Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1

The principal value is $\log z=\log |z|+i\mathrm{Arg}z$, where $-\pi <\mathrm{Arg}z\le \pi$ and $z\ne 0$.

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Since $1+i=\sqrt{2}{e}^{i\pi /4}$, the principal value is $\frac{1}{2}\log 2+i\pi /4$. All values are $\frac{1}{2}\log 2+i(\pi /4+2\pi k)$, $k\in \mathbb{Z}$.

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Write $2=2e^{2\pi ik}$. Then $z=i+2^{1/5}{e}^{2\pi ik/5}$ for $k=0,\dots ,4$: the vertices of a regular pentagon of radius $2^{1/5}$ centred at $i$.

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Let $P$ have the old domain basis as columns, $C$ the old codomain basis, and $D$ the new codomain basis. Change of coordinates gives $A\prime =D^{-1}CA{P}^{-1}$, hence

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The radius is $R=\sup \{r\ge 0:\sum a_n{z}^n\text{ converges whenever }|z|<r\}$; equivalently $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The same definition applies over $\mathbb{R}$ and $\mathbb{C}$.

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Here $a_{2n}=(-1{)}^n$ and $a_{2n+1}=0$, so $\sum a_n{z}^n=\sum (-z^2{)}^n=(1+z^2{)}^{-1}=((z-i)(z+i){)}^{-1}$ for $|z|<1$.

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Along real $z=x\uparrow 1$, the modulus is $1/(1+x^2)$ and stays bounded, but along $z=iy$ with $y\uparrow 1$ it is $1/(1-y^2)\to \infty$. Since $\mathrm{limsup}|a_n{|}^{1/n}=1$, $R=1$.

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No. For example ${\sum }_{n\ge 0}(-1{)}^n{x}^n=(1+x{)}^{-1}$ has radius $1$ and is bounded by $1$ on $(0,1)$.

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Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

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$a_n\to a$ means that for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$.

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If $(a_n)$ is bounded, apply Bolzano–Weierstrass to $(\Re a_n,\Im a_n)$ (or successively to both coordinates). If it is unbounded, recursively choose $n_k>n_{k-1}$ with $|a_{n_k}|\ge k^2$. The alternatives are not exclusive: $0,1,0,4,0,9,\dots$ has both kinds of subsequence.

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$x\cdot y={\sum }_i{x}_i{y}_i$ and $\Vert x\Vert =(x\cdot x{)}^{1/2}$.

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Cauchy-Schwarz inequality states $|x\cdot y|\le \Vert x\Vert \Vert y\Vert$. For $y\ne 0$, positivity of $\Vert x-ty{\Vert }^2$ at $t=(x\cdot y)/\Vert y{\Vert }^2$ gives $\Vert x{\Vert }^2-(x\cdot y{)}^2/\Vert y{\Vert }^2\ge 0$; $y=0$ is immediate.

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$(b-a)\times (c-a)=(-1,0,0)\times (-1,-1,1)=(0,1,1)$. Thus one choice is $\hat{n}=(0,1,1)/\sqrt{2}$ and $\beta =\sqrt{2}$.

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The distance is $|d\cdot \hat{n}-\beta |=|3/\sqrt{2}-\sqrt{2}|=1/\sqrt{2}$.

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The centre-to-plane distance is $|g-2|/\sqrt{2}$. Intersection occurs iff this is at most $2\sqrt{2}$, i.e. $-2\le g\le 6$. Equality gives one tangent point; strict inequality gives a circle of radius $\sqrt{8-(g-2{)}^2/2}$.

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$A(A^2-4A+5I)=2I$, so $A$ has inverse.

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$A^{-1}=\frac{1}{2}(A^2-4A+5I)$.

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The relation $A^3=4A^2-5A+2I$ reduces $A^k$ inductively to a linear combination of $I,A,A^2$.

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$Bx=0⟺ASx=0⟺x\in S^{-1}(\ker A)$; invertibility of $S$ preserves dimension.

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Rank-nullity theorem gives $\mathrm{rank}T+\dim \ker T=n$. Equal nullities therefore give equal ranks.

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$B^k=S^{-1}{A}^{k}S$, so the preceding argument applied to $A^k,B^k$ gives equal ranks.

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Cayley-Hamilton theorem says ${\chi }_B(B)=0$. Since $B$ is nilpotent, all eigenvalues and hence every non-leading coefficient of ${\chi }_B(t)$ vanish, so ${\chi }_B(t)=t^n$ and $B^n=0$.

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From $Hv=\mu v$, $v^{†}Hv=\mu \Vert v{\Vert }^2$ is real, so $\mu \in \mathbb{R}$. For distinct eigenvalues, ${\mu }_i{v}_i^{†}{v}_j=v_i^{†}H{v}_j={\mu }_j{v}_i^{†}{v}_j$, hence orthogonality. Normalising the eigenvectors and taking them as the rows of $U$ gives $H=U^{†}\Lambda U$.

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${\chi }_H(\lambda )=\det (H-\lambda I)={\prod }_i({\mu }_i-\lambda )$ for this sign convention.

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The subspace $y^{\perp }$ is fixed by $M=I-x{y}^{†}$, while the remaining determinant factor is $1-y^{†}x$. Thus $\det M=1-y^{\ast }\cdot x$.

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Apply the matrix determinant lemma to $HP-\lambda I=(H-\lambda I)-Hu,u^T$: its determinant is ${\chi }_H(\lambda )[1-u^T(H-\lambda I{)}^{-1}Hu]$.

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$Pu=0$, hence $HPu=0$. In the diagonal basis, $f(\lambda )=1-{\sum }_j{u}_j^2{\mu }_j/({\mu }_j-\lambda )=-\lambda {\sum }_j{u}_j^2/({\mu }_j-\lambda )$. Between consecutive ${\mu }_j$ its nonzero factor is strictly monotone from one infinite sign to the other, giving one root; zero weights give equality at endpoints. Hence ${\mu }_1\le {\sigma }_1\le {\mu }_2\le \cdots \le {\sigma }_{n-1}\le {\mu }_n$.

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All three Pauli matrices are Hermitian. Direct multiplication gives ${\sigma }_a^2=I$ and, for $a\ne b$, ${\sigma }_a{\sigma }_b=i{\epsilon }_{abc}{\sigma }_c$, which combines as the displayed identity.

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Because $(n\cdot \sigma {)}^2=I$, even and odd powers sum separately: $R=I\cos \theta +i(n\cdot \sigma )\sin \theta$. Then $R{R}^{†}=I$.

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$A=\left(\begin{array}{cc}{x}_3& x_1-i{x}_2\\ x_1+i{x}_2& -x_3\end{array}\right)$, so $\mathrm{tr}A=0$ and $\det A=-|x{|}^2$. Unitary conjugation preserves Hermiticity, trace, and determinant; every traceless Hermitian $2\times 2$ matrix is $x\prime \cdot \sigma$ with real $x\prime$, and $|x\prime |=|x|$.

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Using $(a\cdot \sigma )(b\cdot \sigma )=(a\cdot b)I+i(a\times b)\cdot \sigma$ givesThe sign follows from the convention $R=e^{i\theta n\cdot \sigma }$.

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For a polynomial of degree $d$, Taylor expansion about $x_0$ terminates: $P(x_0+h)={\sum }_{n=0}^d{P}^{(n)}(x_0)h^n/n!$.

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If $f$ has $n+1$ derivatives near $x_0$, then $f(x)={\sum }_{k=0}^n{f}^{(k)}(x_0)(x-x_0{)}^k/k!+f^{(n+1)}(\xi )(x-x_0{)}^{n+1}/(n+1)!$ for some intermediate $\xi$.

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Choose a compact interval around $x_0$ and its constant $C$. For $|h|<1/C$, the Lagrange remainder after order $n$ is at most $(C|h|{)}^{n+1}\to 0$. Thus the taylor series with $a_n=f^{(n)}(x_0)/n!$ equals $f(x_0+h)$.

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$e^x$ is real analytic because $e^{x_0+h}=e^{x_0}{\sum }_{n\ge 0}{h}^n/n!$; it is not a polynomial since all its derivatives are nonzero.

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If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then some $c\in (a,b)$ has $f\prime (c)=0$. A maximum or minimum is attained; unless $f$ is constant, one extremum lies inside, where the derivative vanishes.

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The mean value theorem gives $f(b)-f(a)=(b-a)f\prime (c)>0$.

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Since $f\prime \prime \prime <0$, $f\prime \prime$ decreases from $0$, so $f\prime$ strictly decreases from $1$ and $f(x)<x$. If $f\prime$ never vanished, its decreasing positive limit would make $f$ stay positive, but $f\prime \prime <0$ bounded away from zero after any fixed point forces $f\prime$ negative eventually. Thus $f$ rises to one maximum and then decreases strictly to $-\infty$, crossing zero exactly once; it is positive before that crossing.

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$\sin x={\sum }_{n\ge 0}(-1{)}^n{x}^{2n+1}/(2n+1)!$ and $\cos x={\sum }_{n\ge 0}(-1{)}^n{x}^{2n}/(2n)!$. Here $\sin \prime \prime \prime =-\cos$ is negative while $\cos >0$ near zero; applying the preceding shape argument until the first zero of $\cos$ and then the addition identities establishes a first positive zero $x_0$ with positivity before it.

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From the addition formula and $\sin x_0=0$, $\cos x_0=-1$, so $2x_0$ is a period. Repeatedly subtract $2x_0$ from a positive period $2x_1$; the remainder in $[0,2x_0)$ is also a period, and positivity plus the addition identities force it to be $0$. Thus $x_1=n{x}_0$; the converse follows from periodicity and oddness.

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For continuous $f$, $F(x)={\int }_a^{x}f(t)dt$ is differentiable with $F\prime =f$. Conversely, if $F\prime$ is continuous, ${\int }_a^{b}F\prime =F(b)-F(a)$.

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Put $h(x)=|f(0)|+{\int }_0^{x}g(t)|f(t)|dt$. Then $|f|\le h$ and $h\prime \le gh$. If $h>0$, $(\log h)\prime \le g$; integration gives $h(x)\le h(0)e^{{\int }_0^{x}g}$. For $f(0)=0$, apply the same argument after adding $\epsilon$ and let $\epsilon \downarrow 0$.

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For $h=f_1-f_2$, $h\prime =\sqrt{1+x}(f_1+f_2)h$ and $h(0)=0$. On every compact subinterval the coefficient is continuous, so Gronwall inequality gives $h=0$; hence equality holds throughout $[0,a)$.

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False. Every interval contains rationals and irrationals, so every lower Darboux sum is $0$ and every upper sum is $1$.

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False. Thomae’s function is Riemann integrable and discontinuous at every rational, an infinite countable set.

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False. Continuous functions can converge pointwise to the non-Riemann-integrable Dirichlet function; for example enumerate the rationals and use continuous narrow bumps whose first $n$ peaks cover the first $n$ rationals while widths shrink suitably pointwise off $\mathbb{Q}$.

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False without continuity: the function equal to $1$ at one point and $0$ elsewhere is nonnegative, Riemann integrable with integral $0$, but is not identically zero.

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False. The Cantor function has derivative $0$ outside the Cantor set, hence its derivative extends to the Riemann-integrable zero function, while $F(1)-F(0)=1$.

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True. If $|f(x_0)|>C$, continuity gives an interval of positive length on which $|f|>C+\delta$. Its $L^p$ norm exceeds $(C+\delta ){\ell }^{1/p}$, which tends to $C+\delta >C$ as $p\to \infty$, a contradiction.

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