Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2

Use completing the square rather than computing the eigenvalues of the symmetric matrix. Successively removing the mixed terms givesThus one suitable invertible linear change of coordinates isThe coordinate matrix is upper triangular with determinant one, so the change is an invertible linear map. Its inverse is $z=Z$, $y=Y+4Z/7$, $x=X-2Y+5Z/14$. Hence the diagonal coefficients can be taken as $2,-7,-3/14$. This is a matrix congruence of a quadratic form; an orthogonal transformation was not required.

The fraction field of an integral domain $R$ is a field $F$ containing an identified copy of $R$ in which every element has the form $a/b$ with $a,b\in R$ and $b\ne 0$. Equality means $a/b=c/d$ exactly when $ad=bc$; the operations are the usual addition and multiplication of fractions. We use unital ring homomorphisms.

Since $\phi$ is injective, $\phi (b)\ne 0$ whenever $b\ne 0$. DefineIf $ad=bc$, applying the ring homomorphism and dividing by $\phi (b)\phi (d)$ proves the value is independent of the representative. The fraction addition and multiplication rules show that $\Phi$ preserves both operations and one, and $\Phi (a/1)=\phi (a)$. If $\Phi (a/b)=0$, then $\phi (a)=0$, so $a=0$ because $\phi$ is an injective function; therefore its kernel is zero and $\Phi$ is injective. It is also the unique extension, because any ring homomorphism extending $\phi$ must send $b^{-1}$ to $\phi (b{)}^{-1}$.

For a counterexample, take $R=\mathbb{Z}$, $F=\mathbb{Q}$, and reduction modulo two $\psi :\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$. An extension would give $1=\Psi (2)\Psi (1/2)=0$, a contradiction. Nonzero denominators must become invertible in the target; an arbitrary ring homomorphism need not have that property.

Differentiability here is understood in the sense of the Frechet derivative: there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe map $L=Df(x)$ is the Frechet derivative. In finite dimensions a linear map is bounded, so $\Vert Lh\Vert \le C\Vert h\Vert$ for some $C$. The defining estimate therefore impliesThis is precisely continuity at $x$: differentiability implies continuity.

First suppose each $g_j$ is differentiable at $x_j$. Write $g_j(x_j+t)=g_j(x_j)+v_{j}t+r_j(t)$, where $v_j\in {\mathbb{R}}^m$ and $r_j(t)=o(|t|)$. Define the linear map $Lh={\sum }_j{v}_j{h}_j$. Given $\epsilon >0$, choose a common sufficiently small neighbourhood so that $\Vert r_j(h_j)\Vert \le \epsilon |h_j|$ for all $j$, with $r_j(0)=0$. ThenThis proves differentiability, withConversely, suppose $f$ is differentiable with Frechet derivative $L$. Along the $j$th coordinate direction $e_j$, the other summands cancel exactly:Thus $g_j$ is differentiable at $x_j$, with $g_j^{\prime }(x_j)=L{e}_j$. The equivalence works because each summand varies in only one coordinate; it is stronger than merely having partial derivatives.

The function $g(t)=|t{|}^{3/2}$ is differentiable everywhere: for $t\ne 0$, its derivative is $(3/2)\mathrm{sgn}(t)\sqrt{|t|}$, while at zero the difference quotient is $\mathrm{sgn}(t)\sqrt{|t|}\to 0$.

The function $k(t)=|t{|}^{1/2}$ is differentiable for $t\ne 0$, with derivative $\mathrm{sgn}(t)/(2\sqrt{|t|})$. At zero its difference quotient is $\mathrm{sgn}(t)/\sqrt{|t|}$, which has no finite limit. Applying the coordinatewise-sum argument givesAt those points,where the first coefficient is zero at $x=0$. At any point with $y=0$, the restriction to the vertical coordinate line already fails to be differentiable.

Let $K$ be a compact set of a Hausdorff space $X$. For any $x\notin K$ and each $y\in K$, the Hausdorff property gives disjoint open sets $U_y\ni y$ and $V_y\ni x$. The sets $U_y$ cover $K$; compactness supplies a finite subcover $U_{y_1},\dots ,U_{y_r}$. The open set $V={\bigcap }_i{V}_{y_i}$ contains $x$ and misses every $U_{y_i}$, hence misses $K$. Thus $X\setminus K$ is open, and every compact subset of a Hausdorff space is closed. The empty-set case is immediate.

Since $\mathbb{R}$ with its usual topology is Hausdorff, its compact sets are closed. The open cover $\{(-n,n):n\ge 1\}$ of $\mathbb{R}$ has a finite subcover of any compact set $K$. Taking the largest index in that subcover gives $K\subset (-N,N)$, so $K$ is a bounded set. This proves the required direction of the Heine-Borel theorem without assuming its converse.

Take $X$ to be the set $\mathbb{Q}$, equipped with the discrete metricEvery Cauchy sequence is eventually constant: its terms must eventually have mutual distance in this metric less than $1/2$. Consequently $X$ is a complete metric space. Its topology is the discrete topology, since every singleton is a ball of radius $1/2$ and every subset is a union of such balls.

For the codomain use the usual subspace topology on $\mathbb{Q}$. The inverse image of any open set there is a subset of the discrete $X$, hence is open. Therefore $f$ is continuous and plainly surjective. A continuous image of a complete metric space need not be complete. The two occurrences of $\mathbb{Q}$ have different domain and codomain metrics.

For the method of characteristics, parameterize by $x$; the characteristic curves satisfyThus $x^2+y=C$ and $u-\sin x$ is constant on each characteristic curve. The general form in the quadrant is $u=\sin x+F(x^2+y)$. The boundary condition gives $F(s^2)=\cos s-\sin s$ for every $s\ge 0$.

The characteristic curve through $(x,y)$ meets the boundary at $(s,0)$ with $s=\sqrt{x^2+y}\ge x$. Integrating backwards from that boundary value yieldsIndeed ${\partial }_x-2x{\partial }_y$ annihilates $x^2+y$ and sends $\sin x$ to $\cos x$, while on $y=0$ the expression reduces to $\cos x$. Every characteristic in the quadrant meets the prescribed boundary, so its value is determined. The displayed solution is smooth away from the origin. At the origin the boundary is characteristic and the solution is continuous but has a singular inward $y$ derivative; the differential equation is interpreted in the interior, with the boundary data attained continuously.

In SI units the Ampère-Maxwell equation is $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$. Since $\mathbf{E}=0$, the displacement-current term vanishes and $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$. Apply Stokes theorem to an oriented surface $S$ with positively oriented boundary $C$:Hence the integral form of Ampère's law isHere ${\mu }_0$ is vacuum magnetic permeability, $\mathbf{J}$ is current density, and positive enclosed electric current is defined by the orientation of $S$.

For the field generated by the infinitely long, axially symmetric shell currents, axisymmetry and the right-hand rule give $\mathbf{B}=B_{\varphi }(r){\mathbf{e}}_{\varphi }$. A circular loop has $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi r{B}_{\varphi }(r)$. Its enclosed electric current is zero inside the inner shell, $I$ between the shells and $2I$ outside both. Ampère's law therefore givesThe positive ${\mathbf{e}}_{\varphi }$ direction is counterclockwise viewed from positive $z$. The ideal current sheets make the tangential magnetic field discontinuous at each shell; these formulas specify the three open regions, with no additional imposed background field.

Only the signed enclosed electric current changes. Between the shells it remains $I$, but outside both it is $I-I=0$. The same circular-loop application of Ampère's law givesOpposite currents confine the generated magnetic field to the annulus, as in an ideal infinitely long coaxial cable. The two shell current sheets account for the field jumps.

For this steady unidirectional flow, the convective acceleration $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanishes because $u$ depends only on $y$. The axial Navier-Stokes equation reduces to $0=-p_x+\mu u^{\prime \prime }$, with $p_x=-G$. The no-slip boundary conditions therefore giveThis is a superposition of Couette flow and pressure-driven Poiseuille flow.

Integrate the axial ordinary differential equation twice: $u=-G{y}^2/(2\mu )+C_{1}y+C_2$. The no-slip boundary condition at zero gives $C_2=0$, and the one at $h$ gives $C_1=U/h+Gh/(2\mu )$. HenceThe first term is Couette flow; the second is Poiseuille flow driven in positive $x$ when $G>0$. Their addition is valid because the reduced Stokes equation is linear.

The bottom-wall shear stress vanishes exactly when $u^{\prime }(0)=0$. Differentiating the velocity profile givesWith this value the velocity profile becomes $u(y)=-G{y}^2/(2\mu )$, so the volume flux per unit transverse width isFor $G>0$, the plate must move against the pressure-driven direction, and the resulting signed volume flux is negative. Zero bottom-wall shear does not mean zero flow.

Under the null hypothesis the three outcome probabilities are the same in the two independent samples. The pooled row totals are $50,20,30$, and both column totals are $50$, out of a total of $100$. Thus the expected counts in each column are $25,10,15$. These exceed the usual small-count threshold for the asymptotic Pearson chi-squared test of homogeneity.

The Pearson chi-squared statistic for contingency tables isUnder the null, its asymptotic chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. Equivalently, the two independent three-category samples have four probability parameters unrestricted and two under the common-probability null. The $1\%$ upper-tail critical value is $9.21$.

Since $9.12<9.21$, do not reject the equal-effect hypothesis at the 1% level. For two degrees of freedom, the approximate p-value is $e^{-9.12/2}\simeq 0.01046$, just above $0.01$. This is a decision at the specified significance level, not evidence that the two outcome distributions have been proved identical.

Introduce nonnegative slack variables $s_1,s_2,s_3$ and start the simplex algorithm at $x=y=z=0$, with slacks $9,8,2$. The objective's largest positive entering coefficient belongs to $y$. Its ratio test is $9/3=3$, $8/2=4$, $2/1=2$, so $s_3$ leaves and $y$ enters.

Solve that pivot row for $y$ and substitute into the other rows and the objective $v$:This is a feasible simplex dictionary at nonbasic variables $x=z=s_3=0$. All their objective coefficients are negative, so no feasible increase can improve the objective. ThusThe negative reduced coefficients also prove uniqueness of this optimum.

The dual linear program, with nonnegative variables $u_1,u_2,u_3$, isThe feasible choice $\boxed{(u_1,u_2,u_3)=(0,0,5)}$ has objective $10$. Weak duality proves that it and the primal point are optimal. Complementary slackness also forces $u_1=u_2=0$ because the first two primal slacks are positive, and then the positive $y$ forces $u_3=5$.

Group the matrix entries by columns. Each column of $BA$ is $B$ times the corresponding column of $A$. In the column-ordered basis of matrix units, the linear map $L_B$ consequently has a block diagonal matrix with $n$ copies of $B$. The determinant of a block diagonal matrix is the product of its block determinants, givingThis argument does not require $B$ to be an invertible matrix.

Right multiplication by $B^{\ast }$ acts separately on the rows. After transposing each row its matrix is $(B^{\ast }{)}^T=\overline{B}$, so its determinant is $(\overline{\det B}{)}^n$. Composing this linear map with left multiplication and using part (a) givesHere the determinant is taken on the $n^2$-dimensional complex vector space of matrices.

If $A$ is a Hermitian matrix, then $(BA{B}^{\ast }{)}^{\ast }=BA{B}^{\ast }$, so $M_B$ preserves $W$. Addition and multiplication by real scalars commute with this operation, making $T_B$ a real linear map.

A real basis of $W$ consists of $E_{ii}$ and, for $i<j$, $E_{ij}+E_{ji}$ and $i(E_{ij}-E_{ji})$. It has $n^2$ elements and is also a complex basis of all matrices. Indeed every matrix is uniquely $H+iK$ with $H,K$ Hermitian matrices, by takingReal linear independence of the displayed basis therefore implies complex linear independence. Since its images under $M_B$ are Hermitian matrices, the matrix of $M_B$ in this complex basis has exactly the same real entries as the matrix of $T_B$ in the real basis. Their determinants are equal. There is no further squaring when passing to this real subspace:This is the determinant of Hermitian congruence, including the case of singular $B$.

Write $|G|=p^{r}m$ with $p$ prime and $p\nmid m$. The Sylow theorems assert that a subgroup of order $p^r$ exists; every p-group subgroup is contained in a conjugate of a Sylow subgroup; all Sylow subgroups are conjugate; and their number $n_p$ satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

For the containment and conjugacy assertions, fix a Sylow subgroup $P$ and let any p-group subgroup $Q$ act by left multiplication on the $m$ left cosets $G/P$. By the orbit-stabilizer theorem, every orbit has size a power of $p$. Since $p\nmid m$, at least one orbit has size one. If $gP$ is such a fixed point, then $qgP=gP$ for all $q\in Q$, equivalently $g^{-1}Qg\subseteq P$. Thus $Q\subseteq gP{g}^{-1}$. If $Q$ too has order $p^r$, this inclusion is equality, proving conjugacy. This is Sylow containment from a coset fixed point.

Since $351=3^3\cdot 13$, the number of Sylow subgroups for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper normal subgroup.

In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot 12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every Sylow subgroup for $p=3$ has order $27$, and all its elements lie among these remaining elements, by Lagrange's theorem. It must therefore be precisely that set. The Sylow subgroup for $p=3$ is unique and hence normal. In either case $G$ is not a simple group.

For the general linear group over ${\mathbb{F}}_{31}$, choose a nonzero first column in $31^2-1$ ways and a second column outside its span in $31^2-31$ ways. ThusSince $5^3=125\equiv 1(\mathrm{mod}31)$ and $5\ne 1$, the residue $5$ has order three. The nine diagonal matrices $\mathrm{diag}(5^i,5^j)$, with $0\le i,j<3$, form a subgroup isomorphic to $C_3\times C_3$. It is an abelian group and a Sylow subgroup for $p=3$. Every other Sylow subgroup for $p=3$ is conjugate to it by the Sylow theorems, hence is also Abelian.

A norm on a real or complex vector space $V$ is a function $\Vert \cdot \Vert :V\to [0,\infty )$ with $\Vert v\Vert =0$ exactly when $v=0$, $\Vert \lambda v\Vert =|\lambda |\Vert v\Vert$, and the triangle inequality $\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$. These are positivity, absolute homogeneity and the triangle inequality.

For a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$, define the operator norm byIt is finite. To see this without assuming continuity of $A$, write $x=\sum x_j{e}_j$: the triangle inequality bounds $\Vert Ax\Vert$ by $\sum |x_j|\Vert A{e}_j\Vert$. Every norm in finite dimensions is equivalent to the Euclidean norm. One proof first uses this same coordinate bound to establish continuity of the norm on the Euclidean unit sphere; compactness then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\Vert x\Vert$.

Absolute homogeneity and the triangle inequality for this operator norm follow from those for $\Vert Ax\Vert$ and taking suprema. If $\Vert A\Vert =0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:{\mathbb{R}}^{\ell }\to {\mathbb{R}}^m$, the defining bound $\Vert Ax\Vert \le \Vert A\Vert \Vert x\Vert$ givesThis is submultiplicativity of the operator norm.

The function $X\mapsto \det (I-X)$ is a polynomial in the matrix entries, hence continuous. Consequently $U$ is the inverse image of the open set $\mathbb{R}\setminus \{0\}$ and is an open set.

If $\Vert X\Vert <1$ and $(I-X)v=0$, then $v=Xv$ implies $\Vert v\Vert \le \Vert X\Vert \Vert v\Vert$. This forces $v=0$. The square matrix $I-X$ is therefore injective, hence an invertible matrix, soThe argument works with any of the induced operator norms in part (b).

Put $r=\Vert X\Vert <1$. By submultiplicativity of the operator norm, $\Vert X^k\Vert \le r^k$, so the Neumann series converges absolutely in the complete finite-dimensional matrix space. Its partial sums satisfyTaking limits gives the inverse and a quantitative remainder bound:To establish two actual Frechet derivatives, use the resolvent identityFor fixed $X\in U$ and sufficiently small $H$, factor $I-X-H=(I-X)(I-f(X)H)$ and apply the Neumann series to the second factor. This proves that $X+H\in U$, that $f(X+H)$ is locally bounded, and that $f(X+H)-f(X)=O(\Vert H\Vert )$. Subtracting $f(X)Hf(X)$ in the identity leaves $O(\Vert H{\Vert }^2)$. Thus $Df(X)[H]=f(X)Hf(X)$ is the Frechet derivative.

Near zero, $f(X)=I+X+O(\Vert X{\Vert }^2)$, uniformly in the operator norm. Hence, as a linear operator in $H$,This proves differentiability of $Df$ at zero, givingThe order of multiplication matters: the second Frechet derivative is the symmetric bilinear map $HK+KH$, not $2HK$.

Let $a=N+\frac{1}{2}$. The useful identity on the square contour isOn the vertical edges the first term equals one; on the horizontal edges $|y|=a$ and $\sinh \pi a>1$. Thus $|1/\sin \pi z|\le 1$ on every edge. Also $|z|\ge a$ and the contour has length $8a$, so the estimation lemma givesAt every nonzero integer $k$, the residue is $(-1{)}^k/(\pi k^2)$. At zero the Laurent series isso the residue there is $\pi /6$. The positive orientation and the residue theorem therefore giveLetting $N\to \infty$ yields the alternating reciprocal-square sum

In the hyperbolic law of cosines, the angle opposite $a$ being at least $\pi /2$ makes its cosine nonpositive. ThusSince $\cosh$ is increasing on $[0,\infty )$, the opposite side is longest: $\boxed{a\ge b,a\ge c}$. For a nondegenerate hyperbolic triangle both comparisons are strict.

At an interior point $w$ of the geodesic segment from $z_1$ to $z_2$, the two angles made with the segment toward $w^{\prime }$ add to $\pi$. At least one is at least $\pi /2$. In the corresponding hyperbolic triangle, part (a) says that its opposite side $w^{\prime }{z}_i$ is at least the adjacent side $w^{\prime }w$. ThereforeIf $w$ is an endpoint the assertion is immediate. The same inequality also holds when $w^{\prime }$ lies on the complete geodesic through $z_1,z_2$: parameterize that geodesic by arclength and use the elementary inequality $|t-s|\le \max (|t-a|,|t-b|)$ for $a\le s\le b$. This extension will allow repeated use of the result in part (c).

Consider the filled hyperbolic triangle with vertices $A,B,C$. Every point $P$ is on some geodesic segment $Aw$ with $w\in BC$: for instance, in the Beltrami-Klein model these are the ordinary straight segments filling a convex triangle. For any other point $Q$ of the triangle, apply part (b) first on $Aw$ and then on $BC$:Applying the same argument to $Q$, with each vertex held fixed, bounds every term on the right by the maximum of the three side lengths. The endpoints of a longest side attain that bound. Thus the diameter is exactly the longest side length:

The arclength element is $ds=\sqrt{1+y^{\prime 2}+z^{\prime 2}}dx$. For uniformly small slopes, the Taylor expansion $\sqrt{1+s}=1+s/2+O(s^2)$ givesThus to quadratic order in the slopes the excess length is $\boxed{\ell -L\simeq \frac{1}{2}{\int }_0^L(y^{\prime 2}+z^{\prime 2})dx}$. This is the small-slope approximation used in the subsequent elastic energy.

Vary $y$ by $y+\epsilon \eta$, where the clamped endpoint conditions require $\eta ={\eta }^{\prime }=0$ at both endpoints. The first variation of the quadratic elastic energy isBoth integrations by parts have zero boundary terms. Since the variation is arbitrary in the interior, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Varying $z$ independently gives the second equation:The clamped endpoint values accompany these equations. If $A$ or $B$ depends on $x$, it must remain inside the derivatives.

With unit constant bending coefficients and unit length, the quadratic elastic energy isTo find the first loss of positivity, put $v=y^{\prime }$. The clamped conditions give $v(0)=v(1)=0$ and ${\int }_0^{1}vdx=0$. Extend $v$ periodically with period one. The sharp periodic Wirtinger inequality gives ${\int }_0^1{v}^{\prime 2}dx\ge 4{\pi }^2{\int }_0^1{v}^{2}dx$. For completeness, expand the mean-zero periodic $v$ in its Fourier series ${\sum }_{n\ge 1}(a_n\cos 2\pi nx+b_n\sin 2\pi nx)$. Parseval identity giveswhich proves the inequality and its equality condition. The same holds for $z^{\prime }$. ConsequentlyFor $F<4{\pi }^2$, the straight filament is the unique minimum. Equality in the periodic Wirtinger inequality has $v=a\cos 2\pi x+b\sin 2\pi x$; the endpoint value $v(0)=0$ eliminates $a$. Integration and the position boundary conditions therefore give the first buckling load and modes:At this load every such pair has zero quadratic elastic energy and solves the Euler-Lagrange equations. For $F>F_c$ these modes give negative energy, signaling instability of the straight configuration. The quadratic small-slope model does not select a finite post-buckling amplitude; nonlinear terms would be needed for that. This is Euler buckling of an elastic filament.

The Neumann boundary conditions select the spatial eigenfunctions $\cos nx$, $n=0,1,\dots$, of the separated wave equation. Indeed separation gives $X^{\prime \prime }+\lambda X=0$ with $X^{\prime }(0)=X^{\prime }(\pi )=0$. Integration by parts gives $\lambda \int X^2=\int X^{\prime 2}\ge 0$. The zero value yields the constant mode; for positive $\lambda$, the first boundary condition removes the sine term and the second requires $\sqrt{\lambda }=n\in \mathbb{N}$. For $n>0$, the temporal factor is $\cos nct$: the sine factor vanishes because the initial velocity is zero. The zero mode has constant displacement, with no term linear in time for the same reason.

The constant Fourier cosine series coefficient is the mean ${\pi }^{-1}{\int }_0^{\pi }bxdx=b\pi /2$. The remaining coefficients areOnly odd modes survive. Thus the displacement isThe series for displacement converges absolutely and uniformly. The initial velocity is zero in the finite-energy sense, and the initial displacement is $bx$. There is an endpoint compatibility qualification: when $b\ne 0$, the initial slope does not satisfy the Neumann boundary conditions at $t=0$. The answer is therefore a finite-energy weak solution, piecewise classical between the propagating corners, rather than a globally twice continuously differentiable solution up to the initial endpoints. The mode expansion implements reflection at those endpoints.