Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1

By Euler's formula, $z=e^{i\theta }$. Factoring out the half-angle givesSince $0\le \theta <\pi$, the cosine is positive. ThereforeOn the Argand diagram, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta /2$; resolving either side along the diagonal gives its length $2\cos (\theta /2)$.

Similarly,Thus, for $0<\theta <\pi$,At $\theta =0$, $1-z=0$, whose complex argument is undefined. Geometrically, $|1-z|$ is the chord joining the two unit-circle points $z$ and $1$. The isosceles triangle with vertex angle $\theta$ has half-chord $\sin (\theta /2)$, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

Componentwise,soRepeated application gives $(A^k{)}^T=(A^T{)}^k$. Transposing the matrix exponential term by term therefore yieldsMultiplication of the two power series givesHenceIf $e^{tA}$ is an orthogonal matrix for every real $t$, thenIts coefficient of $t$ must vanish, sothat is, $A$ is a skew-symmetric matrix.

Because $(a_n)$ is a monotone sequence, every one of its first $n$ terms is at most $a_n$, soOn the other hand, every term from $a_n$ through $a_{2n}$ is at least $a_n$, and henceBoth the lower bound and upper bound tend to $x$ because the Cesaro means converge to $x$. The squeeze theorem gives

Convergence at $z_0$ implies that the terms $a_n{z}_0^n$ form a bounded sequence: $|a_n{z}_0^n|\le M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, thenComparison with the convergent geometric series $\sum M{r}^n$ proves absolute convergence. The case $z_0=0$ is vacuous.

The radius of convergence is the number $R\in [0,\infty ]$ such that a power series converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.

TakeThen $\sum v^n/n$ is the convergent alternating harmonic series, whereas $\sum w^n/n$ is the divergent harmonic series. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that

The cross product $x\times y$ is the vector perpendicular to $x$ and $y$, of magnitude $|x||y|\sin \alpha$ and orientation fixed by the right-hand rule. In suffix notation,where ${\epsilon }_{ijk}$ is the Levi-Civita symbol. ThereforeUsinggivesHence the vector triple product identity is

The identity in part (a), together with $|a|=1$, givesFor every $n\ge 1$, $x_n$ is perpendicular to $a$, because it is a scalar multiple of $a\times x_{n-1}$. ThereforeAlso $|x_1|=\lambda |a\times x_0|$ and $|x_2|={\lambda }^2|a\times x_0|$. Induction using the two-step recurrence yields

The recurrence $x_{n+2}=-{\lambda }^2{x}_n$ shows that every odd-indexed position vector is a scalar multiple of $x_1$, while every positive even-indexed position vector is a scalar multiple of $x_2$. HenceMoreover $x_2=\lambda a\times x_1$, so these two straight lines through the origin are perpendicular.

Let $d_n=x_{n+1}-x_n$ be the direction vector of the segment $X_n{X}_{n+1}$. Consecutive vectors $x_n,x_{n+1}$ are perpendicular and $|x_{n+1}|=\lambda |x_n|$. Using $x_{n+2}=-{\lambda }^2{x}_n$,Thus adjacent segments are perpendicular. AlsosoIf $0<\lambda <1$, the norm formula in part (i) gives $x_n\to 0$. If $\lambda =1$, then $x_{n+2}=-x_n$: the points $X_1,X_2,-X_1,-X_2$ repeat cyclically and form a square centered at the origin.

Put $L=|x_n|$ and choose perpendicular unit vectors $e,f$ such that $x_n=Le$ and $x_{n+1}=\lambda Lf$. Direction vectors for the two lines areandTheir scalar product and norms giveTherefore the acute angle between the straight lines satisfies

A matrix and its matrix transpose have the same characteristic polynomial, becauseThus they have the same eigenvalues, including algebraic multiplicities.

The proposed sum identity is false. Forboth eigenvalues ${\lambda }_i$ are zero, whereashas eigenvalues $0,1$. HenceEquivalently, the two sides are generally $\mathrm{tr}(A^{T}A)$ and $\mathrm{tr}(A^2)$, which need not agree.

The product identity always holds, sinceTherefore

If $p=m\times n$, then $n^{T}p=0$, and hence $Bp=p$. Also $Bm=m$ because $m\perp n$. Thus $p$ and $m$ are linearly independent eigenvectors with eigenvalue one.

For every vector $v$,so $B-I$ is nonzero but has square zero:Consequently all three eigenvalues are one. The eigenspace iswhich has dimension two. It cannot supply three linearly independent eigenvectors, so

Multiplication givesThe vector $p=m\times n$ is an eigenvector with eigenvalue one. On the plane spanned by the orthonormal pair $m,n$, the matrix isIts eigenvalues are the roots of ${\mu }^2-3\mu +1$, namelyFor either root, an eigenvector is $m+({\mu }_{\pm }-1)n$. ThereforeThese positive eigenvalues are the squared singular values of $B$.

If $A^T=-A$, then the scalar $x^{T}Ax$ equals its own transpose and henceso it vanishes.

Conversely, suppose $x^{T}Ax=0$ for every real $x$. Taking $x=e_i$ gives $A_{ii}=0$. Taking $x=e_i+e_j$ then gives $A_{ij}+A_{ji}=0$. Thus every entry satisfies $A^T=-A$, and

A real skew-symmetric matrix satisfies $A^{†}=A^T=-A$, so it is a skew-Hermitian matrix. If $Ax=\lambda x$, thenTaking the complex conjugate and using $A^{†}=-A$ shows that this scalar is the negative of its conjugate. Henceso every eigenvalue is purely imaginary or zero.

If $Ax=\lambda x$ and $Ay=\mu y$, thenSince ${\lambda }^{\ast }=-\lambda$, this becomes $(\mu -\lambda )x^{†}y=0$. Distinct eigenvalues therefore have orthogonal eigenvectors:

In odd dimension,so $\det A=0$. Since $A$ is real, its kernel contains a nonzero real vector $a$.

The plane $a^{\perp }$ is invariant under $A$, because $a\cdot Ab=-(Aa)\cdot b=0$. On this two-dimensional plane, $Ab$ is perpendicular to $b$. Put $|Ab|=\theta |b|$ with $\theta >0$. Then $A^{2}b$ is parallel to $b$, andsoThe matrix exponential fixes the kernel vector:Separating even and odd powers in its series and using $A^{2j}b=(-1{)}^j{\theta }^{2j}b$ givesThus $e^A$ acts as a rotation through angle $\theta$ on $a^{\perp }$ and fixes its axis $\mathbb{R}a$.

Write $y=(y_1,y_2,y_3{)}^T$. From the coordinate formula for the cross product,If $y=c_1{e}_1$, this becomesThus $c_1\ne 0$ gives rank three and a zero-dimensional kernel, while $c_1=0$ gives rank one and kernel dimension two.

If $y\cdot e_1=0$, then $y_1=0$. When $y\ne 0$, the equations $Tx=0$ force $x_1=0$ and $y_3{x}_2-y_2{x}_3=0$, soWhen $y=0$, again $\mathrm{rank}T=1$ and $\dim \ker T=2$. These dimensions also agree with the rank-nullity theorem.

First computeThe vector $\ell =(1,-2,4{)}^T$ lies in the kernel of $(AB{)}^T$, since ${\ell }^{T}AB=0$. If $ABx=d$ is solvable, multiplying by ${\ell }^T$ gives the necessary compatibility conditionTherefore

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then there exists $c\in [a,b]$ such that $f(c)=y$.

A function is differentiable at $a$ when the finite limitexists. A derivative need not be continuous. For example,is differentiable everywhere and $f^{\prime }(0)=0$, but for $x\ne 0$,which has no limit at zero.

The mean value theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfies

Set $h(x)=f(x)-yx$. ThenThe extreme value theorem gives a minimum of $h$ on $[a,b]$. If it occurs in $(a,b)$, the two-sided difference quotient gives $h^{\prime }(c)=0$. If it occurs at $a$, either $h^{\prime }(a)=0$ or a negative right derivative contradicts minimality; the endpoint $b$ is analogous. Hence some $c\in [a,b]$ satisfiesThis proves the Darboux theorem for derivatives in the stated case without assuming that $f^{\prime }$ is continuous.

Because a differentiable function is continuous, defineThe fundamental theorem of calculus gives $F^{\prime }(x)=f^{\prime }(x)+f(x)$. At the endpoints,whileApplying the result just proved to $F$ yields $d\in [a,b]$ with

The function $f$ is continuous at $y_0$ when for every $\epsilon >0$ there is a $\delta >0$ such thatSuppose $f$ has a local minimum at $c$ and is differentiable there. For sufficiently small $h>0$,Both one-sided limits equal $f^{\prime }(c)$, so it is both nonnegative and nonpositive. Therefore

The upper bound follows immediately from convexity, sinceFor the lower bound, writeApplying convexity and rearranging givesChoose a fixed $r_0>0$ with $[y_0-r_0,y_0+r_0]\subset (a,b)$. Applying these bounds with $r=r_0$ and $r=-r_0$ shows that the difference $f(y_0+h)-f(y_0)$ is trapped between quantities of order $|h|/r_0$. Both tend to zero, so every convex function on an open interval is continuous.

Now suppose $c$ is a local minimum but some $x$ satisfies $f(x)<f(c)$. For sufficiently small $\lambda >0$, the point $z=(1-\lambda )c+\lambda x$ lies in the neighbourhood on which $c$ is minimal, whereas convexity givesa contradiction. ThusDifferentiability at the minimizer is unnecessary: $f(x)=|x-c|$ is convex, has its global minimum at $c$, and is not differentiable there.

For each $k\ge 0$, monotonicity gives the dyadic-block boundsSumming the upper bounds proves convergence of $\sum x_n$ whenever $\sum 2^k{x}_{2^k}$ converges. Summing the lower bounds gives, up to the first term,so convergence of $\sum x_n$ forces convergence of the condensed series. This is the Cauchy condensation test:

If $s\le 0$, the terms $n^{-s}$ do not tend to zero, so the series diverges by the term test for divergence. If $s>0$, the terms form a nonnegative decreasing sequence, and the Cauchy condensation test gives the condensed seriesThis geometric series converges exactly when $1-s<0$. Hence the p-series criterion is

For $u_n=2^{-n}{n}^k$,Thus for all sufficiently large $n$ the ratio is, for example, at most $3/4$. Comparison with a decaying geometric sequence, equivalently the ratio test, yields

We first prove by induction thatIt holds for $n=1$, and if it holds at $n$, thenConsequentlyPart (c) and the squeeze theorem now give

Let $L$ be the supremum of all lower Darboux sums and $U$ the infimum of all upper Darboux sums. The assumed comparison of arbitrary lower and upper sums gives $L\le U$. For the partition supplied for a given $\epsilon$,Since this holds for every positive $\epsilon$, $U=L$. Thus the Riemann integrability criterion gives

If $f$ is continuous on $[a,b]$, the Heine-Cantor theorem makes it uniformly continuous. Given $\epsilon >0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\epsilon /(b-a)$. A partition of mesh below $\delta$ then satisfiesproving that Continuous functions are Riemann integrable.

For the function defined using $g$, choose $K$ with $|f|\le K$. Given $\epsilon >0$, choose $\delta >0$ so small that $2K\delta <\epsilon /2$. On $[\delta ,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\epsilon /2$. Adding the interval $[0,\delta ]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\epsilon$, and

Finally, for any partition $a=x_0<\cdots <x_n=b$, the mean value theorem supplies ${\xi }_i\in (x_{i-1},x_i)$ such thatSumming telescopes toa Riemann sum for the Riemann-integrable function $f^{\prime }$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common upper and lower Darboux integrals proves the fundamental theorem of calculus conclusion