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The relation $Rm=m{R}^{-1}$ implies $m{R}^j=R^{-j}m$. Move each reflection generator to the right in any word, cancel pairs using $m^2=I$, and reduce exponents modulo $N$. Every element then has a normal form $R^j$ or $R^{j}m$, $0\le j<N$.

For the actual symmetry group of a regular polygon these are $N$ distinct rotations and $N$ distinct reflections. The determinant distinguishes rotations from reflections in the planar action. Hence the source's dihedral group $D_N$ has order $2N$, with its subscript counting polygon vertices. The calculations below use that convention.

For a subgroup $H$ of a finite group $G$, Lagrange's theorem statesHere $[G:H]$ is the number of distinct left cosets. Every coset has $|H|$ members, and the cosets partition $G$, proving the identity. In particular the order of a subgroup divides the order of the group. Divisibility is generally necessary, not sufficient, for a subgroup of a given order to exist.

Write the cyclic factors additively. The order of a group element $(r,s)$ in a direct product of groups is the least common multiple of its component orders, since a power is the identity exactly when it is the identity in both coordinates.

In $C_2\times C_3$, $(1,1)$ has order $\mathrm{lcm}(2,3)=6$, so it generates all six pairs. An explicit group isomorphism isIt is well defined and multiplicative in additive notation; its kernel is zero because an integer divisible by both two and three is divisible by six. Equal finite orders then give bijectivity.

In $C_2\times C_4$, every element has order dividing four. A generator of $C_8$ has order eight, and isomorphisms preserve element orders, soFor positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, attained by a pair of generators. The product is cyclic of order $nm$ exactly when that number is $nm$. Thus cyclicity of a product of two finite cyclic groups givesFor coprime factors the map $[k{]}_{nm}\mapsto ([k{]}_n,[k{]}_m)$ gives the isomorphism, the group form of the Chinese remainder theorem.

The identity commutes with every group element, so it lies in the center of a group. If $z,w$ are central and $g\in G$, then $gw=wg$ implies $g{w}^{-1}=w^{-1}g$. ConsequentlyThus $z{w}^{-1}$ is central. The subgroup test provesEquivalently, the centre is nonempty, closed under products, and closed under inverses. This proof does not require finiteness. In fact the centre is an Abelian subgroup, and it is normal because $gz{g}^{-1}=z$ for each central element.

Resolve the initial displacement into the mode shapes from part (i):There is no component along $(1,0,-1)$: the initial displacement is invariant under exchanging beads 1 and 3, whereas that mode is antisymmetric. Zero initial velocities select cosine time dependence. ThereforeHere ${\omega }_1=\sqrt{\tau (2-\sqrt{2})/(m\ell )}$ and ${\omega }_3=\sqrt{\tau (2+\sqrt{2})/(m\ell )}$. At zero time these expressions give $(0,a,0)$ and all three velocities vanish.

The motion is a superposition of the two symmetric normal modes, with opposite signed modal amplitudes. The outer beads always move together. Equivalently, $z_2=a\cos [({\omega }_1+{\omega }_3)t/2]\cos [({\omega }_3-{\omega }_1)t/2]$, displaying the modulation from the two frequencies. The result describes the linearized string while the initial amplitude and subsequent slopes remain small compared with the bead spacing.

The kinetic energy is $T=m{\sum }_i{\dot{z}}_i^2/2$. Differentiating the quadratic potential energy gives the equations for small oscillations,Put $z=v\cos (\omega t+\phi )$ and $\lambda =m\ell {\omega }^2/\tau$. The eigenvalue problem is $Kv=\lambda v$. Its characteristic polynomial isThus a convenient set of normal modes isEach shape can be multiplied by an arbitrary sinusoidal time amplitude. In the lowest mode all beads move in phase. In the middle mode the central bead is fixed and the outer beads move oppositely. In the highest mode the two outer beads move together against the centre. These vectors are mutually orthogonal and span the three displacement coordinates.

These are the three-mass case of the normal modes of a uniformly loaded finite string, with discrete sine shapes $v_i^{(j)}=\sin (ij\pi /4)$. The sketches use straight string segments between masses and fixed zero displacements at the walls. The small-slope and constant-tension assumptions justify the quadratic energy and linear equations.

Direct differentiation givesThe curl is $(0,0,b-a)$, so the preceding convention givesPut $h=(a+b)/2$. The normalized principal axes and eigenvalues of the symmetric tensor $s$ areThe planar axes lie along the two angle bisectors of the coordinate axes. If $h\ne 0$, one planar direction extends and the perpendicular direction contracts, with no change along the third axis. The labels need not be ordered by size when $h<0$. If $a+b=0$, then $s=0$, all principal values vanish and any orthonormal frame is a principal frame; the flow is pure local rotation in the plane.

Write $A_{ij}={\partial }_j{u}_i$. The trace fixes the isotropic part, and splitting the remaining matrix into its symmetric and antisymmetric parts gives the scalar, symmetric-traceless and axial tensor decomposition:Here $s_{ij}=s_{ji}$ and $s_{ii}=0$. The antisymmetric part is $a_{ij}=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Its associated axial vector isIndeed the contraction identity ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$ gives ${ϵ}_{ijk}{\omega }_k=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Adding the three parts reconstructs $A_{ij}$.

The minus sign is required by the index order $A_{ij}={\partial }_j{u}_i$. The symmetric gradient and curl decomposition written instead for ${\partial }_i{u}_j$ has the opposite axial sign. The divergence supplies $3p$, the trace-free symmetric part describes shape change, and the antisymmetric part describes local rotation.

An axial vector transforms under an orthogonal matrix $Q$ asFor a proper rotation, $\det Q=1$, so this is the usual polar vector rule. Under an orientation-reversing change, the extra sign distinguishes the two. In particular, spatial inversion leaves an axial vector unchanged while reversing a polar vector.

Let $C_i=(\nabla \times u{)}_i={ϵ}_{ijk}{\partial }_j{u}_k$. The determinant identity for the Levi-Civita symbol isIt follows by multiplying the three-index determinant identity by $Q_{ir}$ and using orthogonality. The derivative transformation from part (ii) then givesHence the curl of a polar vector field is an axial vector field. The Levi-Civita determinant factor is exactly the extra sign required under reflections.

Under the same Cartesian change, a vector field has $u_i^{\prime }(x^{\prime })=Q_{ip}{u}_p(x)$ with $x=Q^T{x}^{\prime }$. The chain rule gives $\partial /\partial x_j^{\prime }=Q_{jq}\partial /\partial x_q$. Since $Q$ is constant,This is exactly the order-two tensor component transformation law. Thus the Cartesian vector gradient is a Cartesian second-rank tensor, as in the velocity gradient tensor when $u$ is a velocity field.

The restriction to Cartesian transformations matters. In a general position-dependent basis, differentiating the basis contributes extra terms, so the tensorial derivative is the covariant derivative, not an arbitrary array of partial derivatives.

For a constant orthogonal Cartesian change of coordinates $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law for order $n$ isRepeated indices are summed. In Cartesian orthonormal coordinates the metric identifies upper and lower indices, so the same orthogonal matrix acts on every index.

An isotropic tensor has the same numerical components after every proper orthogonal change of axes: $Q^{\otimes n}T=T$ for every $Q$ with $Q^{T}Q=I$ and $\det Q=1$. For example, ${\delta }_{ij}$ is isotropic because $Q_{ik}{Q}_{jk}={\delta }_{ij}$. Some conventions require invariance under all orthogonal transformations, including reflections; this stronger convention must be distinguished for odd-order tensors, notably the Levi-Civita symbol, which is invariant under proper rotations but changes sign under its ordinary three-index transformation for a reflection.

The initial profile is $f(x)=e^{-|x|}$, with the absolute value confirmed in the original PDF. Its Fourier transform isUsing ${\tilde{u}}_t(k,0)=0$ in the regular cosine-sine form giveswith the quotient interpreted as $t$ when $ck=0$. Put $v_0=|c|>0$. Inverting, or applying the D'Alembert formula to $e^{\gamma t}u$, whose initial velocity is $\gamma f$, givesFor an explicit expression at $t\ge 0$, let $r=|x|$ and $s=v_{0}t$. The integral is $2e^{-r}\sinh s$ for $r\ge s$ and $2(1-e^{-s}\cosh r)$ for $r\le s$. ThereforeThe branches agree at $r=v_{0}t$. At zero time, the integral vanishes and the translated profiles coincide, recovering $f$. Their time derivatives cancel, while the integral contributes $\gamma f$, cancelling the derivative of the prefactor, so $u_t(x,0)=0$.

For $c=0$ the limiting answer is $u=(1+\gamma t)e^{-\gamma t}{e}^{-|x|}$. The cusp in $f$ means the solution is understood as a weak solution, classically smooth away from the characteristic fronts $x=\pm v_{0}t$; it is not globally twice differentiable. No assumption that $\gamma$ is positive is needed for this initial-value solution.

The characteristic polynomial is $(r+\gamma {)}^2+c^2{k}^2$, with roots $r=-\gamma \pm ick$. For $ck\ne 0$, the transformed general solution isThe repeated-root case $ck=0$ is $e^{-\gamma t}[A_0+B_{0}t]$, rather than two independent copies of the same exponential. An alternative basis regular at $k=0$ is $e^{-\gamma t}[C(k)\cos (ckt)+D(k)\sin (ckt)/(ck)]$, using the limit $t$ for the last factor at zero.

For $c\ne 0$, the Fourier shift theorem turns the two exponential factors into translations. Equivalently, set $v=e^{\gamma t}u$; then $v_{tt}=c^2{v}_{xx}$. The D'Alembert formula gives the general formwhere $F,G$ are arbitrary functions with the required regularity and decay. The arbitrary profiles can accommodate a spatially transformed zero-frequency velocity by their integral components; the isolated repeated root must not be used to discard that degree of freedom. If $c=0$, the equation instead has $u=e^{-\gamma t}[F(x)+tG(x)]$.

This is the exponentially conjugated wave equation. Positive $\gamma$ produces damping, but the algebra also covers negative $\gamma$, which gives exponential amplification. The sign of $c$ is immaterial because the differential equation contains only $c^2$.

The Fourier transform of $u_{xx}$ is $-k^2\tilde{u}$: two integrations by parts have no boundary terms under the stated decay assumption. Time differentiation commutes with the spatial transform. ThereforeFor every fixed spatial wavenumber $k$, this is a constant-coefficient ordinary differential equation in time.

Use the principal branch of the complex logarithm for $\mathrm{Log}(z+i)$. Its branch cut is the ray $z=-i-s$, $s\ge 0$, lying below the real axis. Thus it is analytic in the upper half-plane, where the only pole of the integrand is $z=i$.

Close the real interval by a large upper semicircle. The estimation lemma bounds the arc integral by $O(\ln R/R)$, so it vanishes. The simple-pole residue isThe residue theorem then givesFor real $x$, $\mathrm{Re}\mathrm{Log}(x+i)=\frac{1}{2}\ln (x^2+1)$. Taking real parts and then using evenness of the real integrand yieldsBoth endpoints converge: the integrand is $O(x^2)$ near zero and $O(\ln x/x^2)$ at infinity. This also agrees with the logarithmic parameter integral with Cauchy weight at parameter one.

Put $d=\sqrt{a^2-b^2}>0$. The two roots areSince $(a-d)(a+d)=b^2$, their moduli are $b/(a+d)<1$ and $(a+d)/b>1$. Thus exactly one root is inside the unit circle, and neither lies on it.

For the real integral use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The change of variables givesThe contour is counterclockwise. At its single interior simple pole, the residue isBy the residue theorem, the reciprocal affine sine integral is thereforeThe condition $a>b>0$ also makes the real denominator strictly positive, consistent with the positive answer and the absence of contour poles on the unit circle.

Fix the evaluation point $r^{\prime }$ inside $V$ and put $\psi (r)=G(r,r^{\prime })$ in Green's second identity. The two $k^2$ terms cancel because ${\nabla }^2\varphi =-k^2\varphi$ and ${\nabla }^{2}G={\delta }^{(3)}-k^{2}G$. The remaining volume integral is $\int \varphi (r){\delta }^{(3)}(r-r^{\prime })dV=\varphi (r^{\prime })$. ThusThe Dirac delta has a positive sign in this problem's Green-function equation, fixing the order of the two terms. All boundary derivatives act on the integration variable $r$.