Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

The two diagonal $2\times 2$ blocks allow the characteristic polynomial to factor immediately:Thus the characteristic equation isSolving the two block eigenvalue equations gives the following nonzero eigenvectors and corresponding eigenvalues:For example, the lower block maps $(1,i{)}^T$ to $(2i,-2{)}^T=2i(1,i{)}^T$, and maps $(1,-i{)}^T$ to zero. The four eigenvalues are distinct. Alternatively, the determinant of the matrix with columns $a,b,c,d$ is $(-2)(-2i)=4i\ne 0$. Hence they form a basis and span the complex vector space ${\mathbb{C}}^4$. Their nonzero scalar multiples are all the corresponding eigenvectors.

For any eigenvector $v$ with eigenvalue ${\lambda }_v$, its one-dimensional vector subspace satisfies $A(sv)={\lambda }_{v}sv$. If ${\lambda }_v\ne 0$, every $tv$ has preimage $(t/{\lambda }_v)v$, so the restricted linear map is onto that same subspace. ConsequentlyThe last line is the kernel of a linear map direction: multiplication by $A$ collapses that line to the zero-dimensional subspace.

A real function is differentiable at an interior point $x$ when the finite limitexists, with $h$ restricted so $x+h$ lies in its domain. In quantified form, for every $\epsilon >0$ there is $\delta >0$ such that $0<|h|<\delta$ implies that the difference quotient differs from $f^{\prime }(x)$ by less than $\epsilon$.

To prove differentiability implies continuity, the convergent difference quotient is bounded by $|f^{\prime }(x)|+1$ for sufficiently small nonzero $h$. Thus $|f(x+h)-f(x)|\le (|f^{\prime }(x)|+1)|h|\to 0$. This is exactly continuity at $x$.

At zero, the particular function has difference quotient $h\sin (1/h)$. The bound $|h\sin (1/h)|\le |h|$ and the squeeze theorem giveFor $x\ne 0$, the product rule and chain rule giveAt $x_k=1/(2\pi k)$ this equals $-1$, although $x_k\to 0$ and $f^{\prime }(0)=0$. At $y_k=1/((2k+1)\pi )$ it equals $+1$. Hence the derivative has no limit at zero and is not continuous there. This demonstrates that differentiability does not imply continuity of the derivative, even though the function itself is continuous.

The radius of convergence of a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ is the unique $R\in [0,\infty ]$ for which the series converges absolutely when $|z|<R$ and diverges when $|z|>R$. When $R=\infty$ it converges for every complex $z$; when $R=0$ it converges only at the centre. The definition makes no assertion about convergence on $|z|=R$, so boundary points must be checked separately.

The absolute ratio of successive terms is $|z|n^2/(n+1{)}^2\to |z|$. The ratio test gives convergence for $|z|<1$ and divergence for $|z|>1$, so the radius of convergence is $\boxed{R=1}$. On the boundary, $|n^{-2}{z}^n|=n^{-2}$ andThe positive partial sums are bounded, and thus converge. By the comparison test for series, the original power series converges absolutely at every point of $|z|=1$.

The coefficient ratio isThus the ratio test again gives radius of convergence $\boxed{R=1}$. On $|z|=1$, however, the terms have absolute value $n+2^{-n}$, which does not tend to zero. The necessary term condition for a convergent series fails at every boundary point. Therefore this power series diverges everywhere on $|z|=1$.

A positive, counterclockwise rotation takes the standard basis vectors $(1,0{)}^T$ and $(0,1{)}^T$ to $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. These images are the columns of its rotation matrix, soIt fixes the origin, preserves the dot product and has determinant one.

Since $A{A}^T=I$ and $\det A=1$, $A$ is an orthogonal matrix in the special orthogonal group. It also satisfies $A^{T}A=I$ and thus preserves the dot product. Normalize the fixed eigenvector to $e_3=v/\Vert v\Vert$. For $u\cdot e_3=0$,so the orthogonal complement of the fixed axis is invariant. Choose an oriented orthonormal basis $e_1,e_2,e_3$. In this basis the matrix has form $\mathrm{diag}(B,1)$, where $B$ is a real $2\times 2$ orthogonal matrix of determinant one.

Write the first column of $B$ as $(\cos \theta ,\sin \theta {)}^T$. Its unit second column must be perpendicular to the first; positivity of the determinant selects $(-\sin \theta ,\cos \theta {)}^T$. Thus $B=R(\theta )$, and $A$ fixes the axis through $v$ while rotating its perpendicular plane by $\theta$. Since trace is invariant under change of basis,This proves that $A$ is the required axial rotation. The trace determines the unsigned angle; the action in the oriented perpendicular plane determines its sign. The identity case has angle zero and any axis.

Use the orthogonal matrix identity $A(I-A^T)=A-I$. The product rule for determinants, $\det A=1$, and invariance of determinant under matrix transpose givewhere the last sign comes from dimension three. Hence $\det (A-I)=0$. The real linear map $A-I$ has a nonzero real vector in its kernel, so $A$ has an eigenvector with eigenvalue one. This is the odd-dimensional special orthogonal transformation has a fixed vector property, here proved directly without first classifying the other eigenvalues.

For a linear map $\alpha$, its kernel of a linear map and image of a linear map areBy the definition of the image, $\alpha x=y$ has a solution exactly when $y\in I$. If $x_0$ is one solution, every other solution is $x_0+k$ for some $k\in K$: subtracting the two equations proves necessity, and linearity proves sufficiency. Thus a solvable equation has a unique solution exactly when $K=\{0\}$.

Write the columns of the given matrix as $c_1=(1,0,1{)}^T$, $c_2=(1,t,t{)}^T$ and $c_3=(t,-2b,0{)}^T$. Its determinant isThe columns $c_1,c_2$ are linearly independent for every real $t$: proportionality would require the proportionality factor to be one from the first coordinate, then simultaneously $t=0$ from the second and $t=1$ from the third. Thus the rank is always at least two.

If $D\ne 0$, the matrix is invertible andIf $D=0$, the rank is exactly two. A direct substitution verifies the nonzero kernel vector $k=(-t^2,t,t-1{)}^T$: its first and third row products vanish identically and its second is $D$. The rank-nullity theorem then givesIn that case the image is the plane spanned by $c_1,c_2$. Their cross product is $(-t,1-t,t{)}^T$, so an equivalent description isThese formulas include the singular case $t=b=0$, giving kernel $\mathbb{R}(0,0,1)$ and image $y_2=0$; no exceptional case was lost by dividing by $t$.

Finally, when $0<b<2$,Therefore the equation has a unique solution for every $t$ and every $y\in {\mathbb{R}}^3$, in particular for every $y\in I$. Outside this parameter range, the singular cases are precisely $t=b\pm \sqrt{b(b-2)}$ when that square root is real.

For a group action of $G$ on a set $X$ and $x\in X$, the orbit-stabilizer theorem statesIf $G$ is finite, this becomes $\boxed{|G|=|Gx||G_x|}$. The underlying bijection sends the left coset $g{G}_x$ to $gx$. Equality of two image points is equivalent to $g_2^{-1}{g}_1\in G_x$, so this map is both well-defined and injective, and it is surjective by the definition of the group orbit.

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Every symmetry fixes its centre, and permutes the three perpendicular pairs of face normals, so it is a signed permutation matrix. Conversely every such matrix preserves the cube. The symmetry group of a cube acts transitively on its eight vertices: independent coordinate sign changes take $(1,1,1)$ to any vertex.

A symmetry fixing $(1,1,1)$ must permute its three incident edges. This gives an injective map from its stabilizer subgroup into $S_3$, since those three edge directions determine the linear transformation. All six permutations are realized by coordinate permutation matrices, which fix $(1,1,1)$. Thus the vertex stabilizer has order six. The orbit-stabilizer theorem now yieldsThe use of all symmetries includes reflections as well as the twenty-four elements of the rotational symmetry group of a cube.

There are nontrivial proper normal subgroups. The determinant defines a surjective group homomorphism $G\to \{1,-1\}$. Its kernel is the rotational symmetry group of a cube, a normal subgroup of order twenty-four. Also $\{I,-I\}$ is a central subgroup of order two: central inversion preserves the cube and commutes with every matrix. Both provide affirmative examples.

For the body diagonal through $(1,1,1)$, write a general cube symmetry as $D_{\epsilon }P$, with $P$ a permutation matrix and $D_{\epsilon }=\mathrm{diag}({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$, each ${\epsilon }_j=\pm 1$. Since $P(1,1,1{)}^T=(1,1,1{)}^T$, preserving the unoriented diagonal requires $({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$ to be either $(1,1,1)$ or $(-1,-1,-1)$. ThusThe map $(\pi ,\epsilon )\mapsto \epsilon P_{\pi }$ from $S_3\times C_2$ to $H$ is a group homomorphism, because central inversion commutes with every coordinate permutation. It is surjective by the displayed description and injective because a permutation matrix cannot equal the negative of another permutation matrix. Consequently the body-diagonal stabilizer in the cube symmetry group isThis also agrees with the orbit-stabilizer theorem for the transitive action on the four body diagonals. The line is preserved as a set, so exchanging its opposite endpoints is allowed.

Two nonzero vectors are linearly independent when $ax+by=0$ implies $a=b=0$. Their span then has dimension two. If they are linearly dependent, one is a scalar multiple of the other, so their span is a nonzero line and has dimension one. Thus the requested dimensions are two and one, respectively.

The dot product and corresponding Euclidean norm areThe Cauchy-Schwarz inequality states $\boxed{|x\cdot y|\le \Vert x\Vert \Vert y\Vert }$. If $y=0$ this is immediate. Otherwise the nonnegativity of a squared norm, with $t=(x\cdot y)/\Vert y{\Vert }^2$, givesMultiplying by $\Vert y{\Vert }^2$ and taking square roots proves the inequality. Equality holds precisely when the vectors are linearly dependent. Expanding another squared norm and using the inequality givesand hence the triangle inequality $\boxed{\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert }$.

In ${\mathbb{R}}^3$, the two vectors lie in a common plane. If their directions make angle $\alpha$, the law of cosines gives $\Vert x-y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2-2\Vert x\Vert \Vert y\Vert \cos \alpha$. Comparing with the dot-product expansion yieldsThe sketch shows the same relation as a projection: multiply the length of $x$ by the signed component of $y$ in its direction. That component is negative for obtuse $\alpha$.

A unit vector is a vector of Euclidean norm one, equivalently $u\cdot u=1$. In particular the dot product of two unit vectors is the cosine of their mutual angle.

Put $q=u\cdot v$ and $d=u+v$. The linear independence assumption implies $d\ne 0$. Since $w$ is a unit vector, the Cauchy-Schwarz inequality givesEquality holds exactly when $w$ points opposite to $d$. Its required unit length then fixes the scale:Here $\Vert u+v{\Vert }^2=2+2q$ because both vectors have unit norm. The minimizing vector is unique, and the minimum pairwise dot-product sum of three unit vectors for these fixed $u,v$ is

The triangle inequality for the two unit vectors gives $0<\Vert u+v\Vert \le 2$. Taking reciprocals and using the negative sign in $\lambda =-1/\Vert u+v\Vert$ yieldsIn fact equality would require $u=v$, by the equality condition in the Cauchy-Schwarz inequality. Since the two vectors here are linearly independent, the inequality is strict for every admissible pair.

For the stated angle, $q=\cos (2\pi /3)=-1/2$, so $\Vert u+v\Vert =\sqrt{2+2q}=1$. The minimizing unit vector is therefore $w=-(u+v)$, andIndeed all three pairwise dot products equal $-1/2$, so the vectors lie in one plane, separated by $120$ degrees. This is also the global minimum pairwise dot-product sum of three unit vectors: for any three unit vectors,so $S\ge -3/2$, with equality exactly when $u+v+w=0$.

To prove the Archimedean property, suppose the positive integers were bounded above in $\mathbb{R}$. By the least upper bound axiom they would have a finite supremum $s$. Since $s-1$ is not an upper bound, some positive integer $k$ satisfies $k>s-1$. Then the positive integer $k+1>s$, contradicting the definition of $s$. Thus the positive integers are unbounded. Equivalently, for any positive real $a,b$, some positive integer $n$ satisfies $na>b$, by applying unboundedness to $b/a$.

For fixed $m$, put $r_m={\cos }^2(m!\pi x)$. The standard properties needed are $0\le {\cos }^{2}t\le 1$, and ${\cos }^{2}t=1$ exactly when $t\in \pi \mathbb{Z}$. If $0\le r<1$, then $r^n\to 0$; for $0<r<1$ this also follows from the Bernoulli inequality, since $r^{-n}\ge 1+n(r^{-1}-1)\to \infty$. Therefore the inner limit isIf $x=p/q$ is a rational number, with $q\ge 1$, then $q$ divides the factorial $m!$ whenever $m\ge q$. Hence $m!x$ is an integer for all sufficiently large $m$, and the outer limit is one. This includes the endpoints $x=0,1$. If $x$ is an irrational number, $m!x$ cannot be an integer for any $m$, since division by $m!$ would make $x$ rational. Then every inner limit is zero. The factorial cosine iterated limit detects rationality result is consequentlyEach inner limit was taken with $m$ fixed; the argument does not interchange the two limits or assert a joint limit as both indices grow.

The integral test for convergence applies when $a_n=h(n)$ for all sufficiently large $n$, where $h$ is nonnegative, continuous and nonincreasing on $[N,\infty )$. It states that ${\sum }_{n=N}^{\infty }{a}_n$ converges if and only if the improper integral ${\int }_N^{\infty }h(x)dx$ is finite. Its comparison bounds areNo conclusion about an arbitrary nonnegative sequence follows without such a monotone comparison function; finite initial terms do not affect convergence.

For $h(x)=x^{-\alpha }$ with $\alpha >0$, all the hypotheses hold. If $\alpha \ne 1$,whereas for $\alpha =1$ the integral is $\log R$. The first expression has a finite limit exactly when $\alpha >1$. Thus the p-series satisfies

On $[3,\infty )$, $h(x)=1/(x\log x)$ is positive, continuous and decreasing, since its denominator has positive derivative $\log x+1$. Substituting $u=\log x$ givesThe integral test for convergence therefore shows that the series is divergent.

Because $3>e$, each of $x$, $\log x$ and $\log \log x$ is positive and increasing for $x\ge 3$. Hence $h(x)=1/[x\log x(\log \log x{)}^2]$ is positive, continuous and decreasing. With $u=\log \log x$, $du=dx/(x\log x)$, soThe integral test for convergence implies that the series is convergent.

Compare with the divergent series from part (i). The ratio of its new summand to $1/(n\log n)$ isFor completeness, $\log n={\int }_1^n{x}^{-1}dx\le {\int }_1^n{x}^{-1/2}dx=2(\sqrt{n}-1)$, so $(\log n)/n\to 0$. The limit comparison test now gives divergence. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent p-series with a fixed exponent greater than one.

For a partition $a=x_0<\cdots <x_k=b$ and tags ${\xi }_j\in [x_{j-1},x_j]$, form the Riemann sum ${\sum }_{j=1}^{k}f({\xi }_j)(x_j-x_{j-1})$. The Riemann integral is the unique real number $I$ such that for every $\epsilon >0$ there is $\delta >0$ for which every tagged partition with mesh ${\max }_j(x_j-x_{j-1})<\delta$ has its sum within $\epsilon$ of $I$. Then $I={\int }_a^{b}f(x)dx$. The stated continuity ensures existence, which need not be proved here.

The properties required are linearity, the integral of a constant, and monotonicity of the Riemann integral. Monotonicity follows directly because a nonnegative function has nonnegative Riemann sums, and therefore a nonnegative integral. Integrating $f-m\ge 0$ and $M-f\ge 0$ givesSince $g\ge 0$, multiplication preserves the pointwise inequalities: $mg\le fg\le Mg$. The continuous products are integrable, so the same monotonicity and linearity giveNeither $m$ nor $f$ needs to be nonnegative for this argument.

For the first limit, set $r_n=n^{-1/2}$, $g_n(x)=n{e}^{-nx}$, and ${\epsilon }_n={\max }_{0\le x\le r_n}|f(x)-f(0)|$. By continuity at zero, ${\epsilon }_n\to 0$. On this shrinking interval $f(0)-{\epsilon }_n\le f(x)\le f(0)+{\epsilon }_n$, andApplying the weighted bounds just proved givesBoth bounding expressions tend to $f(0)$, so the squeeze theorem proves the first limit.

For the full interval, continuity on $[0,1]$ gives a finite bound $B={\max }_{[0,1]}|f|$ by the extreme value theorem. The remaining tail has absolute value at mostAdding this to the already evaluated near-end integral provesThese are instances of a one-sided exponential approximate identity: the positive weight has total mass tending to one, concentrated in a shrinking neighborhood of zero. No interchange of limit and integral has been assumed.

For real functions $u,v$ that are continuously differentiable on a closed interval $[a,b]$, integration by parts statesIt follows from the product rule and the fundamental theorem of calculus. With the convention ${\int }_a^b=-{\int }_b^a$, the identity also applies to reversed limits. These regularity assumptions ensure that all the proper integrals exist.

For the smooth $f$ in the problem, define the integral remainder in Taylor theorem byThe fundamental theorem of calculus gives $f(t)=f(0)+R_1(t)$. Applying integration by parts to $R_n$, with $u=f^{(n)}(x)$ and antiderivative $v=-(t-x{)}^n/n!$ of $(t-x{)}^{n-1}/(n-1)!$, givesInduction starting from $n=1$ therefore provesThe same calculation holds for negative $t$: all integrals are oriented, and the segment joining $0$ and $t$ lies in $(-1,1)$. Thus the proof covers every stated $t$ and every positive integer $n$.

Now take $f(x)=\log (1-x)$. For $k\ge 1$, differentiation gives $f^{(k)}(x)=-(k-1)!/(1-x{)}^k$, and $f(0)=0$. With $n=N+1$ and $t=1/2$, the Taylor formula with integral remainder becomesFor $0\le x\le 1/2$, $0\le (1/2-x)/(1-x)\le 1/2$ and $(1-x{)}^{-1}\le 2$. Hence $0\le J_N\le 2^{-N}$ after integrating over an interval of length $1/2$. Thus the logarithmic series from an integral Taylor remainder satisfiesThe explicit vanishing remainder establishes convergence and identifies the sum, without using a pre-existing logarithmic power series.