solved by gpt-5.6-sol high.

The values of $\log i$ are $i(\pi/2+2\pi n)$. Hence
$$
i^{-2i}=\exp\{-2i\log i\}
=\exp((4n+1)\pi),\qquad n\in\mathbb Z.
$$
These are infinitely many distinct positive real numbers.

Solved by gpt-5.6-sol high.

= c
{parent=1a}
{scope}

= Solution
{parent=c}

Since
$$
\mathcal F^{-1}[e^{ika}]=\delta(x+a),
\qquad
\mathcal F^{-1}[e^{-ika}]=\delta(x-a),
$$
the <inverse Fourier transforms of phase factors> give
$$
\boxed{
\mathcal F^{-1}[\cos(ka)]
=\frac12\bigl[\delta(x+a)+\delta(x-a)\bigr]
}
$$
and
$$
\boxed{
\mathcal F^{-1}[\sin(ka)]
=\frac1{2i}\bigl[\delta(x+a)-\delta(x-a)\bigr].
}
$$

Solved by gpt-5.6-sol high.

From $z=\tan w=-i(e^{2iw}-1)/(e^{2iw}+1)$ one obtains
$$
e^{2iw}=\frac{1+iz}{1-iz},
\qquad
w=\frac1{2i}\log\frac{1+iz}{1-iz}.
$$
For $z=(2\sqrt3-3i)/7$ the quotient is $1+i\sqrt3=2e^{i\pi/3}$. Its principal logarithm is $\log2+i\pi/3$, so the principal value is
$$
\tan^{-1}z=\frac\pi6-\frac i2\log2.
$$

Solved by gpt-5.6-sol high.

= 2C
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= Solution
{parent=2C}

The Hermitian conjugate is $A^\dagger=\overline A^{,T}$. A <matrix> is unitary when $A^\dagger A=AA^\dagger=I$, and Hermitian when $A^\dagger=A$.

Solved by gpt-5.6-sol high.

= a
{parent=2c}
{scope}

= Solution
{parent=a}

For $B=A^{-1}A^\dagger$,
$$
B^\dagger B=A(A^\dagger)^{-1}A^{-1}A^\dagger.
$$
Multiplying $B^\dagger B=I$ on the left by $A^{-1}$ and on the right by $(A^\dagger)^{-1}$ shows that it is equivalent to $(A^\dagger)^{-1}A^{-1}=A^{-1}(A^\dagger)^{-1}$. Taking inverses gives $A^\dagger A=AA^\dagger$. Thus $B$ is unitary exactly when $A$ is a <normal matrix>.

Solved by gpt-5.6-sol high.

= b
{parent=2c}
{scope}

= Solution
{parent=b}

Normality gives
$$
|Cx|^2=x^\dagger C^\dagger Cx=x^\dagger CC^\dagger x=|C^\dagger x|^2.
$$
Therefore one norm vanishes exactly when the other does.

Solved by gpt-5.6-sol high.

= c
{parent=2c}
{scope}

= Solution
{parent=c}

Apply part (b) to the normal <matrix> $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one has
$$
(D^\dagger-\overline\lambda I)e=C^\dagger e=0.
$$
Thus $e$ is an <eigenvector> of $D^\dagger$ with <eigenvalue> $\overline\lambda$.

Solved by gpt-5.6-sol high.

= 3E
{parent=Paper 1}
{scope}
{title2=Analysis}

= Solution
{parent=3E}

If $f$ is <differentiable> at $a$ and $g$ at $f(a)$, then $g\circ f$ is <differentiable> at $a$ and
$$
(g\circ f)'(a)=g'(f(a))f'(a).
$$
Write $g(f(a)+u)-g(f(a))=u\{g'(f(a))+\varepsilon(u)\}$, where $\varepsilon(u)\to0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the <limit> proves the formula.

Solved by gpt-5.6-sol high.

= i
{parent=3e}
{scope}

= Solution
{parent=i}

No. Take $f(x)=x^2$, $a=0$, and $g(y)=|y|$. The <function> $g$ is not <differentiable> at $f(0)=0$, but $g(f(x))=x^2$ is <differentiable>. Neither <function> is constant on an interval.

Solved by gpt-5.6-sol high.

= ii
{parent=3e}
{scope}

= Solution
{parent=ii}

No. Take $f(x)=|x|$, $a=0$, and $g(y)=y^2$. Then $f$ is not <differentiable> at zero, but $g\circ f=x^2$ is <differentiable> there.

Solved by gpt-5.6-sol high.

= iii
{parent=3e}
{scope}

= Solution
{parent=iii}

No. Take
$$
f(x)=\begin{cases}x,&x\geq0,\\2x,&x<0,\end{cases}
\qquad
g(y)=\begin{cases}2y,&y\geq0,\\y,&y<0.
\end{cases}
$$
Both <functions> are nonconstant on every interval and nondifferentiable at zero, but $g(f(x))=2x$ is <differentiable>.

Solved by gpt-5.6-sol high.

= 4E
{parent=Paper 1}
{scope}
{title2=Analysis}

= Solution
{parent=4E}

The comparison test says that if $0\leq b_n\leq c_n$ eventually and $\sum c_n$ converges, then $\sum b_n$ converges. If $\sum a_nz_0^n$ converges, its terms are bounded: $|a_nz_0^n|\leq M$. For $|z_1|<|z_0|$,
$$
|a_nz_1^n|\leq M\left|\frac{z_1}{z_0}\right|^n,
$$
so comparison with a geometric <series> proves absolute convergence.

The radius $R$ is the number for which the power <series> converges absolutely for $|z|<R$ and diverges for $|z|>R$. Given $r_i<R_i$, both $\sum|a_n|r_1^n$ and $\sum|b_n|r_2^n$ converge, so their terms are bounded, say $|a_n|r_1^n\leq M$. Then
$$
\sum|a_nb_n|(r_1r_2)^n
\leq M\sum|b_n|r_2^n<\infty.
$$
Thus $R\geq r_1r_2$; letting $r_i\uparrow R_i$ gives $R\geq R_1R_2$.

Solved by gpt-5.6-sol high.

= 5A
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=5a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

This is the plane through $a$ perpendicular to the nonzero <normal vector> $n$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

This is the affine plane through $b,d,f$, provided $d-b$ and $f-b$ are independent. The <scalars> $\lambda,\mu$ are affine coordinates in those two directions.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

This is the sphere with centre $c$ and radius $\rho>0$.

Solved by gpt-5.6-sol high.

= b
{parent=5a}
{scope}

= Solution
{parent=b}

The cross product of the plane normals is proportional to $(1,-1,-1)$, so
$$
m=\frac1{\sqrt3}(1,-1,-1).
$$
The point on the intersection perpendicular to $m$ is
$$
u=\frac19(11,4,7),
$$
which satisfies both plane equations and $u\cdot m=0$. Hence the line is $r\times m=u\times m$.

The equally inclined line has direction $d=(1,1,1)/\sqrt3$. The distance between the two skew lines is
$$
\frac{|u\cdot(m\times d)|}{|m\times d|}
=\frac1{3\sqrt2}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=5a}
{scope}

= Solution
{parent=c}

The signed displacement of the sphere centre from the plane is $g\cdot n-p$. Orthogonal projection gives the circle centre
$$
h=g+(p-g\cdot n)n.
$$
Pythagoras gives
$$
R=\sqrt{A^2-(p-g\cdot n)^2}.
$$
A real circle exists when $|p-g\cdot n|\leq A$: equality gives tangency and radius zero, while strict inequality gives a genuine circle.

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=6c}
{scope}

= Solution
{parent=a}

The characteristic <polynomial> is
$$
(\lambda-3)(\lambda-2)^2.
$$
For $\lambda=2$, the eigenspace is spanned only by $(1,0,0)^T$, so its geometric multiplicity is one although its algebraic multiplicity is two. Hence $M$ is not diagonalisable.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

If $B=S^{-1}AS$, then
$$
\det(tI-B)=\det(S^{-1}(tI-A)S)=\det(tI-A),
$$
so the characteristic <polynomials>, <eigenvalues>, and algebraic multiplicities agree. The converse is false: $I_2$ and $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ have the same characteristic <polynomial>, but the first is diagonalisable and the second is not, so they cannot be similar.

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= Solution
{parent=c}

The <Cayley-Hamilton theorem> states that a square <matrix> satisfies its characteristic <polynomial>. If a $2\times2$ <matrix> is diagonalisable, $A=S\operatorname{diag}(\lambda_1,\lambda_2)S^{-1}$; applying $p(t)=(t-\lambda_1)(t-\lambda_2)$ to the diagonal <matrix> gives zero and hence $p(A)=0$.

If $B^k=0$, every <eigenvalue> $\lambda$ satisfies $\lambda^k=0$, so all <eigenvalues> are zero and the characteristic <polynomial> is $t^n$. Cayley--Hamilton then gives $B^n=0$.

Solved by gpt-5.6-sol high.

= 7B
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=7b}
{scope}

= Solution
{parent=a}

For $x\ne0$,
$$
x^\dagger Gx=|Ax|^2>0,
$$
so the Hermitian <matrix> $G$ has positive real <eigenvalues>. If $Ge_i=\lambda_i e_i$, then
$$
H(Ae_i)=AA^\dagger Ae_i=A(Ge_i)=\lambda_iAe_i.
$$
Moreover $|Ae_i|^2=\lambda_i|e_i|^2$, so $|f_i|/|e_i|=\sqrt{\lambda_i}$.

Solved by gpt-5.6-sol high.

= b
{parent=7b}
{scope}

= Solution
{parent=b}

Choose an orthonormal eigenbasis $e_i$ of $G$ and set $u_i=e_i$, $v_i=Ae_i/\sqrt{\lambda_i}$. Part (a) shows that the $v_i$ form an orthonormal eigenbasis of $H$. If $U$ and $V$ have these <vectors> as columns, then they are unitary and
$$
V^\dagger AU=\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n}).
$$
This is the <singular value decomposition>.

Solved by gpt-5.6-sol high.

= c
{parent=7b}
{scope}

= Solution
{parent=c}

Here
$$
A^\dagger A=\frac12\begin{pmatrix}5&-3\\-3&5\end{pmatrix},
$$
with <eigenvalues> $4,1$ and corresponding normalized <vectors> $(1,-1)^T/\sqrt2$ and $(1,1)^T/\sqrt2$. A compatible choice is
$$
U=\frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix},
\qquad
V=\begin{pmatrix}1&0\\0&-1\end{pmatrix},
\qquad
D=\begin{pmatrix}2&0\\0&1\end{pmatrix}.
$$
Direct multiplication gives $V^\dagger AU=D$.

Solved by gpt-5.6-sol high.

= 8B
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= Solution
{parent=8B}

A real <matrix> is orthogonal when $Q^TQ=I$. If $Qv=\lambda v$, norm preservation gives $|\lambda|=1$. If $Qv=\lambda v$ and $Qw=\mu w$, preservation of the Hermitian <inner product> gives $(1-\overline\lambda\mu)v^\dagger w=0$; distinct unit-modulus <eigenvalues> therefore have orthogonal <eigenvectors>.

The nonreal <eigenvalues> of a real $3\times3$ <matrix> occur in conjugate pairs. Since their product is one and $\det Q=-1$, the remaining real <eigenvalue> is $-1$. If $x\cdot n=0$, then $(Qx)\cdot(Qn)=x\cdot n=0$, and $Qn=-n$, so $Qx\cdot n=0$: the plane $\Pi$ is invariant.

The restriction to $\Pi$ is a planar orthogonal map whose <determinant> is $(-1)/(-1)=1$, hence a rotation through some $\theta$. In an orthonormal <basis> adapted to $n$,
$$
Q\sim\operatorname{diag}(-1,R_\theta).
$$
Therefore $\operatorname{tr}Q=-1+2\cos\theta$ and
$$
\det(Q-I)=(-2)\det(R_\theta-I)=4(\cos\theta-1).
$$

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 1}
{scope}
{title2=Analysis}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

The arithmetic--geometric mean inequality gives $x_n\geq1$ from $n=2$ onward. For $x>1$,
$$
1<\frac12(x+x^{-1})<x,
$$
so the tail decreases to a <limit> $L\geq1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.

For the subadditive <sequence>, $0\leq x_n/n\leq x_1$, so it is bounded. The <Fekete lemma> can be proved directly here as follows. Let $\alpha=\inf_kx_k/k$. Fix $k$ and write $n=qk+r$, $0\leq r<k$. Then
$$
\frac{x_n}{n}\leq\frac{qx_k+x_r}{qk+r}\longrightarrow\frac{x_k}{k}.
$$
Thus $\limsup x_n/n\leq\alpha$, while the definition gives $\liminf x_n/n\geq\alpha$. Hence $x_n/n\to\alpha$.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

Let $S_n=\sum_{i\leq n}a_i$ and $T_n=\sum_{i\leq n}|a_i|$. Conditional convergence gives $S_n\to S$ and $T_n\to\infty$. Since $P_n=T_n+S_n$ and $N_n=T_n-S_n$,
$$
\frac{P_n}{N_n}=\frac{1+S_n/T_n}{1-S_n/T_n}\longrightarrow1.
$$

The alternating-series test says that $b_n\downarrow0$ implies convergence of $\sum(-1)^nb_n$. The hypothesis implies $b_n/b_{n+1}>1$ eventually, so $b_n$ is eventually decreasing. For some $c>0$, eventually $b_n/b_{n+1}\geq1+c/n$; the divergent product of these factors forces $b_n\to0$. The test now applies.

Solved by gpt-5.6-sol high.

= 10E
{parent=Paper 1}
{scope}
{title2=Analysis}

= Solution
{parent=10E}

The <intermediate value theorem> says that a continuous $f:[a,b]\to\mathbb R$ takes every value between $f(a)$ and $f(b)$. For $f(a)<y<f(b)$, the closed sets where $f\leq y$ and $f\geq y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\leq y\}$ and using continuity gives a point with value $y$.

Set $\phi(a)=0$ and $\phi(x)=\sin(1/(x-a))$ for $x>a$. Every interval $[a,b]$ contains a zero $c<b$; continuity on $[c,b]$ makes $\phi$ take every value between $0=\phi(a)$ and $\phi(b)$, yet $\phi$ is discontinuous at $a$.

A monotone <function> can be discontinuous only by a jump. If it had a jump at $c$, any number strictly between the left and right <limits> would lie between $f(a)$ and $f(b)$ but would not be attained, contrary to the hypothesis. Thus it is continuous.

For the last assertion, pass to subsequences with $x_n,y_n$ both within $1/n$ of $a$ and $g(x_n)$ near $l$, $g(y_n)$ near $L$. For $\lambda\in(l,L)$, the intermediate value theorem on the interval joining $x_n$ and $y_n$ gives $z_n$ with $g(z_n)=\lambda$; then $z_n\to a$. The endpoint cases use the original <sequences>.

Solved by gpt-5.6-sol high.

= 11E
{parent=Paper 1}
{scope}
{title2=Analysis}

= a
{parent=11e}
{scope}

= Solution
{parent=a}

The mean value theorem says that for continuous $f$ on $[a,b]$, <differentiable> inside, $f(b)-f(a)=f'(c)(b-a)$ for some $c$. Applied to $\log$ on $[b,a]$,
$$
\log(a/b)=\frac{a-b}{c},\qquad b<c<a,
$$
which gives $(a-b)/a<\log(a/b)<(a-b)/b$.

Solved by gpt-5.6-sol high.

= b
{parent=11e}
{scope}

= Solution
{parent=b}

The ordinary mean value theorem gives $\Delta_hf(a)=hf'(b_1)$ for some $b_1\in(a,a+h)$. More generally, applying it to the difference of two translated <functions> shows inductively that
$$
\Delta_h^kf(a)=h^kf^{(k)}(b_k)
$$
for some $b_k\in(a,a+kh)$. Taking $k=n$ proves the result.

Solved by gpt-5.6-sol high.

= c
{parent=11e}
{scope}

= Solution
{parent=c}

No. The <Darboux theorem> says that every <derivative> has the intermediate-value property. Choose $\epsilon<1/3$. In a deleted neighborhood of $a$, the assumed <limit> puts $\phi(x)$ within $\epsilon$ of $\phi(a)+1$. Darboux's theorem applied between $a$ and any such $x$ would require values near $\phi(a)+1/2$, which that neighborhood excludes. This contradiction shows that $\phi$ cannot be a <derivative>.

Solved by gpt-5.6-sol high.

= 12E
{parent=Paper 1}
{scope}
{title2=Analysis}

= a
{parent=12e}
{scope}

= Solution
{parent=a}

For a <bounded function>, the lower and upper <integrals> are the supremum of lower Darboux sums and infimum of upper Darboux sums. It is Riemann integrable when these agree.

Here $u(x)=1/x-\lfloor1/x\rfloor$ for $x>0$, so $0\leq u<1$. On $[\delta,1]$ it has only finitely many discontinuities and is piecewise continuous, hence integrable. On $[0,\delta]$ its upper-minus-lower contribution is at most $\delta$. Taking $\delta\downarrow0$ proves integrability on $[0,1]$.

Solved by gpt-5.6-sol high.

= b
{parent=12e}
{scope}

= Solution
{parent=b}

Changing variables gives
$$
\frac1h\int_a^x(f(t+h)-f(t))dt
=\frac1h\left(\int_x^{x+h}f(s)ds-\int_a^{a+h}f(s)ds\right).
$$
The average of a <continuous function> over $[y,y+h]$ tends to $f(y)$ as $h\to0$. The two terms therefore tend to $f(x)$ and $f(a)$, proving the stated <limit>.

Solved by gpt-5.6-sol high.

= c
{parent=12e}
{scope}

= Solution
{parent=c}

The <continuous approximation of a Riemann-integrable function> follows here by choosing partitions $P_n$ with upper-minus-lower sum below $1/n$. On each subinterval choose the midpoint of the infimum and supremum to form a step <function> $s_n$. Then
$$
\int_a^b|g-s_n|<\frac1n.
$$
Replace each of the finitely many jumps of $s_n$ by a linear transition on intervals of sufficiently small total length, obtaining a continuous $\phi_n$ with $\int|s_n-\phi_n|<1/n$. Hence $\int|g-\phi_n|<2/n$, and uniformly for every subinterval $[\alpha,\beta]$,
$$
\left|\int_\alpha^\beta(g-\phi_n)\right|\leq\int_a^b|g-\phi_n|\to0.
$$

Solved by gpt-5.6-sol high.