Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1

Write $z=x+iy$ and $\alpha =p+iq$. The complex conjugates in the PDF giveSince $(p,q)\ne (0,0)$, the locus is the straight line through the origin perpendicular to the position vector of $\alpha$ in the complex plane. Equivalently,

Write $z=x+iy$. Dividing by two givesThe left side is nonnegative, so squaring is reversible only when $x+a\ge 0$. Squaring yields $y^2=4ax$.

If $a>0$, this equation implies $x\ge 0$, so the sign condition holds automatically. The locus is the parabolawith vertex at the origin, focus of a conic $(a,0)$ and directrix of a conic $x=-a$. Indeed the original equation equates distance from the focus of a conic with distance from the directrix of a conic on its right side.

If $a<0$, the squared equation forces $x\le 0$, making $x+a<0$. This violates the sign condition. Thus the locus is empty for $a<0$; squaring without checking that condition would introduce a spurious parabola.

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.

For a complex matrix, its Hermitian conjugate is obtained by transposing and taking entrywise complex conjugates:The defining conditions areFor a square matrix, either one of the two inverse identities in the unitary matrix condition implies the other. A Hermitian matrix may have complex entries, but its diagonal entries are real.

Since $A$ is an invertible matrix, so is its Hermitian conjugate, andIf $A$ is a normal matrix, substitute $A^{†}A=A{A}^{†}$ into the second identity to obtain $B{B}^{†}=I$, so $B$ is a unitary matrix. Conversely, if $B$ is a unitary matrix, multiply $B{B}^{†}=I$ on the left by $A$ and on the right by $A^{†}$. This gives $A^{†}A=A{A}^{†}$, the defining normal matrix identity. Thus $B$ is a unitary matrix exactly when $A$ is a normal matrix.

Using the complex Euclidean norm and the normal matrix identity,Both sides are nonnegative. Therefore $\Vert Cx\Vert =0$ if and only if $\Vert C^{†}x\Vert =0$. This proves the stronger reusable fact that normal matrices have equal adjoint norms.

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.

Rationalizing avoids cancellation between the two large terms of the sequence:By continuity of the square root and the reciprocal at the limiting values,

Separate off the integer part of the phase. Set $d_n=\sqrt{n^2+n}-n$. Rationalization givesSince squaring removes the sign in $\cos (\pi n+u)=(-1{)}^n\cos u$, the given sequence satisfies $b_n={\cos }^2(\pi d_n)$. The continuity of the cosine now gives

The key fact is that convergence of a power series at one nonzero point forces absolute convergence at every smaller modulus. Indeed, if $\sum a_n{w}^n$ converges and $w\ne 0$, its terms form a bounded sequence, say $|a_n{w}^n|\le M$. For $|z|<|w|$,by convergence of a geometric series.

Let $S$ be the set of moduli of points at which the power series converges, and put $R=\sup S$, allowing $R=\infty$. The set contains zero, so its supremum is defined in $[0,\infty ]$. Whenever $|z|<R$, the definition of supremum provides a convergence point $w$ with $|w|>|z|$. The preceding estimate gives absolute convergence at $z$. Whenever $|z|>R$, convergence at $z$ would put $|z|$ in $S$, contradicting its upper bound. This proves the asserted radius of convergence property, including $R=0$ and $R=\infty$. Nothing in the argument decides convergence on $|z|=R$.

For the requested boundary behavior, take the power seriesIts coefficients are $a_{2k}=1/k$ for $k\ge 1$, $a_{2k+1}=0$ and $a_0=0$. At either $z=1$ or $z=-1$ it is the divergent harmonic series. At either $z=i$ or $z=-i$ it is ${\sum }_{k\ge 1}(-1{)}^k/k$, which converges by the alternating series test. In fact its radius of convergence is one, since inside the unit circle the terms are bounded by a geometric series, and outside it they do not tend to zero.

Use suffix notation, summing repeated indices from one to three. The contraction identity for the Levi-Civita symbol givesHere $a\cdot a=1$ because $a$ is a unit vector. For the second cross product identity, use ${ϵ}_{ijk}{ϵ}_{j\ell m}={\delta }_{im}{\delta }_{k\ell }-{\delta }_{i\ell }{\delta }_{km}$:The first term vanishes because $a_{\ell }{a}_n$ is symmetric while the Levi-Civita symbol is antisymmetric in those indices. The last equality uses cyclicity and antisymmetry of the scalar triple product.

On the unit sphere, a great circle has radius one, so the spherical distance is its shorter arc length, equal to its central angle in $[0,\pi ]$. The dot product of its endpoint unit vectors is the cosine of that angle. HenceFor the angular formulas, take the ordinary nondegenerate convex minor spherical triangle: no pair of vertices is antipodal and the three vertices do not lie on one great circle. These are needed for the displayed denominators and interior angles to be defined. Merely requiring three distinct points, as the PDF does, does not exclude degeneracy.

Put $p=\delta (A,B)$, $q=\delta (A,C)$ and $r=\delta (B,C)$. The outgoing unit tangent vectors at $A$ areTheir dot product is $\cos \alpha$. Equivalently, they are obtained from the normalized plane normals $a\times b$ and $a\times c$ by the same right-angle rotation in the tangent plane: $u=(a\times b)\times a/|a\times b|$, and similarly for $v$. Rotation preserves the dot product, givingSince $|a\times b|=\sin p$ and $|a\times c|=\sin q$, the first proved identity becomes the spherical law of cosines:For the spherical law of sines, take norms in the second proved identity. Writing $\tau =a\cdot (b\times c)$, and using $0<\alpha <\pi$, givesCyclic permutation gives the corresponding formulas at $B$ and $C$. Dividing each by the positive product $\sin p\sin q\sin r$ provesFinally, if all sides equal $s$, the spherical law of cosines givesbecause $0<s<\pi$ makes $1+\cos s>0$ and $\cos s<1$. Cosine is strictly decreasing on $(0,\pi )$, so every angle of a nondegenerate equilateral spherical triangle is greater than $\pi /3$.

For any linear system $Ax=c$, a solution exists exactly when $c$ belongs to the image of a linear map represented by $A$. If $x_0$ is one solution, all others are exactlysince $A(x-x_0)=0$, and conversely adding a kernel vector preserves the equation. Thus there are no solutions if $c\notin \mathrm{im}A$, one if $c\in \mathrm{im}A$ and $\ker A=\{0\}$, and infinitely many if $c\in \mathrm{im}A$ and the kernel is nontrivial: every real multiple of a nonzero kernel vector gives a distinct solution. Any two solutions differ by a kernel vector. Equivalently, consistency means the augmented matrix and $A$ have equal rank of a matrix; a consistent square linear system is unique exactly when its determinant is nonzero.

For the given matrix, set $d=b-a$ and $s=2a+b$. Perform the column operation $C_1\leftarrow C_1+C_2+C_3$, then the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-R_1$. Factoring the first column givesThese operations are determinant-preserving, and the displayed polynomial identity remains valid even when a factored expression vanishes.

To find all kernels and images of a linear map, putIf $J$ is the all-ones matrix, then $A=aJ+dP$. The direct sum ${\mathbb{R}}^3=\mathbb{R}u\oplus H$ is invariant under $A$: on $\mathbb{R}u$ it acts by multiplication by $s$, and on $H$ by $dP$. Since $P^3=I$, its restriction to $H$ is invertible.

For $d\ne 0$ and $s\ne 0$, the kernel is $\{0\}$ and the image of a linear map is all of ${\mathbb{R}}^3$. Decompose the right-hand side asApplying the inverse on each invariant summand gives the unique solutionFor $d\ne 0$ and $s=0$, the kernel is $\mathbb{R}u$ and the image of a linear map is $H$. A solution exists exactly when $c+2=0$. For $c=-2$, all solutions areFor $d=0$ and $a\ne 0$, so $a=b$, the kernel is $H$ and the image of a linear map is $\mathbb{R}u$. A solution exists exactly when $c=1$. All such solutions areFinally, for $a=b=0$, the kernel is ${\mathbb{R}}^3$ and the image of a linear map is $\{0\}$. The right-hand side has first entry one, so there is no solution for any $c$. These four cases exhaust all parameter values.

Let $v\ne 0$ be an eigenvector of the Hermitian matrix $A$, with eigenvalue $\lambda$. Multiplying $Av=\lambda v$ by $v^{†}$ givesThe denominator is the positive real number $\Vert v{\Vert }^2$, whileThus the numerator is real, and every eigenvalue of a Hermitian matrix is real. This argument does not require distinct eigenvalues.

Take eigenvectors $v_i,v_j$ for distinct eigenvalues ${\lambda }_i,{\lambda }_j$. Since the eigenvalues are real and $A$ is a Hermitian matrix,Hence $({\lambda }_j-{\lambda }_i)v_i^{†}{v}_j=0$, so eigenvectors for distinct eigenvalues are orthogonal vectors. They are therefore linearly independent; after normalization the given $n$ eigenvectors form an orthonormal basis of ${\mathbb{C}}^n$.

Use the conventionfor the characteristic polynomial. Choosing $\det (A-tI)$ instead multiplies it by $(-1{)}^n$ and leaves the requested zero identity unchanged. For each eigenvector $v_j$, induction gives $A^r{v}_j={\lambda }_j^r{v}_j$, including $r=0$. Thusbecause ${\lambda }_{j}I-A$ has the nonzero kernel vector $v_j$ and therefore has zero determinant. Part (a) supplies a basis consisting of these $n$ eigenvectors. A linear map vanishing on every basis vector vanishes everywhere. ConsequentlyThis proves the Cayley-Hamilton theorem identity directly in the distinct-eigenvalue case assumed here.

Expand $x={\sum }_j{c}_j{v}_j$ in the orthonormal basis from part (a). ThenFor $x\ne 0$, the weights $|c_j{|}^2/{\sum }_k|c_k{|}^2$ are nonnegative and sum to one. The Rayleigh quotient therefore lies between the smallest and largest eigenvalues. Conversely, normalized extreme eigenvectors $v_{\min },v_{\max }$ give, for $0\le t\le 1$,Thus the exact Hermitian Rayleigh quotient range isWhen $n=1$, the range is the singleton containing the sole eigenvalue.

There is no such universal constant. For a real parameter $M$, consider the real matrixIts characteristic polynomial is $(t-1{)}^2$, so both eigenvalues are real and ${\lambda }_{\max }(B_M)=1$. ButFor any proposed finite $C$, choosing $M>2(C-1)$ violates the proposed bound. This is an explicit illustration that nonnormal eigenvalues do not bound a quadratic quotient.

Preserving the Euclidean scalar product means $(Ax{)}^T(Ay)=x^{T}y$ for every pair of real vectors $x,y$. Since these bilinear expressions agree on all pairs of basis vectors, this is equivalent to $A^{T}A=I$: the columns of $A$ form an orthonormal basis.

The first column is a unit vector, so it is $(\cos \theta ,\sin \theta {)}^T$ for some real $\theta$. A perpendicular unit vector is one of the two choices $\epsilon (-\sin \theta ,\cos \theta {)}^T$, where $\epsilon \in \{1,-1\}$. Hence all the required orthogonal matrices areHere $R(\theta )$ is a planar rotation, and $\theta$ is determined modulo $2\pi$. To check the group laws explicitly, $D_{\epsilon }R(\varphi )=R(\epsilon \varphi )D_{\epsilon }$, soThis proves closure. The identity is $R(0)D_1=I$, the inverse of $R(\theta )D_{\epsilon }$ is $R(-\epsilon \theta )D_{\epsilon }$, and associativity is inherited from matrix multiplication. Thus the set is the orthogonal group $O(2)$.

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

A member of both groups satisfies $A^{T}A=I$ and $A^{T}JA=J$. Multiplication of the second identity by $A$ on the left, using $A{A}^T=I$, gives $JA=AJ$. A real two-by-two matrix commuting with $J=\mathrm{diag}(1,-1)$ must be diagonal, since the two off-diagonal entries change sign in this equation. Its diagonal entries are $\pm 1$ by the orthogonal matrix identity, and the restriction $a>0$ forces the first to be $1$. Therefore the intersection is the two-element group

Put $s=\sin x$, so $-1\le s\le 1$. The printed summand is $(3s^n+s^{2n})/n$. If $|s|<1$, its absolute value is at most $3|s{|}^n+|s{|}^{2n}$, and the two geometric series converge. Thus there is absolute convergence.

For $s=1$ the series becomes $\sum 4/n$, a divergent multiple of the harmonic series. For $s=-1$ it becomesThe first series converges by the alternating series test, while the second diverges to $+\infty$, so their partial sums tend to $+\infty$. In particular, alternation of some terms does not imply convergence here. The complete classification isThere are no conditionally convergent cases.

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

For fixed real $x$, define $u_0=x$ and $u_n=\sin (0.99u_{n-1})$. This makes $u_n$ exactly the summand containing $n$ nested sines. The elementary inequality $|\sin t|\le |t|$ gives inductivelyThe fact that an iterated contraction gives a summable orbit applies here. Comparison with a geometric series therefore givesThus the nested-sine series converges absolutely for every real $x$; there are no divergent or merely conditionally convergent cases.

The intermediate value theorem says that if $f:[a,b]\to \mathbb{R}$ is a continuous function, every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.

For a proof, endpoint values are immediate, and replacing $f$ by $-f$ reduces the remaining case to $f(a)<y<f(b)$. The set $S=\{x\in [a,b]:f(x)\le y\}$ is nonempty and bounded above; let $c=\sup S$ by the least-upper-bound property. Continuity and the strict endpoint inequalities ensure $a<c<b$: near $a$ there are points in $S$, and near $b$ there are none. If $f(c)>y$, continuity gives a neighborhood of $c$ containing no points of $S$, contradicting the defining approximation property of the supremum. If $f(c)<y$, continuity gives points to the right of $c$ still in $S$, contradicting its being an upper bound. Hence $f(c)=y$, proving the intermediate value theorem.

For the distance problem, define the continuous functionIt satisfies $d(x)\ge |x-a|$, so it is a coercive function: $d(x)\to \infty$ as $|x|\to \infty$. Choose $R>|a|+d(0)+1$. Outside $[-R,R]$, the distance is greater than $d(0)$. On $[-R,R]$, the extreme value theorem gives a minimizer $x_0$, with $A=d(x_0)\le d(0)$. It is therefore a global minimizer and $A\ge 0$.

For any $y>A$, choose $X>x_0$ so large that $d(X)>y$. Apply the intermediate value theorem to $d$ on $[x_0,X]$ to obtain a point whose distance is $y$. The distance $A$ itself is attained at $x_0$, and no smaller value is possible. This is the range of a continuous coercive real function. Thus the distance set is exactlyIn particular $A=0$ precisely when the fixed point lies on the graph.

Rolle's theorem states that a continuous function $f:[a,b]\to \mathbb{R}$, differentiable on $(a,b)$ with $a<b$ and $f(a)=f(b)$, has $f^{\prime }(c)=0$ for some $c\in (a,b)$. If $f$ is constant this is immediate. Otherwise the extreme value theorem supplies a maximum or minimum with value different from the common endpoint value, so one such extremum occurs in $(a,b)$. At an interior maximum, the difference quotients for positive increments are nonpositive and those for negative increments are nonnegative. Their common limit, the derivative, must be zero. At a minimum the inequalities reverse, with the same conclusion. This proves Rolle's theorem.

For the two-function identity, write $\Delta f=f(b)-f(a)$ and $\Delta g=g(b)-g(a)$, and set $\varphi (x)=g(x)\Delta f-f(x)\Delta g$. This is a continuous function on the closed interval and is differentiable inside it. Its endpoint values agree:They need not be zero, but Rolle's theorem only requires equality. At the resulting point $c$,which proves the Cauchy mean value theorem identityFor the final endpoint limit, first state the denominator convention explicitly. In the usual zero-over-zero version of L'Hopital rule, $g^{\prime }(x)\ne 0$ on some whole interval $(0,\delta )$; this is understood if the displayed derivative quotient is to be a function defined throughout a deleted right neighborhood of zero. Define $f(0)=g(0)=0$, making both functions continuous at zero. For $0<x<\delta$, Rolle's theorem implies $g(x)\ne 0$, since otherwise $g$ would have equal zero values at $0$ and $x$ and a zero derivative between them. Apply the proved Cauchy mean value theorem on $[0,x]$ to obtain $c_x\in (0,x)$ withGiven $\epsilon >0$, the hypothesis on the derivative quotient supplies $\eta >0$ such that its distance from $\ell$ is less than $\epsilon$ whenever $0<t<\eta$. If $0<x<\min (\eta ,\delta )$ then $0<c_x<x<\eta$, so the same bound holds for $f(x)/g(x)$. ThereforeNo continuity of the derivatives is needed.

The PDF does not explicitly state the nonvanishing condition on $g^{\prime }$. If “the limit exists” permits a limit only along the natural domain where $g^{\prime }\ne 0$, the assertion is false, even with $g>0$ everywhere. The distinction concerns derivative quotient limits on partial domains. Here is a concrete counterexample to that weaker reading. Put $x_n=2^{-n}$ and $h_n=4^{-n}$ for $n\ge 0$. On each interval $[x_n/2,x_n]$, split off the plateau $[3x_n/4,x_n]$. On the plateau setOn the transition interval $[x_n/2,3x_n/4]$ setExtend $g=1,f=0$ for $x\ge 1$. The endpoints fit because $S(0)=0$, $S(1)=1$, while both $S^{\prime }$ and $q^{\prime }$ vanish at zero and one and $q$ also vanishes there. Thus $f,g$ are differentiable for all $x>0$ and tend to zero at zero. The derivative $g^{\prime }$ is nonzero exactly inside the transition intervals, where $f^{\prime }=0$, so the derivative quotient has natural-domain limit zero. But $f(x_n)/g(x_n)=0$, whereas $f(7x_n/8)/g(7x_n/8)=1$. Hence $f/g$ has no limit. The preceding qualified proof is the intended valid form of L'Hopital rule.

A Riemann integrable function on a closed interval is bounded and has equal upper and lower Darboux integrals. More explicitly, for a partition of an interval $P:a=x_0<x_1<\cdots <x_m=b$, define the Darboux sumsThen Riemann integrability meansand this common value is the Riemann integral. Equivalently, the Riemann integrability criterion requires that for every $\epsilon >0$ some partition of an interval satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, the common partition refinement of two partitions increases each lower Darboux sum and decreases each upper Darboux sum, so equality of the two extremal values is equivalent to arbitrarily small gaps.

Let $|f|\le M$. If $M=0$ the result is immediate, so take $M>0$. Given $\epsilon >0$, choose $0<\delta <\epsilon /(4M)$ and the finitely many intervals supplied by the hypothesis. On each $[a_i,b_i]$, choose a partition of an interval whose Darboux sum gap is less than $\epsilon /(2n)$, using Riemann integrability there.

Combine all these partitions with $0,1$ and all interval endpoints to obtain a global partition of an interval $P$. On the chosen intervals the gaps add to less than $\epsilon /2$. The complementary intervals have total lengthand on each the difference between the supremum and infimum is at most $2M$. ConsequentlyThe Riemann integrability criterion proves $f$ is Riemann integrable on $[0,1]$. Shared endpoints cause no problem: the Darboux sums use closed intervals, and the complementary interval estimate includes every endpoint value. This proves Riemann integrability from almost full interval coverage.

Let $K$ be the set where the indicator function $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\dots ,w_N)\in \{2,7{\}}^N$ gives the intervalThere are $2^N$ such intervals, each of length $10^{-N}$, and they cover $K$. Every point of $K$ is strictly inside the interval for its prefix, since its remaining tail lies between $2/(9\cdot 10^N)$ and $7/(9\cdot 10^N)$. Thus none of the interval endpoints belongs to $K$. The intervals are disjoint and separated: their distinct prefix integers cannot be consecutive because their final digits are only $2$ and $7$.

The complement consists of finitely many closed intervals on which $f=0$, of total length $1-2^{N}10^{-N}=1-5^{-N}$. Part (b) therefore already proves Riemann integrability, since this length tends to one.

To compute the Riemann integral directly, make a partition of an interval using $0,1$ and all the endpoints of the $I_w$. On complementary cells the upper Darboux sum contribution is zero, and on the $I_w$ it is at most their total length. Moreover every nonempty cell contains a terminating decimal, which is not in $K$: neither its terminating expansion with a zero tail nor its alternative expansion with a nine tail consists entirely of twos and sevens. Such decimals are dense, so every cell has infimum zero. ThereforeLetting $N\to \infty$ gives the exact answerThis is an example of an indicator of a restricted-digit decimal set: infinitely many, indeed uncountably many, points can support an indicator function whose Riemann integral is zero.