Past exam of the mathematics course of the University of Cambridge / 2011 / ia

For a nonzero complex number, write $z=r{e}^{i\theta }$ with $r=|z|>0$ and principal complex argument $-\pi <\theta \le \pi$. The principal complex logarithm and corresponding complex exponentiation areThe value at zero is not defined by this formula. The choice of complex argument matters: these are single-valued principal powers, rather than all values of a multivalued complex logarithm.

For the requested sketch, $(1+i)\mathrm{Log}z=(\log r-\theta )+i(\log r+\theta )$. Its complex exponential has complex modulus $e^{\log r-\theta }$, soThis is one turn of a logarithmic spiral, with increasing radius as the polar angle increases. It passes through $1$ at $\theta =0$, approaches the excluded endpoint $-e^{-\pi }$ from below the negative real axis, and ends at the included endpoint $-e^{\pi }$ from above that axis. The endpoint distinction follows from the chosen principal complex argument.

The principal complex exponentiation gives $z^i=e^{-\theta }{e}^{i\log r}$. Equality to one requires its complex modulus to be one, hence $\theta =0$, and then its complex argument requires $\log r\in 2\pi \mathbb{Z}$. Thus all solutions are positive real, namelySubstitution verifies every value, and the modulus condition excludes every nonreal value.

Away from the negative real axis, the complex conjugate has principal complex argument $-\theta$. ConsequentlyTaking the complex modulus gives $\cosh \theta =1$, hence $\theta =0$. The remaining equation is $e^{i\log r}=i$, soOn the negative real axis, both $z$ and $\overline{z}$ have principal complex argument $\pi$, so the left-hand side has modulus $2e^{-\pi }$, not two. There are no additional solutions there. The printed second term is the principal power of $\overline{z}$, not the complex conjugate of $z^i$.

Use change of basis matrices whose columns are the respective basis vectors in standard coordinates:The coordinate relation is $[\Phi (x){]}_C=A[x{]}_B$. Therefore the standard-coordinate matrix of a linear map is $T=S_{C}A{S}_B^{-1}$, and the new coordinate matrix is $A^{\prime }=S_{C^{\prime }}^{-1}T{S}_{B^{\prime }}$. Multiplying givesOne can also read off the columns: the two new domain basis vectors map to twice the first and four times the second new codomain basis vector. All four change of basis matrices are invertible. In particular the printed codomain vectors are three-dimensional; retaining their third coordinates is essential.

The Bolzano-Weierstrass theorem states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. More generally, a bounded sequence in ${\mathbb{R}}^d$ has a subsequence converging to a point of ${\mathbb{R}}^d$.

Take $a_n=n$. Every subsequence has the form $a_{n_k}=n_k$, with $n_k\ge k$, so it tends to positive infinity and cannot converge to a real number. Thus this sequence has no convergent subsequence.

Define $a_{2k}=0$ and $a_{2k-1}=k$. The odd-indexed terms are unbounded, while the even-indexed subsequence is identically zero and hence converges to zero. This gives an unbounded sequence with a convergent subsequence.

The floor function has even value on indices congruent to zero or one modulo four, and odd value on indices congruent to two or three. Accordingly the displayed sequence has terms $2+1/n$ in the first two residue classes and $-1/n$ in the other two. In particular,Every term is within $1/n$ of the set $\{0,2\}$. If a subsequence converged to $c\notin \{0,2\}$, its eventual distance from that set would be bounded below by a positive constant, contradicting this bound. The complete set of subsequential limits is $\{0,2\}$.

For the power series coefficient $c_n=n^2$, the Cauchy-Hadamard theorem givesThus the radius of convergence is $\boxed{R=1}$. The power series converges absolutely for $|z|<1$ and diverges for $|z|>1$. On $|z|=1$ its terms have modulus $n^2$, so they do not tend to zero and the series diverges there as well.

The original PDF has coefficient $c_n=n^{n^{1/3}}$. Its $n$th root isThe Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$. Again the power series converges absolutely inside the unit circle, while on that circle its terms do not tend to zero. The converted TeX loses the outer exponent and writes $n^{1/3}$; the calculation here uses the printed coefficient.

In the parametric description, $a$ is a point of the straight line and the nonzero vector $b$ gives its direction. The real parameter runs through every point of that line. For the cross product description, take another cross product with $c$ and use the vector triple product identity:Hence every solution has the representationHere $a\cdot b=0$ and $|b|=|c|$, as required. Conversely,Thus the perpendicularity assumption is sufficient, and the solutions form a straight line parallel to $c$. When $d\ne 0$, the vector $d$ is perpendicular both to $c$ and to every position vector on the line; it is the normal to the plane through the origin containing that line. The point $a=c\times d/|c{|}^2$ is the perpendicular foot from the origin. Its distance from the origin is $|d|/|c|$, since $c\cdot d=0$.

For a general parametric straight line, minimize the squared Euclidean distanceThis strictly convex quadratic function has its unique minimum at $\lambda =b\cdot (y-a)/|b{|}^2$, giving the orthogonal projectionFor the cross product description, $a\cdot c=0$, so this becomes

For the two planes, put $s=m\cdot n$ and $c=m\times n$. Nonparallel unit normals give $|s|<1$ and $|c{|}^2=1-s^2$. Seek the perpendicular foot in their span, $a=um+vn$. Its two dot products impose $u+sv=\mu$ and $su+v=\nu$. Solving gives both requested descriptions of the intersection:andIndeed the vector triple product identity gives $x\times (m\times n)=m(x\cdot n)-n(x\cdot m)$. Conversely, since $m,n$ are linearly independent, equality to $\nu m-\mu n$ forces both plane equations. Thus neither description introduces extra points.

The vector triple product identity gives $n\times (x\times n)=x-(n\cdot x)n$. Decompose $x=x_{\perp }+x_{\parallel }$, where $x_{\parallel }=(n\cdot x)n$ and $x_{\perp }\cdot n=0$. ThenThe linear map acts as the identity on $n^{\perp }$ and as multiplication by $\alpha$ along $n$. Consequently it is invertible exactly when $\alpha \ne 0$, withApplying $\Phi$ to this expression recovers the perpendicular and parallel components of $y$ separately.

At $\alpha =0$, $\Phi$ is the orthogonal projection onto the plane perpendicular to $n$. Its image of a linear map and kernel of a linear map areEvery perpendicular vector is unchanged and every parallel vector is annihilated. The image plane is perpendicular to the kernel axis by the definition of $n^{\perp }$; their dimensions two and one also agree with the rank-nullity theorem.

Expanding the formula from part (a) in components gives $y_i=x_i+(\alpha -1)n_i{n}_j{x}_j$, with repeated indices summed. Hence the matrix of a linear map and its inverse, when $\alpha \ne 0$, have entriesHere ${\delta }_{ij}$ is the Kronecker delta. To check the matrix inverse, write $P=n{n}^T$; $P^2=P$ because $|n|=1$. Thus $A=(I-P)+\alpha P$ and $B=(I-P)+{\alpha }^{-1}P$, whose product is the identity matrix.

For the specified unit vector,For a cyclic matrix with first row $(a,b,c)$, its determinant is $a^3+b^3+c^3-3abc$. Here it givesTherefore $Ax=Cx$ has only the zero solution if $\alpha \ne 1$. At $\alpha =1$, $A=I$ and $A-C$ has kernel consisting of the equal-coordinate vectors. The complete answer isGeometrically, $C^{T}C=I$, $\det C=1$ and $Cn=n$, so $C$ is a rotation matrix about the $n$ axis. Its matrix trace is two, giving $\cos \theta =(\mathrm{tr}C-1)/2=1/2$; its skew part selects angle $-\pi /3$ about the oriented axis $n$. It has no fixed nonzero perpendicular vector. Meanwhile $A$ preserves perpendicular components and scales the axial component by $\alpha$. Comparing these components first forces $x_{\perp }=0$, and then $(\alpha -1)x_{\parallel }=0$, exactly matching the determinant calculation, including the noninvertible case $\alpha =0$.

Expanding the characteristic polynomial givesThus the eigenvalues are $1,2,3$. Solving the corresponding homogeneous linear systems givesEvery nonzero vector in the displayed eigenspace is an eigenvector. Direct multiplication verifies the three eigenvalue equations. The distinct eigenvalues imply these eigenvectors are linearly independent and form a basis of ${\mathbb{R}}^3$.

The equation is solvable exactly when $b\in \mathrm{im}A$, the image of a linear map, equivalently when the rank of a matrix satisfies $\mathrm{rank}A=\mathrm{rank}[A\mid b]$. If it is solvable and $x_0$ is one solution, every solution is preciselyIndeed differences of solutions lie in the kernel of a linear map, and adding a kernel vector preserves the equation. Hence there are no solutions when $b\notin \mathrm{im}A$; one solution when $A$ is invertible; and infinitely many solutions when $b\in \mathrm{im}A$ and $A$ is singular. In the last case the rank-nullity theorem gives a nonzero kernel vector $v$, and $x_0+tv$ gives distinct solutions for every real $t$. This proves that a finite number greater than one is impossible.

Use the eigenvector basis $v_1=(-1,0,1{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$. The forcing has decomposition $b=-v_1+3v_3$. If $x=c_1{v}_1+c_2{v}_2+c_3{v}_3$, the linear system reduces toFor $\lambda =0$, all coefficients are determined, givingFor $\lambda =1$, the first equation would say $0=-1$, so there are no solutions. For $\lambda =2$, $c_1=1$, $c_3=3$, and $c_2$ is arbitrary. Thus all solutions areThese outcomes illustrate respectively an invertible matrix, incompatible forcing outside its image of a linear map, and a compatible forcing with a nontrivial kernel of a linear map.

Let $Mv=\lambda v$ with $v\in {\mathbb{C}}^n$ nonzero. For a real symmetric matrix, $M^{\ast }=M$, where the star denotes the conjugate transpose. Consequently $v^{\ast }Mv$ equals its complex conjugate and is real. The eigenvalue equation givessince $v^{\ast }v>0$. Allowing a complex eigenvector in this argument is essential: reality of an eigenvalue must be proved, rather than assumed in advance.

Write a complex eigenvector as $v=u+iw$, where $u,w\in {\mathbb{R}}^n$. Since $M$ and its eigenvalue $\lambda$ are real,At least one of $u,w$ is nonzero, and it is a real eigenvector for the same eigenvalue. More generally every complex eigenvector is a complex linear combination of real ones: its real and imaginary parts lie in the real eigenspace. Thus each eigenspace has a basis chosen over the reals. The conclusion does not assert that every arbitrary complex eigenvector is itself real.

Choose real eigenvectors $u,v$ with distinct eigenvalues $\lambda ,\mu$. Symmetry givesTherefore $(\lambda -\mu )u^{T}v=0$, and $\boxed{u\cdot v=0}$. For complex choices, replace transpose by conjugate transpose; the reality proved in part (i) gives the same conclusion for the Hermitian inner product. Thus different eigenspaces are orthogonal.

For a real antisymmetric matrix, $A^T=-A$, so $(A^2{)}^T=A^2$. The previous result makes every eigenvalue $\eta$ of $A^2$ real, with a real nonzero eigenvector $v$. Moreover,Thus every eigenvalue of $A^2$ is real and nonpositive.

Take real unit eigenvectors $u,w$ of $A^2$ for the respective distinct nonzero eigenvalues $-{\lambda }^2,-{\mu }^2$, where the real parameters $\lambda ,\mu$ are nonzero. DefineThe displayed quadratic identity gives $|u^{\prime }|=|w^{\prime }|=1$. Also $u\cdot Au=0$ and $w\cdot Aw=0$, because a real antisymmetric matrix has zero quadratic form. Thus $u\perp u^{\prime }$ and $w\perp w^{\prime }$. Since $A$ commutes with $A^2$, $u^{\prime }$ lies in the same eigenspace of $A^2$ as $u$, and $w^{\prime }$ in the same one as $w$. Distinct eigenspaces of the real symmetric matrix $A^2$ are orthogonal, so all four vectors are orthonormal. Finally,Thus on each of the two perpendicular planes, $A$ acts as a scaled quarter-turn. This constructs the required vectors, rather than merely counting dimensions.

For a positive-term series, the ratio test says that if $a_{n+1}/a_n\to L<1$, the series converges; if the ratio tends to $L>1$, including positive infinity, it diverges. More generally $\mathrm{limsup}{a}_{n+1}/a_n<1$ suffices for convergence and $\mathrm{liminf}{a}_{n+1}/a_n>1$ for divergence. A limiting ratio of one gives no conclusion.

The P-series $\sum n^{-2}$ converges and the harmonic series $\sum n^{-1}$ diverges. Both have strictly decreasing positive terms, and their successive ratios, $(n/(n+1){)}^2$ and $n/(n+1)$, tend to one. These supply the two requested possibilities.

The function $1/x$ is decreasing and positive. Therefore on $[k,k+1]$ it is at most $1/k$, and the Riemann integral givesSumming from $k=1$ to $n-1$ yieldsFor $n=1$ both sides are zero; for $n>1$ the inequality is actually strict. The same integral test for convergence gives ${\sum }_{k=N}^{n-1}1/k\ge \log (n/N)$.

First, for $0<x<1$, the derivative of $h(x)=\log (1-x)+x$ is $-x/(1-x)<0$, while $h(0)=0$. Hence $\log (1-x)<-x$.

Choose $N>c$ large enough that the given ratio bound holds for every $n\ge N$. Positivity permits taking logarithms, and the preceding inequality givesSumming, and using the integral test for convergence from part (b), gives for $n>N$Thus $a_n\le a_N{N}^c{n}^{-c}$. Since $c>1$, the P-series $\sum n^{-c}$ converges, so the comparison test for series provesThe finitely many initial terms do not affect convergence. This power-law comparison from a refined ratio bound supplies information when the ordinary ratio test can be inconclusive.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous, then every real value between $f(a)$ and $f(b)$ is attained at some $c\in [a,b]$. Here is a proof using completeness of the real numbers. Endpoint values are immediate, so suppose $f(a)<y<f(b)$; reversing the sign of $f$ handles the other ordering. LetThe set is nonempty and bounded above. Continuity at $a$ puts points to the right of $a$ in $S$, while continuity at $b$ excludes an entire interval to the left of $b$, so $a<c<b$. If $f(c)<y$, continuity puts some point larger than $c$ in $S$, contradicting its supremum. If $f(c)>y$, continuity excludes $S$ in a neighborhood of $c$, contradicting the existence of points of $S$ arbitrarily close to its supremum from below. Therefore $f(c)=y$, proving the theorem.

For the fixed-point assertion, put $g(x)=f(x)-x$. This is continuous, with $g(0)=f(0)\ge 0$ and $g(1)=f(1)-1\le 0$. If either endpoint value is zero it already gives a fixed point; otherwise the intermediate value theorem gives an interior zero. HenceThe proof needs both continuity and the closed-interval endpoint information; the following counterexamples separate those hypotheses.

No. The continuous function $f(x)=(x+1)/2$ maps $(0,1)$ into $(0,1)$, but $f(x)-x=(1-x)/2>0$ throughout the domain. Its only possible fixed point is the excluded endpoint one.

No. The continuous function $f(x)=x+1$ is a self-map of $\mathbb{R}$, but the fixed point equation would require $x+1=x$, which is impossible.

No. Define $f(x)=1$ for $0\le x<1$ and $f(1)=0$. This self-map of $[0,1]$ is discontinuous at one. Every point smaller than one is sent to a different point, and one is sent to zero, so it has no fixed point.

Yes: a tangential zero and a crossing zero can give exactly two fixed points. One explicit continuous piecewise linear example isAdjacent formulas agree at each junction. Their endpoint values lie in $[0,1]$, so linear interpolation proves that the whole image lies in $[0,1]$; also $f(0)=1$ and $f(1)=0$. On the four pieces, $f(x)-x$ is respectively $1-3x$, $3x/2-1/2$, $1-3x/2$, and $2-3x$. Their only zeros on their respective intervals are $1/3$ and $2/3$. Thus the complete fixed point set is $\boxed{\{1/3,2/3\}}$. Continuity forces at least one zero, but does not force every zero to change sign.

For the first additional construction, modify the rational/irrational example toAt zero, $|g(h)/h|=|h|\to 0$, so $g^{\prime }(0)=0$. At any nonzero point, density of the rational numbers and density of the irrational numbers give incompatible limiting values $x^2$ and $-x^2$. Thus $g$ is discontinuous, and hence not differentiable, there. Its differentiability set is exactly $\{0\}$.

For the second construction, use a summable absolute-value cusp series:On $[-R,R]$, each summand is bounded by $2^{-n}(R+1/2)$. The Weierstrass M-test therefore gives uniform convergence on every bounded interval, and the uniform limit theorem makes $G$ continuous everywhere.

To justify differentiability, each summand's difference quotient has absolute value at most $2^{-n}$, by the reverse triangle inequality. The sum of these bounds is finite, uniformly in the increment. Thus the tail of the difference quotient is uniformly small, and limits may be passed through the sum by first retaining finitely many terms and then making the tail small. If $x\ne 1/n$ for every $n\ge 2$, each individual summand is differentiable, givingIn particular $G^{\prime }(0)=-{\sum }_{n=2}^{\infty }{2}^{-n}=-1/2$. The accumulation of cusp locations at zero does not destroy the derivative, because their weights are summable.

At $x=1/m$, the same tail argument applies to both one-sided derivatives. Every term except the $m$th has the same derivative from both sides; the $m$th has derivatives $-2^{-m}$ and $2^{-m}$. ConsequentlyTherefore $G$ fails to be differentiable exactly at $1/2,1/3,1/4,\dots$, and is differentiable everywhere else.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.