Past exam of the mathematics course of the University of Cambridge / 2007 / ia

The three spherical basis vectors form an orthonormal basis: their displayed Cartesian components have unit lengths and pairwise zero dot products. For an orthonormal basis, a vector is the sum of its projections onto the basis vectors. Taking the dot product of each Cartesian unit vector with each spherical unit vector therefore gives the inverse change of basis:Equivalently, the change-of-basis matrix is orthogonal, so its matrix inverse is its transpose. These formulas use the stated angular convention for spherical polar coordinates; at a pole the tangential basis vectors still depend on the chosen azimuth.

Use suffix notation, summing repeated indices, and the Levi-Civita symbol convention $(A\times B{)}_i={ϵ}_{ijk}{A}_j{B}_k$. Contracting the outer symbol with the symbol from the second cross product gives the epsilon-delta identitywhere $\delta$ is the Kronecker delta. ThusThe second term is zero because $A_p{A}_{\ell }$ is symmetric in $p,\ell$ and the Levi-Civita symbol is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the scalar triple product $A\cdot (B\times C)$. HenceThis cross products with a common vector identity requires no independence assumption, so it also covers zero or parallel vectors.

The coefficient matrix has determinantA square linear system has a unique solution for every right-hand side precisely when its coefficient matrix is an invertible matrix. Here a unique solution exists exactly when $p\ne 1$, for every real $t$.

For an explicit Gaussian elimination, subtract twice the second equation from the third, then use the second equation, obtaining $x=2z+2t-t^2$ and $y=(t^2-t)/2-3z$. The first equation determinesand the preceding formulas determine $x,y$. When $p=1$ this last coefficient vanishes; the next parts determine compatibility rather than dividing by zero.

For $p=1$, the third coefficient row equals three times the second minus twice the first. The first two rows are independent, so the coefficient matrix rank is two. The same dependence must hold for the right-hand side, givingThus infinitely many solutions occur exactly for $p=1$ and $t\in \{1,2\}$. The null space has dimension one by the rank-nullity theorem, and an explicit parameterization, with $s\in \mathbb{R}$, isSubstitution verifies all three equations in each family.

When $p=1$ but $t^2\ne 3t-2$, the dependent coefficient rows would require an impossible equality between their right-hand sides. Thus the augmented matrix has larger matrix rank than the coefficient matrix, and the linear system is inconsistent. For $p\ne 1$ part (i) already supplies a unique solution. Consequently there is no solution exactly when $p=1$ and $t\notin \{1,2\}$.

The two-variable arithmetic-geometric mean inequality follows from a square:Hence $2\sqrt{ab}\le a+b$, with equality exactly when $a=b$.

For the series, apply that same inequality with $a=a_n$ and $b=n^{-2}$ to obtainThe comparison series converges: its first summand has finite sum by hypothesis, and the inverse-square series converges, for example by the telescoping estimate $n^{-2}\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. Thus the comparison test for series provesEquivalently, its positive partial sums are increasing and bounded above by $\frac{1}{2}\sum a_n+\frac{1}{2}\sum n^{-2}$, so the bounded monotone sequence theorem proves convergence directly. This is the case $p=1$ of weighted square roots of a summable sequence.

First establish a comparison fact for a power series. If it converges at $w\ne 0$, its terms $a_n{w}^n$ are bounded, say $|a_n{w}^n|\le C$. For $|z|<|w|$,by the geometric series. Thus convergence at one point implies absolute convergence at every point of smaller modulus.

Let $S$ be the set of moduli of points at which the power series converges. It contains zero. Define the radius of convergence $R=\sup S\in [0,\infty ]$. If $|z|<R$, the definition of supremum gives a convergent point $w$ with $|w|>|z|$; the comparison just proved gives absolute convergence at $z$. This reasoning also applies when $R=\infty$. If $|z|>R$ and the power series converged there, its modulus would belong to $S$, contradicting the definition of $R$. ThereforeThe boundary $|z|=R$ is intentionally undecided by this argument. If $R=0$, only the exterior assertion is nonvacuous.

For the particular coefficients $a_n=1/n$ for $n\ge 1$, comparison with $\sum |z{|}^n$ gives absolute convergence whenever $|z|<1$. If $|z|=r>1$, then $r^n/n$ fails to tend to zero: its successive ratio is $rn/(n+1)$, eventually greater than a fixed number bigger than one. The necessary term condition for convergence of a series therefore fails. HenceFor instance the positive boundary point gives the divergent harmonic series; no assertion of universal boundary convergence is needed.

Decompose the position vector into its components parallel and perpendicular to the unit vector $\mathbf{u}$:This is an orthogonal projection, and the Pythagorean theorem gives $|\mathbf{r}{|}^2=s^2+|\mathbf{v}{|}^2$. The two surfaces are distinguished by fixing an angle in part (a), and fixing distance from an axis in part (b).

For $\mathbf{r}\ne 0$, the dot product equation says that the angle $\theta$ between $\mathbf{r}$ and the oriented unit vector $\mathbf{u}$ satisfies $\cos \theta =\alpha$. Thus $\theta =\arccos \alpha$ is fixed. More explicitly, using the orthogonal projection from the introductory solution,Consequently the surface is one infinite nappe of a right circular cone, with apex at the origin, including the apex itself. The cone's axis points along $\mathrm{sgn}(\alpha )\mathbf{u}$, and its acute semi-angle is $\arccos |\alpha |$. Equivalently, $\alpha$ is the cosine of the angle from the oriented $\mathbf{u}$ axis. Squaring the original equation without retaining the sign of $s$ would incorrectly add the opposite nappe of a double circular cone. No base disk is included in this infinite lateral surface.

The vector $\mathbf{r}-(\mathbf{r}\cdot \mathbf{u})\mathbf{u}$ is the perpendicular component of $\mathbf{r}$ relative to the axis $\mathbb{R}\mathbf{u}$, so its length is the distance from that axis. Hence the surface is an infinite circular cylinder of radius $\beta$ about the line through the origin parallel to $\mathbf{u}$. Here $\beta$ is the radius and $\mathbf{u}$ fixes the axis direction; reversing $\mathbf{u}$ leaves the same circular cylinder.

If ${\mathbf{v}}_1,{\mathbf{v}}_2$ are an orthonormal basis of the plane perpendicular to $\mathbf{u}$, a parameterization isThis verifies that all heights and all angular positions, but no interior points, belong to the surface.

The sphere-plane intersection is governed by the orthogonal projection of the sphere's centre onto the plane. Since $\mathbf{n}$ has unit length, let $h=p-\mathbf{n}\cdot \mathbf{c}$, the signed distance from $\mathbf{c}$ to the plane along $\mathbf{n}$. The following parts determine the centre and radius and retain this sign; the radii alone cannot distinguish the two possible sides of the oriented plane.

Set ${\mathbf{b}}_0=\mathbf{c}+h\mathbf{n}$, where $h=p-\mathbf{n}\cdot \mathbf{c}$. Then $\mathbf{n}\cdot {\mathbf{b}}_0=p$, so ${\mathbf{b}}_0$ lies in the plane. For any position $\mathbf{r}$ in that plane, $\mathbf{r}-{\mathbf{b}}_0$ is perpendicular to ${\mathbf{b}}_0-\mathbf{c}=h\mathbf{n}$. The Pythagorean theorem therefore givesOn the sphere the left side is $a^2$. Thus the intersection is the circle in the plane centred at ${\mathbf{b}}_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf{b}={\mathbf{b}}_0$, andIn terms of the radii, $\lambda =\pm \sqrt{a^2-{\rho }^2}$, with the sign fixed by $\lambda =p-\mathbf{n}\cdot \mathbf{c}$. In the central-plane case $\rho =a$, both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.

Substitute the signed displacement $\lambda =p-\mathbf{n}\cdot \mathbf{c}$ into the Pythagorean theorem relation from part (a):A genuine circle requires $|p-\mathbf{n}\cdot \mathbf{c}|<a$. Equality gives a tangent plane and a single point, interpreted as radius zero; a larger distance gives no intersection. Under the stipulated circle intersection, the radicand is positive. Its centre is $\mathbf{c}+(p-\mathbf{n}\cdot \mathbf{c})\mathbf{n}$.

Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

In suffix notation, $x_i^{\prime }=a{x}_i+b{ϵ}_{ikj}{n}_k{x}_j$, so the matrix entries areHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The diagonal determines $a=2$. Comparing the off-diagonal entries withgives $b\mathbf{n}=(1,2,2)$. As $\mathbf{n}$ is a unit vector and $b>0$, $b=3$ and $\mathbf{n}=(1,2,2)/3$. ThusIn this example the transverse scaling is $\sqrt{13}$ and the axial scaling is two; the rotation angle is $\arctan (3/2)$.

Let $P\mathbf{x}=(\mathbf{n}\cdot \mathbf{x})\mathbf{n}$ be the axial orthogonal projection and $N\mathbf{x}=\mathbf{n}\times \mathbf{x}$. Then $NP=PN=0$, while the vector triple-product identity gives $N^2=P-I$. The map is $aP+a(I-P)+bN$. Its inverse is $P/a$ on the axis. On the perpendicular plane,Thus the inverse of an axis-angle linear map isor, in the requested vector form,The denominators are nonzero because $a,b>0$. Multiplication on each invariant component verifies the matrix inverse without choosing a coordinate frame.

For the Cauchy-Schwarz inequality, take $\lambda =(\mathbf{x}\cdot \mathbf{y})/(\mathbf{y}\cdot \mathbf{y})$ and use positivity of the inner product:Multiplication by $\mathbf{y}\cdot \mathbf{y}>0$ provesEquality holds exactly when $|\mathbf{x}-\lambda \mathbf{y}{|}^2=0$. Positive definiteness of the inner product makes this equivalent to $\mathbf{x}=\lambda \mathbf{y}$. Conversely, substitution of a scalar multiple gives equality. Thus equality holds precisely for linearly dependent nonzero vectors.

Apply the Cauchy-Schwarz inequality to $(x,y,z)$ and $(1,1,1)$ in the Euclidean inner product on ${\mathbb{R}}^3$. It givesAfter expansion and division by two,The equality criterion from the introductory proof says $x=y=z$. One can also see both the inequality and equality case from $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$.

Set $s=x+y+z$ and $Q=x^2+y^2+z^2$. The expression in the equation is $4(Q+4)-(s+2{)}^2$. Apply the Cauchy-Schwarz inequality in ${\mathbb{R}}^4$ toTheir dot product is $s+2$, and $|\mathbf{e}{|}^2=4$. Thus the expression is nonnegative, and it is zero precisely when equality holds in Cauchy-Schwarz. That requires $\mathbf{v}=\lambda \mathbf{e}$. The fourth coordinate forces $\lambda =2$, and the other three coordinates then giveSubstitution gives $3(4+4+4+4)-2(4+4+4)-4(2+2+2)=48-24-24=0$, so this unique possible choice really is a solution.

Write a straight line in Cartesian form $Ax+By+C=0$, where $A,B,C$ are real and $(A,B)\ne (0,0)$. With $z=x+iy$, set $c=(A+iB)/2$ and $r=C$. Using the complex conjugate,Thus the real linear equation of a line in the complex plane isConversely, writing $c=u+iv$ gives $2ux+2vy+r=0$, an ordinary straight line with normal vector $(u,v)$. The nonzero condition on $c$ matters: $c=0$ would give either the whole complex plane or the empty set.

The quotient is defined only when $z+\overline{u}\ne 0$. Its unit-modulus condition then gives $|z+u{|}^2=|z+\overline{u}{|}^2$. On expanding with complex conjugates,Hence at least one imaginary part is zero, and either $z$ or $u$ is real. Conversely, either reality condition makes the two squared moduli equal; provided the denominator is nonzero, the quotient has modulus one. No conclusion is drawn from an undefined quotient.

For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.

Let $F(x)=\sqrt{2+\sqrt{x}}$ for $x\ge 0$. This is an increasing continuous function. Starting at $a_1=\sqrt{2}$, induction gives $\sqrt{2}\le a_n\le 2$: if $a_n\le 2$, then $a_{n+1}\le \sqrt{2+\sqrt{2}}<2$, and every iterate is at least $\sqrt{2}$.

The first step has $a_2>\sqrt{2}=a_1$. Since $F$ is increasing, $a_n\ge a_{n-1}$ implies $a_{n+1}=F(a_n)\ge F(a_{n-1})=a_n$. Thus this sequence is increasing and bounded above. Let $L_{-}={\sup }_n{a}_n$. For every $\epsilon >0$, a term exceeds $L_{-}-\epsilon$; all later terms lie between it and $L_{-}$. This proves convergence, the elementary bounded monotone sequence theorem.

For the second starting value, write the iterates as $b_1=4$ and $b_{n+1}=F(b_n)$. Here $b_2=2<4$. Monotonicity of $F$ propagates this inequality, so $b_n$ is decreasing. It is bounded below by $\sqrt{2}$ because every positive iterate is at least that value. Hence it converges to its infimum $L_{+}$. From $b_2=2$ onward all terms lie in $[\sqrt{2},2]$.

Continuity of $F$ permits passing to each limit in the recurrence, giving $L_{\pm }^2=2+\sqrt{L_{\pm }}$. To prove that these fixed points agree, suppose $r,s\ge \sqrt{2}$ both satisfy that equality. Subtracting the two equations and rationalizing givesThe second factor is positive: $r+s\ge 2\sqrt{2}$, whereas $(\sqrt{r}+\sqrt{s}{)}^{-1}\le 1/(2\sqrt{\sqrt{2}})<2\sqrt{2}$. Thus $r=s$. ThereforeExistence has been supplied by the two convergent sequences, and the subtraction argument gives the required uniqueness in their common interval.

The mean value theorem states that a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, has a point $\xi \in (a,b)$ such thatFor the proof, subtract the secant line and put $h(t)=f(t)-f(a)-(t-a)[f(b)-f(a)]/(b-a)$. Then $h(a)=h(b)=0$. If $h$ is identically zero, every interior point works. Otherwise it has a positive maximum or a negative minimum. The extreme value theorem supplies that extremum, and its nonzero value makes its location interior. At an interior differentiable extremum, the difference quotients on the two sides have opposite weak signs, so their common limit is zero. Hence $h^{\prime }(\xi )=0$, proving the formula. This also gives the Rolle theorem argument rather than merely citing it.

Because $f^{\prime \prime }$ exists everywhere, $f^{\prime }$ is differentiable everywhere and therefore continuous everywhere; continuity of $f^{\prime \prime }$ is not required. For $x<y$, apply the mean value theorem to $f^{\prime }$ on $[x,y]$. It gives $\xi \in (x,y)$ withFor $x=y$ equality is immediate. Hence $f^{\prime }$ is nondecreasing: $x\le y$ implies $f^{\prime }(x)\le f^{\prime }(y)$.

Put $q=\lambda a+(1-\lambda )b$. Since $0<\lambda <1$ and $a<b$, $a<q<b$. Apply the mean value theorem separately to $[a,q]$ and $[q,b]$. There exist $x\in (a,q)$ and $y\in (q,b)$ such thatThese are the requested points, and $x<y$. Part (i) makes $f^{\prime }(x)\le f^{\prime }(y)$, so the corresponding secant slopes satisfyMultiply by the positive denominators and collect $f(q)$ to obtainThis is precisely the defining chord inequality for a convex function. The proof uses only existence and nonnegativity of $f^{\prime \prime }$, with no additional regularity hypothesis on that derivative.

To prove boundedness on $[a,b]$, suppose instead that $f$ is unbounded and choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence converging to $x_{\ast }\in [a,b]$. Continuity then gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j$. Hence $f$ is bounded.

Let $M={\sup }_{x\in [a,b]}f(x)$ and $m={\inf }_{x\in [a,b]}f(x)$. Choose $d_n$ with $M-1/n<f(d_n)\le M$. Another application of Bolzano-Weierstrass theorem yields a subsequence tending to $d\in [a,b]$, and continuity gives $f(d)=M$. Applying the same argument to $-f$ gives $c\in [a,b]$ with $f(c)=m$. ThusThis proves the extreme value theorem here, including both attainment assertions.