Past exam of the mathematical methods I course of the University of Oxford / 2017 (A10285W1, CP3)

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Since $(1+i{)}^2=2i$, the two square roots are $1+i$ and $-1-i$. The principal square root is $1+i$, whose real part isIf the notation is meant to include both square-root values, their real parts are $+1$ and $-1$. Stating the principal-branch convention removes that ambiguity.

For $\delta \notin 2\pi \mathbb{Z}$, the finite geometric series givesThe exponential prefactor has modulus one. ThusAt an excluded phase every summand is one, which supplies the removable limiting value directly. Here $N$ is a positive integer.

From the exponential definition, $\cos (iu)=\cosh u$. ThereforeA cosine of a nonreal argument need not lie between $-1$ and $1$; that bound applies to real arguments.

An integrating factor for $u^{\prime }+P(x)u=Q(x)$ is a nonzero function $\mu (x)=\exp (\int P(x)dx)$, chosen so that $(\mu u{)}^{\prime }=\mu Q$. Integrating this total derivative reduces the first-order linear ordinary differential equation to one quadrature.

For the given nonlinear equation, let $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. Multiplication by $2y$ gives $u^{\prime }+u=x^2+2x$. Using the integrating factor $e^x$, or recognizing the particular solution $u=x^2$, yields $u=x^2+C{e}^{-x}$. The initial value gives $C=1$ and selects the positive branch:The radicand is strictly positive for every real $x$. This solution never reaches the singular denominator $y=0$ of the original equation, so multiplication by $y$ has introduced no forbidden crossing. The negative branch would violate $y(0)=1$.

Let $Hv=\lambda v$ with $v\ne 0$ and $H^{\ast }=H$. Then $v^{\ast }Hv=\lambda v^{\ast }v$, while taking the complex conjugate gives $v^{\ast }Hv=\overline{\lambda }{v}^{\ast }v$. Since $v^{\ast }v>0$, every eigenvalue of a Hermitian matrix is real.

If $Hv=\lambda v$ and $Hw=\mu w$, thenDistinct eigenvalues therefore force $v^{\ast }w=0$: their eigenvectors are orthogonal in the Hermitian inner product. A degenerate eigenvalue occurs more than once in the spectrum, counted with multiplicity. For a Hermitian matrix, its algebraic multiplicity equals its geometric multiplicity, so degeneracy gives an eigenspace of dimension greater than one; an orthonormal basis can be chosen within it.

A unitary matrix preserves the Hermitian inner product and hence the norm. If $Uv=\lambda v$ with $v\ne 0$, thenDividing by the positive squared norm givesUnlike the Hermitian case, this constrains eigenvalues to the unit circle, not necessarily the real axis.

Subtracting the second equation from the first gives $2x-\lambda z=-4$. With $\lambda =2$ this is $x-z=-2$, whereas the third equation requires $x-z=-2\mu$. Consistency therefore requiresAt that value one parameter remains free:Thus the solution is not unique. For any other $\mu$ the two required values of $x-z$ conflict and the linear system has no solution.

For $\lambda =1$ and $\mu =1/2$, the third equation gives $x=z-1$. Combining it with $2x-z=-4$ gives $z=-2$ and $x=-3$, after which the second equation gives $y=12$. HenceThe coefficient matrix has determinant $\lambda -2=-1\ne 0$, proving uniqueness. Direct substitution gives the required right-hand sides $1$, $5$ and $-1$.

For a unit vector, the radial coordinate is $r=1$ and the spherical coordinates giveThe Pauli matrices therefore giveThe diagonal entries are real and the off-diagonal entries are complex conjugates, so $H^{\ast }=H$. Its trace is zero and its determinant is $-1$, giving eigenvalues $+1$ and $-1$. A convenient orthonormal eigenbasis isThe half-angle identities verify $H{v}_{\pm }=\pm v_{\pm }$, $v_{\pm }^{\ast }{v}_{\pm }=1$ and $v_{+}^{\ast }{v}_{-}=0$. Multiplication of either vector by any unit-modulus phase gives an equivalent normalized eigenvector. At a pole, the azimuth is undefined geometrically, but the remaining phase freedom makes the chosen expressions consistent with the same one-dimensional eigenspaces.

With the Einstein summation convention, the magnetic force has components $F_i=q{ϵ}_{ijk}{v}_j{B}_k$. Thus its matrix acting on velocity isThe Levi-Civita symbol changes sign on exchanging $i,j$, so $M^T=-M$; the diagonal vanishes and the matrix has zero trace. For nonzero $\mathbf{B}$, $M\mathbf{B}=q\mathbf{B}\times \mathbf{B}=0$, exhibiting a nontrivial kernel and proving noninvertibility. If $\mathbf{B}=0$ or $q=0$, $M$ is the zero matrix and is again noninvertible.

Equivalently, an odd-dimensional real antisymmetric matrix has $\det M=\det M^T=\det (-M)=-\det M$, hence zero determinant. The same antisymmetry gives ${\mathbf{v}}^{T}M\mathbf{v}=0$, expressing that the magnetic force does no work.

For $x>0$, use the Euler differential equation substitution $t=\ln x$, so $x=e^t$, $x{y}^{\prime }=\dot{y}$ and $x^2{y}^{\prime \prime }=\ddot{y}-\dot{y}$. The transformed equation isIts characteristic polynomial factors as $(r-1)(r+2)$. The forcing $e^t$ coincides with a homogeneous solution, so the method of undetermined coefficients requires a particular solution $y_p=At{e}^t$. Substitution gives $3A{e}^t=e^t$, hence $A=1/3$. Transforming back,The conditions at $x=1$ give $C_1+C_2=0$ and $C_1-2C_2+1/3=0$, soDifferentiating confirms both initial values and the original equation. The solution interval containing $x=1$ excludes the singular point $x=0$.

Using the complex forcing method for a real differential equation, $\mathrm{Im}[(4x+2i)e^{ix}]=4x\sin x+2\cos x$, a suitable complex forcing is $f(x)=(4x+2i)e^{ix}$. Solve $u^{\prime \prime }+2u=f$ and take its imaginary part. Writing $u=p(x)e^{ix}$ reduces the equation toA linear polynomial $p=ax+b$ gives $a=4$ and $b=-6i$. ThusFor a direct check, $(4x\sin x{)}^{\prime \prime }+2(4x\sin x)=8\cos x+4x\sin x$, while $(-6\cos x{)}^{\prime \prime }+2(-6\cos x)=-6\cos x$. Their sum is the required forcing. Adding $A\cos (\sqrt{2}x)+B\sin (\sqrt{2}x)$ gives the full real solution if needed.

Let the columns of the given unitary matrix be $u_j$, with $M{u}_j={\lambda }_j{u}_j$. Columnwise, $MU=UD$ for $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Multiplication by $U^{\ast }$ and unitarity giveThe premise includes an orthonormal eigenbasis. It is not true that an arbitrary matrix admits such a basis; over $\mathbb{C}$ it exists precisely for a normal matrix. The real symmetric inertia matrix in part (b) does satisfy the required condition.

In the quadratic form, the two equal off-diagonal contributions together supply the mixed term. Comparing coefficients givesFor a nontrivial physical moment of inertia scale take $I_0>0$; then this inertia tensor is positive definite.

The normalized vectors $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(-1,1{)}^T/\sqrt{2}$ give eigenvalues $I_0$ and $3I_0$. ThereforeThe columns are orthonormal, so $U^{-1}=U^T=U^{\ast }$ and direct multiplication verifies $U^{-1}MU=D$. Signs of individual eigenvectors are free; the displayed signs also make $\det U=+1$.

The matrix has the standard rotation matrix form $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$, with $\cos \theta =\sin \theta =1/\sqrt{2}$. ThusWith $\omega =U{\omega }^{\prime }$, the new first axis points along the original $(1,1)$ direction and is counterclockwise from the original first axis. Specifying this coordinate convention avoids confusing rotation of the axes with the inverse transformation of component columns.

Substituting $\omega =U{\omega }^{\prime }$ and using the diagonalization givesFor $I_0>0$ and energy level one in the chosen units, its ellipse equation isThe major semiaxis has length $a=\sqrt{2/I_0}$ along $u_1=(1,1)/\sqrt{2}$; the minor semiaxis has length $b=\sqrt{2/(3I_0)}$ along $u_2=(-1,1)/\sqrt{2}$. The full axis lengths are $2a$ and $2b$. The original-coordinate curve is the same ellipse tilted by $45^{\circ }$, elongated along ${\omega }_y={\omega }_x$; an orthogonal coordinate change preserves lengths and does not change the ellipse itself.

A map between vector spaces over the same field is linear when $L(av+bw)=aL(v)+bL(w)$ for every $v,w$ and scalars $a,b$. Its kernel is $\ker L=\{v\in V:L(v)=0\}$. It contains zero because $L(0)=0$. If $v,w\in \ker L$, thenThus it is closed under linear combinations and is a vector subspace of $V$. No injectivity or equality of the dimensions of $V,W$ is needed.

Write both cross products with the Levi-Civita symbol. Its contraction identity givesTherefore the vector triple product identity isThe order matters: the cross product is not associative, and exchanging the two nested arrangements changes the expression.

The vector triple product with $|u|=1$ givesThis is the orthogonal projection onto a plane perpendicular to $u$, so it is a linear map. Its image is the plane $u^{\perp }$ through the origin, while its kernel is the normal line $\mathrm{span}\{u\}$. In components,The equation $Ar=r$ is equivalent to $u(u\cdot r)=0$, hence to $u\cdot r=0$. Its solution set is precisely the two-dimensional plane $u^{\perp }$, with infinitely many nonzero vectors, not merely the normal line. These are the eigenvectors with eigenvalue one; the normal direction has eigenvalue zero.

For two nonparallel vectors $v_1,v_2\in {\mathbb{R}}^2$, a nontrivial relation $a{v}_1+b{v}_2=0$ would make one a scalar multiple of the other. They are therefore linearly independent. Two independent vectors in a two-dimensional vector space form a basis.

Likewise, three vectors in ${\mathbb{R}}^3$ that are not contained in one plane through the origin cannot obey a nontrivial linear dependence: such a relation would express one as a combination of the other two, or already make that pair dependent. Thus they are independent and form a basis of ${\mathbb{R}}^3$. The plane condition here is linear coplanarity with the origin, as required for vectors, rather than affine coplanarity of three arbitrary points.

Let $\Delta =v_1\cdot (v_2\times v_3)\ne 0$, since the original vectors form a basis. The reciprocal basis isTaking the dot product with each $v_j$ gives $u^i\cdot v_j={\delta }_{ij}$, using orthogonality of a cross product and cyclic symmetry of the scalar triple product. If ${\sum }_i{a}_i{u}^i=0$, taking its dot product with $v_j$ yields $a_j=0$. The three reciprocal vectors are therefore linearly independent and form a basis of ${\mathbb{R}}^3$.

Equivalently, if $V$ has columns $v_j$ and $U$ columns $u^i$, then $U^{T}V=I$, so $U=V^{-T}$. In particular any vector has coordinates $r={\sum }_i(r\cdot u^i)v_i$, even when the original basis is not orthonormal.

Provided $p,q$ are linearly independent, their linear combinations fill a two-dimensional direction space. Adding $r_0$ translates it to the plane through $r_0$. The unit normal and constant areThen $n\cdot (r_0+t_{1}p+t_{2}q)=c_0$ for every pair of parameters. Conversely a point satisfying $n\cdot r=c_0$ differs from $r_0$ by a vector in $n^{\perp }=\mathrm{span}\{p,q\}$, proving that the two descriptions give exactly the same plane.

For the three points, choose $r_0=(3,1,-2)$, $p=(-4,1,6)$ and $q=(-1,-2,3)$. Their cross product is $(15,6,9)=3(5,2,3)$, which is nonzero. The plane therefore has equationEach supplied point satisfies it. Its unit normal is $(5,2,3)/\sqrt{38}$, up to an overall sign, and $c_0=11/\sqrt{38}$.

Let $s=(x_0,y_0,z_0)$ be the sphere centre. The circle centre is its perpendicular projection onto the plane:For the specific plane this isFor a point on the section, its displacement from $h$ lies in the plane and is perpendicular to $s-h$. The Pythagorean theorem therefore givesA genuine circle requires $d<R$. At $d=R$ the section is a single tangent point, and at $d>R$ the plane misses the sphere. A central cut $d=0$ gives a great circle of radius $R$.