Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1

For the stated angular-frequency Fourier transform convention, the convolution theorem givesThe Fourier inversion theorem and the assumed absolute integrability therefore imply

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

The Fourier transform of a derivative givesSince $p^{\prime }=g$,An arbitrary additive constant in $p$ affects only the zero-frequency Dirac delta function when transforms are interpreted as distributions.

The values of $\log i$ are $i(\pi /2+2\pi n)$. HenceThese are infinitely many distinct positive real numbers.

Sincethe Inverse Fourier transforms of phase factors giveand

From $z=\tan w=-i(e^{2iw}-1)/(e^{2iw}+1)$ one obtainsFor $z=(2\sqrt{3}-3i)/7$ the quotient is $1+i\sqrt{3}=2e^{i\pi /3}$. Its principal logarithm is $\log 2+i\pi /3$, so the principal value is

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

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

Normality givesTherefore one norm vanishes exactly when the other does.

Apply part (b) to the normal matrix $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one hasThus $e$ is an eigenvector of $D^{†}$ with eigenvalue $\overline{\lambda }$.

If $f$ is differentiable at $a$ and $g$ at $f(a)$, then $g\circ f$ is differentiable at $a$ andWrite $g(f(a)+u)-g(f(a))=u\{g^{\prime }(f(a))+\epsilon (u)\}$, where $\epsilon (u)\to 0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the limit proves the formula.

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.

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.

No. TakeBoth functions are nonconstant on every interval and nondifferentiable at zero, but $g(f(x))=2x$ is differentiable.

The comparison test says that if $0\le b_n\le c_n$ eventually and $\sum c_n$ converges, then $\sum b_n$ converges. If $\sum a_n{z}_0^n$ converges, its terms are bounded: $|a_n{z}_0^n|\le M$. For $|z_1|<|z_0|$,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 $|a_n|r_1^n\le M$. ThenThus $R\ge r_1{r}_2$; letting $r_i\uparrow R_i$ gives $R\ge R_1{R}_2$.

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

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.

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

The cross product of the plane normals is proportional to $(1,-1,-1)$, soThe point on the intersection perpendicular to $m$ iswhich 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 $d=(1,1,1)/\sqrt{3}$. The distance between the two skew lines is

The signed displacement of the sphere centre from the plane is $g\cdot n-p$. Orthogonal projection gives the circle centrePythagoras givesA real circle exists when $|p-g\cdot n|\le A$: equality gives tangency and radius zero, while strict inequality gives a genuine circle.

The characteristic polynomial isFor $\lambda =2$, 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.

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

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; 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 $B^n=0$.

For $x\ne 0$,so the Hermitian matrix $G$ has positive real eigenvalues. If $G{e}_i={\lambda }_i{e}_i$, thenMoreover $|A{e}_i{|}^2={\lambda }_i|e_i{|}^2$, so $|f_i|/|e_i|=\sqrt{{\lambda }_i}$.

Choose an orthonormal eigenbasis $e_i$ of $G$ and set $u_i=e_i$, $v_i=A{e}_i/\sqrt{{\lambda }_i}$. 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 andThis is the singular value decomposition.

Herewith eigenvalues $4,1$ and corresponding normalized vectors $(1,-1{)}^T/\sqrt{2}$ and $(1,1{)}^T/\sqrt{2}$. A compatible choice isDirect multiplication gives $V^{†}AU=D$.

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

The nonreal eigenvalues of a real $3\times 3$ 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 $n$,Therefore $\mathrm{tr}Q=-1+2\cos \theta$ and

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

For the subadditive sequence, $0\le x_n/n\le x_1$, so it is bounded. The Fekete lemma can be proved directly here as follows. Let $\alpha ={\inf }_k{x}_k/k$. Fix $k$ and write $n=qk+r$, $0\le r<k$. ThenThus $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition gives $\mathrm{liminf}{x}_n/n\ge \alpha$. Hence $x_n/n\to \alpha$.

Let $S_n={\sum }_{i\le n}{a}_i$ and $T_n={\sum }_{i\le n}|a_i|$. Conditional convergence gives $S_n\to S$ and $T_n\to \infty$. Since $P_n=T_n+S_n$ and $N_n=T_n-S_n$,

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

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 $f\le y$ and $f\ge y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\le y\}$ and using continuity gives a point with value $y$.

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

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

The ordinary mean value theorem gives ${\Delta }_{h}f(a)=h{f}^{\prime }(b_1)$ for some $b_1\in (a,a+h)$. More generally, applying it to the difference of two translated functions shows inductively thatfor some $b_k\in (a,a+kh)$. Taking $k=n$ proves the result.

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

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 $u(x)=1/x-\lfloor 1/x\rfloor$ for $x>0$, so $0\le u<1$. 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 $\delta \downarrow 0$ proves integrability on $[0,1]$.

Changing variables givesThe average of a continuous function over $[y,y+h]$ tends to $f(y)$ as $h\to 0$. The two terms therefore tend to $f(x)$ and $f(a)$, proving the stated limit.

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 $s_n$. ThenReplace each of the finitely many jumps of $s_n$ by a linear transition on intervals of sufficiently small total length, obtaining a continuous ${\varphi }_n$ with $\int |s_n-{\varphi }_n|<1/n$. Hence $\int |g-{\varphi }_n|<2/n$, and uniformly for every subinterval $[\alpha ,\beta ]$,