Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1

De Moivre's theorem says that, for a real angle $\theta$ and an integer $m$,For a complex number $z=\rho (\cos \theta +i\sin \theta )$, the corresponding formula is $z^m={\rho }^m[\cos (m\theta )+i\sin (m\theta )]$, with $z\ne 0$ required for negative powers.

Put $q=e^{i\theta }$. If $\theta \notin 2\pi \mathbb{Z}$, the finite geometric series givesTo extract the real part, multiply numerator and denominator by $1-q^{-1}$. The denominator becomes $2(1-\cos \theta )$, and the numerator is $q-1-q^{n+1}+q^n$. Thus this finite trigonometric sum isAt $\theta \in 2\pi \mathbb{Z}$, the sum equals $n$; the printed quotient is undefined there, though its continuous limiting value is $n$.

Now take $\theta =2p\pi /(n+1)$ with $1\le p\le n$. This is not a multiple of $2\pi$, while $(n+1)\theta =2p\pi$ and $\cos (n\theta )=\cos (2p\pi -\theta )=\cos \theta$. Substitution gives

Taking the conjugate transpose of $U=A+iB$ and using that both parts are Hermitian matrices gives $U^{†}=A-iB$. Addition and subtraction recover the Hermitian parts of a unitary matrix:These depend only on $U$, so any such decomposition has the same $A,B$ and is unique. Conversely the displayed expressions are Hermitian matrices and sum to $U$ in the form $A+iB$, establishing existence as well as uniqueness. Their commutation and quadratic relation are proved in the two labelled parts.

By the unitary matrix identities,Subtracting the two equations yields $2i(AB-BA)=0$. ThereforeThe subtraction is important: entrywise real and imaginary parts cannot generally be used to separate these terms, since the Hermitian matrices need not have real entries.

Adding the two expanded unitary matrix identities instead cancels the commutator and gives $2(A^2+B^2)=2I$. HenceTogether with part (i), this characterizes the commuting Hermitian parts of a unitary matrix.

The ratio test states that, for a series $\sum a_n$ with nonzero terms eventually, if $|a_{n+1}/a_n|\to L<1$, the series converges absolutely; if $L>1$, including $L=\infty$, it diverges because its terms do not tend to zero. At $L=1$ the test is inconclusive: the harmonic series diverges while $\sum n^{-2}$ converges.

For $|x|<1$, the ratio test applied to $x^n/n$ gives absolute convergence when $x\ne 0$, and at $x=0$ every such term vanishes. ThusAt $x=1$ all terms are zero, so the series converges to zero. At $x=-1$, the even terms vanish and the odd terms are $-2/n$, whose partial sums tend to $-\infty$. Finally, if $|x|>1$, then $|x^n-1|/n\ge (|x{|}^n-1)/n\to \infty$, so the term test for divergence applies. Consequently

Fix $x\in (0,1)$. For any sufficiently small nonzero $h$, the additivity of the Riemann integral givesTaking absolute values, for either sign of $h$,Because $f$ is continuous at $x$, the right side tends to zero. Thus the fundamental theorem of calculus here follows directly from the difference quotient:Only continuity at the point under consideration was used.

Without continuity, the assertion is false. Take the Riemann integrable step function $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. Its primitive is $F(x)=0$ for $x\le 1/2$ and $F(x)=x-1/2$ for $x\ge 1/2$. The left and right difference quotients at $1/2$ are respectively zero and one, so the primitive need not be differentiable at every interior point. A function with this single jump is Riemann integrable; changing its value at the jump does not change the integral or the counterexample.

In suffix notation, repeated indices are summed from one to three. The Levi-Civita symbol and Kronecker delta contraction isFor distinct free index pairs this follows from the two possible matching permutations and their signs; if either pair repeats an index both sides vanish. ThereforeRecognizing the dot products proves the vector triple product identity:

Use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{i\ell m}={\delta }_{j\ell }{\delta }_{km}-{\delta }_{jm}{\delta }_{k\ell }$ in suffix notation:Thus the cross product contraction expands as

Put $\tau =a\cdot (b\times c)$. The vector triple product identity givesCyclic invariance of the scalar triple product gives $(a\times b)\cdot c=\tau$. Hence the squared scalar triple product from cyclic cross products isIf an expansion entirely in pairwise dot products is desired, expand the Gram determinant of $a,b,c$:Both expressions remain valid for degenerate, including linearly dependent, vectors.

The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

For two upper triangular matrices and $i>j$,Indeed a potentially nonzero factor $A_{ik}$ requires $k\ge i$, while a potentially nonzero $B_{kj}$ requires $k\le j$. These requirements cannot both hold when $i>j$. Thus their matrix product is upper triangular.

Direct matrix multiplication for the given $A$ givesAdding $A^3+A^2-A$ yields $I$, so $A(A^2+A-I)=(A^2+A-I)A=I$. HenceEvery nonnegative matrix power is in $\mathrm{span}\{I,A,A^2\}$: reduce powers of degree at least three using $A^3=-A^2+A+I$. More formally, repeated reduction shows that products of any two polynomials of degree at most two in $A$ again lie in this span. Since $A^{-1}$ also lies in it, positive powers of $A^{-1}$ prove the same assertion for all negative integers. This is an instance of the integer powers for an annihilating polynomial with roots one and minus one, because $A^3+A^2-A-I=(A-I)(A+I{)}^2=0$.

To find the required entries for every integer, use the block upper-triangular structure. The top-left entry of $A^n$ is $1$ for positive and negative powers. Its lower-right block isThe identity $(I_2-N{)}^n=I_2-nN$ holds for all integers: multiplication adds the coefficients of $N$, and $(I_2-N{)}^{-1}=I_2+N$ proves the negative cases. ThereforeWrite $s=(-1{)}^n$ and $A^n={\alpha }_n{A}^2+{\beta }_{n}A+{\gamma }_{n}I$. Comparing those three entries givesSubtracting the first two equations finds ${\beta }_n$; the third then finds ${\alpha }_n$, and the first finds ${\gamma }_n$. The unique solution isFor clarity, substituting also gives the full matrix power

If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

Let $H$ be a Hermitian matrix and $He=\lambda e$ with $e\ne 0$. The inner product $e^{†}He$ is real because its conjugate equals $e^{†}{H}^{†}e=e^{†}He$. But it also equals $\lambda e^{†}e$, with $e^{†}e>0$. Thus every eigenvalue is real. For two eigenvectors,When ${\lambda }_i\ne {\lambda }_j$, this proves $e_i^{†}{e}_j=0$, so the eigenvectors are orthogonal.

For a real antisymmetric matrix, $(iA{)}^{†}=-i{A}^T=iA$, so $iA$ is a Hermitian matrix. Its real eigenvalues imply that the eigenvalues of $A$ are purely imaginary. To ensure a nonzero one without assuming it, writeExpansion of the characteristic polynomial gives $\det (tI-A)=t(t^2+|w{|}^2)$. Therefore the nonzero eigenvalue $-i\theta$ exists with $\theta =|w|>0$, and a nonzero complex eigenvector $e_1=u+iv$ satisfiesSeparating real and imaginary parts givesNeither real vector can vanish, since either vanishing would force the other to vanish. Antisymmetry yields $0=u\cdot Au=\theta u\cdot v$ andThus $u,v$ are orthogonal and have equal length.

In odd dimension $\det A=\det A^T=\det (-A)=-\det A$, so $\det A=0$. Its real homogeneous system therefore has a nonzero real eigenvector $e_3$ with $A{e}_3=0$. It is perpendicular to $u,v$, because $e_3\cdot Au=-(A{e}_3)\cdot u=0$ and similarly for $Av$. Normalize it and choose its sign so $(u/|u|,v/|v|,e_3)$ is positively oriented; this is an orthonormal basis.

The matrix exponential series converges absolutely, allowing even and odd terms to be grouped. Since $A^{2}u=-{\theta }^{2}u$ and $A^{2}v=-{\theta }^{2}v$, it acts asConsequently the skew-symmetric exponential as an axial rotation has, in the displayed orthonormal basis, the matrixIt is orthogonal with determinant one, fixes its axis and rotates the perpendicular plane. Hence $R=e^A$ is a rotation matrix, with angle $\theta$ about the chosen oriented axis.

The comparison test for series says that an eventual upper bound by a convergent positive series proves convergence, while an eventual lower bound by a divergent positive series proves divergence. The p-series $\sum n^{-s}$ converges exactly when $s>1$.

If $p>1$, then $(\log n{)}^q\ge 1$ for all sufficiently large $n$, soand the series converges. If $0<p<1$, choose $\eta =(1-p)/2>0$. The allowed logarithmic bound, applied with exponent $\eta /q$, gives $(\log n{)}^q<n^{\eta }$ eventually. Thusand the series diverges.

For $p=1$, use the integral test for convergence: for an eventually positive decreasing function $g$, $\sum g(n)$ and $\int g(x)dx$ either both converge or both diverge. Here $g(x)=1/[x(\log x{)}^q]$ is decreasing for $x>1$, and the substitution $u=\log x$ givesThis integral converges exactly when $q>1$. Hence the polynomial-logarithmic series convergence classification isAll other allowed positive $p,q$ give divergence.

For any fixed $r>0$, $\log y=o(y^{1/r})$ as $y\to \infty$. For example, l'Hôpital's rule applied to $\log y/y^{1/r}$ gives the limit zero. With $y=\log n$, this implies $(\log \log n{)}^r<\log n$ eventually. Thereforefor all sufficiently large $n$. The comparison series diverges by the integral test for convergence, sinceThe comparison test for series givesOne may obtain the same growth bound by extending the given logarithmic estimate to real $y$ through $\lceil y\rceil$ and choosing any exponent smaller than $1/r$; no special convergence test for iterated logarithms is needed.

The intermediate value theorem states that, if $f$ is a continuous function on a closed interval $[a,b]$, every value between $f(a)$ and $f(b)$ equals $f(c)$ for some $c\in [a,b]$. If the value is strictly between the endpoint values, $c$ may be chosen in $(a,b)$.

Here is a proof from the supremum property of the real numbers. Suppose $f(a)<y<f(b)$; the reverse inequality is reduced to this case by replacing $f,y$ with $-f,-y$. LetThe set is nonempty because it contains $a$ and is bounded above by $b$. Continuity at $a$ supplies points of $E$ to the right of $a$, while continuity at $b$ excludes a neighborhood of $b$ from $E$. Hence $a<c<b$. By the definition of supremum, there are $t_n\in E$ with $t_n\to c$. Continuity gives $f(c)=\lim f(t_n)\le y$. If $f(c)<y$, continuity would give a point just to the right of $c$ still satisfying $f(t)<y$, contradicting that $c$ is an upper bound for $E$. Thus $f(c)=y$. Values equal to either endpoint are attained at that endpoint, completing the proof.

Let $u,v\in f(I)$ with $u<y<v$. Choose preimages $x_u,x_v\in I$, and let $a=\min (x_u,x_v)$, $b=\max (x_u,x_v)$. The defining property of a real interval ensures $[a,b]\subseteq I$. The restriction of $f$ to this closed interval is continuous, and its endpoint values are $u,v$ in one order or the other. The intermediate value theorem supplies $c\in [a,b]\subseteq I$ with $f(c)=y$. Thus every value between two image values lies in the image, provingThe continuous image of a real interval result does not require $I$ itself to be closed or bounded; only the segment between the chosen preimages is used. Empty and single-point images also satisfy the interval property.

No such continuous function exists. The extreme value theorem makes every continuous function on the closed interval $[0,1]$ bounded and gives it a finite maximum. Its image therefore cannot equal the unbounded real interval $[0,\infty )$.

The continuous functionhas the required image. Its values are nonnegative, and for every $y\ge 0$ the point $x=1/(y+1)\in (0,1]$ satisfies $f(x)=y$. Thus the image is exactly $[0,\infty )$. The missing left endpoint permits unboundedness without a failure of continuity on the domain.

A suitable continuous function isFor $t_k=\pi /2+2\pi k$ and $u_k=3\pi /2+2\pi k$,All these arguments lie in $(0,1]$. Its image is a real interval by the continuous image of a real interval result, and it is unbounded both above and below. Given any real $y$, choose two of the displayed points with function values on opposite sides of $y$, then apply the intermediate value theorem on the segment joining them. Hence the image is all of $\mathbb{R}$. A monotone example would not suffice here; the oscillation gives both directions of unboundedness near the same missing endpoint.

By differentiability implies continuity, $f(x+h)\to f(x)$. Add and subtract $f(x+h)g(x)$ in the difference quotient:The two quotient limits exist by the given differentiability. Taking the limit and using continuity yields the product rule:Thus the product is differentiable at the specified point; differentiability elsewhere is not needed.

At zero, $g(0)=0$ andby continuity of $f$ at zero. Thus $g$ is always differentiable there, with $g^{\prime }(0)=0$, regardless of whether $f^{\prime }(0)$ exists.

At $x\ne 0$, if $f$ is differentiable, the product rule applied to $t^{2}f(t)$ proves that $g$ is differentiable. Conversely, if $g$ is differentiable at $x$, the function $t^{-2}$ is differentiable on a neighborhood of $x$. Apply the product rule to $f(t)=t^{-2}g(t)$ on that neighborhood. This proves the differentiability restored by a quadratic zero equivalence:At a nonzero differentiability point, $g^{\prime }(x)=2xf(x)+x^2{f}^{\prime }(x)$.

For the necessity in the differentiability of the square of a continuous function, suppose $g=f^2$ is differentiable at $x$. If $f(x)\ne 0$, continuity makes $f(x+h)+f(x)$ nonzero for small $h$, andHence $f$ is differentiable at $x$, giving alternative (i).

If $f(x)=0$, $g(x)=0$ is a minimum because $g\ge 0$. The right difference quotients of $g$ are nonnegative and the left ones are nonpositive; since their common limit exists, it must be zero. Thus $g^{\prime }(x)=0$, andTaking square roots of this nonnegative expression shows $f(x+h)/\sqrt{|h|}\to 0$, giving alternative (ii). The two labelled parts below establish sufficiency of the respective alternatives. Together they prove the full stated equivalence, including the zero of $f$ where squaring can remove a failure of differentiability.

If $f$ is differentiable at $x$, the product rule applied to $f\cdot f$ givesThis proves sufficiency of the first alternative. No additional growth condition is required; if $f(x)=0$, the derivative of the square is zero.

Suppose $f(x)=0$ and $f(x+h)/\sqrt{|h|}\to 0$. ThenThereforeThis proves sufficiency of the second alternative even if $f$ itself is not differentiable at $x$. For example, $f(t)=|t{|}^{3/4}$ is continuous and not differentiable at zero, while $g(t)=|t{|}^{3/2}$ is differentiable there with derivative zero.

First establish the interior convergence lemma for a power series. If $\sum a_n{w}^n$ converges at some $w\ne 0$, its terms are bounded: $|a_n{w}^n|\le M$ for a finite $M$. For every $|z|<|w|$,Comparison with a convergent geometric series proves absolute convergence at $z$.

Now letThe set contains zero, since at $w=0$ only the constant term remains. If $|z|<R$, the definition of supremum gives a point of convergence $w$ with $|w|>|z|$. The lemma proves absolute convergence at $z$. If $|z|>R$, convergence would place $|z|$ in the set defining $R$, a contradiction. This proves the existence of a radius of convergence, including $R=0$ and $R=\infty$:The proof leaves behavior on $|z|=R$ open; it need not be uniform around that circle. The two examples illustrate opposite possibilities.

For $z\ne 0$, the ratio of successive absolute terms isThe ratio test therefore gives absolute convergence for $|z|<1$ and divergence for $|z|>1$, while at zero the series is trivially convergent. Hence $R=1$. On the boundary, $|z^n/n^2|=1/n^2$, and the convergent p-series proves

For $|z|=r<1$, the inequality $n!\ge n$ for $n\ge 1$ givesThe first $r$ comes from $0!=1$; both initial terms have exponent one. Thus the factorial-gap power series converges absolutely inside the unit circle. For $|z|>1$, its terms have magnitudes $|z{|}^{n!}\to \infty$, so the term test for divergence proves divergence and $R=1$.

At every point with $|z|=1$, each term has magnitude one, so the terms cannot tend to zero, regardless of their phases. HenceThe large gaps between exponents make this a lacunary power series; cancellation cannot overcome the necessary condition that the terms of a convergent series tend to zero.