Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 1

Squaring first avoids unnecessary expansion: $(2+2i{)}^2=8i$, and squaring again gives the fourth power $\boxed{z^4=-64}$. In the polar form of a complex number, $z=2\sqrt{2}{e}^{i\pi /4}$; multiplying its complex argument by four gives the same negative-real result.

Write a nonzero complex number as $w=r{e}^{i\theta }$. Matching modulus and complex argument gives $r^4=2\sqrt{2}$ and $4\theta =\pi /4+2\pi k$. Thus the four distinct fourth roots areThese angles differ by $\pi /2$, so the values are distinct. Every further integer $k$ repeats one of them; this is multiplication by the fourth roots of unity.

The exponent in the PDF is $2z^2$. A complex exponential equals one precisely when its exponent is an integer multiple of $2\pi i$: its real part must be zero, and its imaginary part must be a full rotation. Therefore $z^2=\pi ik$ for $k\in \mathbb{Z}$. Taking both square roots yields all solutionsThe first nonzero family corresponds to positive exponent index, and the second to negative index. Both signs are required.

For a real line $ux+vy+w=0$, with $(u,v)\ne (0,0)$, choose $B=(u-iv)/2$ and $C=w$. Direct multiplication gives $Bz+\overline{B}\overline{z}=ux+vy$, so every line has the required form with $B\ne 0$.

For a circle of centre $c$ and radius $\rho >0$, expand its modulus equation:Thus the required circle parameters are $\boxed{B=-c,C=|c{|}^2-{\rho }^2}$. Conversely, completing the square in the circle form gives $|z+B{|}^2=|B{|}^2-C$, which describes a genuine circle when $|B{|}^2-C>0$. The complex conjugate terms make both equations real-valued.

An eigenvector of a matrix $A$ is a nonzero vector $v$ for which $Av=\lambda v$. The scalar $\lambda$ is its eigenvalue. For a fixed eigenvalue, the zero vector together with all corresponding eigenvectors forms the eigenspace $\ker (A-\lambda I)$. The requirement $v\ne 0$ prevents every scalar from being an eigenvalue merely because $A0=\lambda 0$.

Expanding the characteristic polynomial givesSolving the three nullspace systems gives the eigenvalues and eigenspacesFor example, at $\lambda =1$ the first two equations force $z=0$ and $x=y$; at $\lambda =2$ they force $x=0$ and $y=-z$; at $\lambda =-1$ they force $x=y=z$.

Putting these eigenvectors in columns proves their linear independence:Hence $A$ is a diagonalizable matrix over the reals.

Call an index $n$ a peak if $a_n\ge a_m$ for every $m>n$. If there are infinitely many peaks, list their indices increasingly. Each selected value is at least every later selected value, so this gives a nonincreasing subsequence.

If there are only finitely many peaks, choose an index beyond the last one. Every index from then onward has a later index with strictly larger value. Select these recursively: $n_{j+1}>n_j$ with $a_{n_{j+1}}>a_{n_j}$. This gives an increasing subsequence. Neither complex argument requires the original sequence to be bounded. Every real sequence therefore has a monotone subsequence, as asserted by the monotone subsequence theorem.

Let $a_n=n!/n^n$. Its consecutive coefficient ratio isThe ratio test for the full terms therefore has limit $|z|/e$: the series has absolute convergence when $|z|<e$ and its terms eventually grow in magnitude when $|z|>e$. Thus the radius of convergence is $\boxed{R=e}$. No boundary decision is needed to determine the radius of convergence.

This is a lacunary power series: the coefficient of $z^{n!}$ is $n^n$, while the other coefficients are zero. Along its nonzero coefficients,The factorial dominates $n\log n$; for example $n!\ge n(n-1)(n-2)(n-3)$ for large $n$ already suffices for this limit. Hence the coefficient limsup in the Cauchy-Hadamard theorem is one. The radius of convergence is $\boxed{R=1}$.

Directly, for $|z|<1$ the consecutive term ratio is $(n+1{)}^{n+1}{n}^{-n}|z{|}^{nn!}$, which tends to zero; for $|z|\ge 1$ the term magnitudes are at least $n^n$ and fail to tend to zero. The factorial is the exponent of $z$, not part of its coefficient.

The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.

The plane through $a_i$ has the constant normal coordinate $r\cdot \hat{n}=a_i\cdot \hat{n}$. Since $\hat{n}$ is a unit vector, the separation of these normal coordinates is a length. Every displacement joining the planes has this fixed normal component, so its length is at least its absolute value; a displacement parallel to $\hat{n}$ attains that bound. Hence the distance between parallel planes isThe absolute value, present in the PDF, is necessary because distance is nonnegative.

Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt{6}$. The planes through the two lines with this common normal have separationThis is a lower bound for the distance between skew lines. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.

View a possible complex eigenvector $v$ using the Hermitian inner product. For a real symmetric matrix, $A^{†}=A$, so $v^{†}Av$ is real. Since $v^{†}Av=\lambda v^{†}v$ and $v^{†}v>0$, every eigenvalue is real.

For eigenvectors $v,w$ with distinct eigenvalues $\lambda ,\mu$, symmetry givesso $v^{†}w=0$. In particular, real eigenvectors are orthogonal in the real dot product.

Use the permitted diagonalizability assumption to take bases of the real eigenspaces whose union spans ${\mathbb{R}}^n$. Apply the Gram-Schmidt process within each eigenspace. Its linear combinations remain eigenvectors of that same eigenvalue; vectors in different eigenspaces are already orthogonal. The resulting union is an orthonormal basis of eigenvectors, giving an orthogonal diagonalization of a real symmetric matrix.

If $Ax=b$ and $h\in \ker A$, symmetry gives $b\cdot h=(Ax)\cdot h=x\cdot Ah=0$, proving necessity. Conversely, expand $b$ in the orthonormal basis from part (i), with $A{y}_i={\lambda }_i{y}_i$. The zero-eigenvalue vectors span the kernel of a linear map. If $b$ is orthogonal to that kernel, then $b\cdot y_i=0$ whenever ${\lambda }_i=0$, so the remaining coordinates can be solved independently.

For a nonzero eigenvalue and any associated nonzero eigenvector $v$, symmetry gives $b\cdot v=\lambda x\cdot v$. Thus the component vector along $v$ isFor the unit vectors $y_i$, this coefficient is $(b\cdot y_i)/{\lambda }_i$. Consequently the general solution isThis proves sufficiency as well as the claimed kernel criterion. If the compatibility condition fails, there is no solution; if it holds, the freedom is exactly an arbitrary kernel vector.

Reading the coefficients of the linear map givesUnder the inverse identification, the four standard real basis vectors map respectively to $(1,0)$, $(i,0)$, $(0,1)$ and $(0,i)$ in ${\mathbb{C}}^2$. Applying $A$ and taking real and imaginary parts therefore gives the columnsThus the real matrix isFor verification, write its blocks as $\left(\begin{array}{cc}{R}_{\theta }& I\\ I& R_{-\varphi }\end{array}\right)$, where $R_{\psi }$ is the planar rotation matrix. Block elimination gives $\det B=\det (R_{\theta -\varphi }-I)=2-2\cos (\theta -\varphi )$. Therefore the determinant identity isThis is an instance of the realification determinant identity. The identification is real-linear; it is not an identification of complex vector-space dimensions.

Expanding the commutator immediately gives $[A,A]=0$, $[A,B]=-[B,A]$, and $[A,\lambda B]=\lambda [A,B]$. For its matrix trace, writeConsequently every matrix commutator is traceless: $\boxed{\mathrm{tr}[A,B]=0}$. This is the trace of a matrix commutator identity, valid for arbitrary square matrices over the complex numbers.

The conjugate transpose reverses multiplication. If $A^{†}=-A$ and $B^{†}=-B$, thenThus the commutator is again a skew-Hermitian matrix. The order reversal is what produces the required minus sign.

For $a=(a_1,a_2,a_3)$ the map has the explicit formIts diagonal entries sum to zero and its conjugate transpose is its negative, because the three coordinates are real. Hence it is traceless and a skew-Hermitian matrix.

Direct multiplication of the given basis matrices yieldsThe reverse products give the negative commutators, and equal-index commutators vanish. Bilinearity then givesThus the cross product is represented by the commutator:This realizes the cross-product model of su(2) with exactly the normalization and cyclic orientation used here.

Every traceless skew-Hermitian matrix of size two has the formso it equals $\mathcal{M}(c)$ for $c=(2p,2\mathrm{Re}w,2\mathrm{Im}w)$. If $c=0$, take $A=B=0$. Otherwise choose a unit vector $u$ perpendicular to $c$ and set $a=u$, $b=c\times u$. The vector triple product gives $a\times b=c$.

By part (iii), choosing $A=\mathcal{M}(a)$ and $B=\mathcal{M}(b)$ gives the required representationIn fact both factors can themselves be chosen as traceless skew-Hermitian matrices. The cross-product model of su(2) explains this geometric commutator construction.

Use $\sin 0=0$, the derivative identity ${\sin }^{\prime }=\cos$, continuity of cosine at zero, and $\cos 0=1$. For $u\ne 0$, the mean value theorem gives $\sin u/u=\cos c_u$ for a point between zero and $u$. Hence $\sin u/u\to 1$ as $u\to 0$. For $x\ne 0$, apply this with $u=x/3^k$ to obtain the limit $\boxed{3^k\sin (x/3^k)\to x}$. For $x=0$ every term is zero.

The additional bound $|\cos t|\le 1$ and the same mean value theorem give $|\sin u|\le |u|$. ThereforeThe right side is summable. By the comparison test for series the stated series has absolute convergence for every real $x$.

Let $r=e^{i\theta }\ne 1$. The geometric series from any starting index is bounded:Using summation by parts,Since the coefficients are nonincreasing and positive, the modulus is at most $K(a_N+a_m-a_N)=K{a}_m\to 0$. The Cauchy criterion proves convergence for every $\theta \notin 2\pi \mathbb{Z}$. This is a direct proof of the Dirichlet test in this setting.

For $a_n=1/n$, the imaginary parts of the convergent partial sums give convergence of the sine series at those angles. If $\theta \in 2\pi \mathbb{Z}$, every sine term is zero. Hence the sine series converges for all real angles.

The mean value theorem states that a function continuous on $[p,q]$ and differentiable on $(p,q)$ has some $c\in (p,q)$ with $f(q)-f(p)=f^{\prime }(c)(q-p)$.

Take any $p<q$ inside the given open interval. Differentiability implies the continuity needed on $[p,q]$, and the derivative is zero throughout it. The mean value theorem therefore gives $f(q)=f(p)$. Since the two points were arbitrary, the function is constant on the interval.

For continuity at any $x$, choose $\delta ={\eta }^{1/\alpha }$ for a desired error $\eta >0$. If $|y-x|<\delta$, the stated bound gives $|f(y)-f(x)|<\eta$. Thus $f$ is continuous, in fact uniformly continuous.

When $\alpha >1$, the difference quotient satisfiesTherefore the derivative exists at every point and is zero. Part (i) proves $f$ is constant. This is the lemma that Hölder exponent greater than one forces constancy; no differentiability was assumed in advance.

Let $D=f_{+}^{\prime }(a)$ and write $g(t)=(f(t)-f(a))/(t-a)$ for $t>a$. By the assumed right derivative, choose $\delta >0$ with $|g(t)-D|<ϵ/2$ whenever $a<t<a+\delta$. Choose $b_0\in (a,\min (b,a+\delta ))$.

Apply the mean value theorem to $f$ on $[a,b_0]$. It gives $x\in (a,b_0)$ with $f^{\prime }(x)=g(b_0)$. Both $g(x)$ and $g(b_0)$ are within $ϵ/2$ of $D$, soThis proves the endpoint secant-tangent approximation without any continuity assumption on $f^{\prime }$. The continuity required by the theorem is only that of $f$ itself.

For $x\ne 0$, put $P(t)={\sum }_{k=0}^{n-1}{f}^{(k)}(0)t^k/k!$ and $K=(f(x)-P(x))/x^n$. The auxiliary function $F(t)=f(t)-P(t)-K{t}^n$ satisfiesRolle's theorem first gives a zero of $F^{\prime }$ strictly between zero and $x$. Apply Rolle's theorem to $F^{\prime }$ between that zero and zero, where $F^{\prime }(0)=0$, to obtain a zero of $F^{\prime \prime }$. Continue in this way. The derivatives through order $n-1$ are continuous, since they are differentiable; hence every application is valid even if $f^{(n)}$ is not continuous. After $n$ applications there is $c$ strictly between zero and $x$ with $F^{(n)}(c)=0$.

But $F^{(n)}(c)=f^{(n)}(c)-n!K$, so $K=f^{(n)}(c)/n!$. Writing $c=\theta x$ yields the Taylor formula with Lagrange remainderThe same construction using repeated applications of Rolle's theorem applies when $x<0$, with intervals ordered from $x$ to zero. For $x=0$ the identity is immediate and any $\theta \in (0,1)$ works. This proves the Taylor theorem with Lagrange remainder using only the allowed theorem and elementary differentiation.

From $f^{\prime \prime }=f$, differentiating successively gives $f^{(2r)}=f$ and $f^{(2r+1)}=f^{\prime }$ whenever those derivatives have been constructed. Starting with twice differentiability, this relation supplies the next two derivatives at each stage. Thus $f$ is infinitely differentiable, and its origin series isFor a fixed $x$, both $f$ and $f^{\prime }$ are bounded on the compact interval between zero and $x$, say by $M_x$. Every higher derivative is one of these two functions. Taylor's theorem, with the remainder just proved, therefore bounds the order-$n$ remainder by $M_x|x{|}^n/n!$, which tends to zero: its consecutive positive-term ratio is $|x|/(n+1)$. Hence the series converges to $f(x)$ for every real $x$. It also gives $f(x)=A\cosh x+B\sinh x$ by the usual hyperbolic-function series.

At any other centre $a$, apply the proved origin result to $g(t)=f(a+t)$, which also satisfies $g^{\prime \prime }=g$. Its even derivatives at zero are $f(a)$, and its odd ones are $f^{\prime }(a)$. ConsequentlyThis proves convergence at every centre without assuming that smoothness alone implies equality to a Taylor series.

For a bounded real function on $[0,1]$, take a partition $P$ with nodes $0=x_0<\cdots <x_N=1$. Its lower Darboux sum and upper Darboux sum areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; their common value is its Riemann integral. Equivalently, for every $\eta >0$ some partition has $U-L<\eta$. Indeed a common refinement of nearly extremizing upper and lower partitions proves necessity, while arbitrarily small gaps force equality of the two integrals. This is the Riemann integrability criterion.

A continuous function on a compact interval is bounded and uniformly continuous. Choose a mesh small enough that its oscillation on each subinterval is less than $\eta$. Then $U-L\le \eta {\sum }_j(x_j-x_{j-1})=\eta$; using $\eta /2$ first makes the desired inequality strict. Thus every continuous function on $[0,1]$ is Riemann integrable.

For a nondecreasing function, the endpoint bounds show it is bounded. On the uniform partition into $N$ intervals, its Darboux sums satisfyFor a nonincreasing function apply the same calculation to $-f$, or use $|f(1)-f(0)|/N$. Thus every monotone function on $[0,1]$ is Riemann integrable.

For the unheaded rational-weight example, each summand is nonnegative, and increasing $x$ only adds indices. Hence its sum is nondecreasing and lies between zero and one, including the separately specified value at zero. It is a left-continuous cumulative function of an atomic measure.

If $q_j\in [0,1)$ and $q_j<x\le 1$, the index $j$ appears in the sum at $x$ but not at $q_j$. All other changes are nonnegative, so $f(x)-f(q_j)\ge 2^{-j}$. This remains true for arbitrarily small positive $x-q_j$, proving discontinuity at $q_j$, including a right discontinuity at zero when that rational is encountered. Every interval of positive length contains such a rational in its interior. The discontinuities are therefore dense, although the function is Riemann integrable by the monotone-function result just proved.

More concretely, truncate after the first $N$ weights. The resulting finite sum of step functions is integrable, and the uniform tail is at most $2^{-N}$. This also proves integrability and gives, if its value is desired,The integral formula follows by integrating the finite sums and using the uniform tail bound. Here “every interval” has the usual nondegenerate-interval meaning; a singleton is not being asserted to contain a discontinuity.