For nonzero $y$ with $G[y]>0$, the first variation of the Rayleigh quotient under $y\mapsto y+\epsilon h$ isChoose fixed admissible homogeneous boundary conditions that make the variation boundary term vanish, such as zero endpoint values or zero endpoint derivatives. Compactly supported $h$ then show that a stationary function satisfies the Sturm-Liouville equation with $\lambda =\Lambda [y]$. Conversely an eigenfunction satisfying those admissible conditions makes the variation zero. Multiplying its differential equation by $y$ and integrating by parts givesso the stipulated energy boundary term yields $\Lambda [y]=\lambda$. The displayed energy condition alone is not a full specification of the admissible variations; it must be paired with boundary conditions for this variational equivalence.

For the final Dirichlet problem take the trial function $y=\sin x$, with $p=1,q=0,w=x$. ThenThe Rayleigh-Ritz variational principle makes this an upper bound for the lowest eigenvalue. It is strictly above the true value: equality would require $\sin x$ itself to be a lowest eigenfunction, but $y^{\prime \prime }+\lambda xy=(-1+\lambda x)\sin x$ cannot vanish throughout $(0,\pi )$ for a constant $\lambda$. ThereforeAlthough the weight vanishes at the left endpoint, it is positive in the interior and its weighted quadratic norm is positive for every nonzero admissible continuous function; this does not obstruct the stated Dirichlet variational comparison.

A group homomorphism satisfies $\phi (g_1{g}_2)=\phi (g_1)\phi (g_2)$; a bijective one is a group isomorphism. Its kernel of a group homomorphism is $K=\{g:\phi (g)=1_H\}$. It is a subgroup, and for $k\in K$, $g\in G$,Thus $gK{g}^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a normal subgroup.

Write $B_n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$. Direct matrix multiplication gives $B_n{B}_m=B_{n+m}$, $B_0=I$ and $B_n^{-1}=B_{-n}$. Consequently these matrices form the integer shear group, a subgroup of $GL(2,\mathbb{R})$. It is an infinite cyclic group, generated by $B_1$, since $B_1^n=B_n$ and no nonzero integer power is the identity.

The map $n\mapsto B_n$ is a bijective group homomorphism from the additive group of integers, proving the required group isomorphism. There are proper subgroups: for example $\{B_{2n}:n\in \mathbb{Z}\}$ omits $B_1$. In fact every subgroup is $\{B_{mn}:n\in \mathbb{Z}\}$ for an integer $m\ge 0$, by the classification of subgroups of an infinite cyclic group; $m=0$ gives the trivial subgroup and $m=1$ the whole group.

Lagrange's theorem states that the group order of a finite group divides the group order; more precisely $|G|=[G:H]|H|$.

If every element squares to the identity element, every element is its own inverse element, and $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a Klein four-group subgroup of order four.

Now let $|G|=2p$ with odd prime number $p$. By Lagrange's theorem, every element order is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid 2p$. Hence there is an element of order $p$.

For a conservative system near a stable equilibrium, the quadratic Lagrangian has the form $L=\frac{1}{2}{\dot{q}}^{T}M\dot{q}-\frac{1}{2}{q}^{T}Kq$, with symmetric mass matrix $M$ and stiffness matrix $K$. A normal mode $q=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem for small oscillations $(K-{\omega }^{2}M)v=0$, so the frequency equation is $\det (K-{\omega }^{2}M)=0$.

To quadratic order, vertical velocities contribute only fourth-order corrections to the kinetic energy, leaving $T=({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)/2$. A string of length $\ell$ with small horizontal difference $d$ shortens its vertical projection by $d^2/(2\ell )$. The upper segment carries all three masses, the middle segment two, and the lower segment one. Hence their increases in gravitational potential energy sum toSubtracting $V$ from $T$ gives the stated approximate Lagrangian. Here $M=I$ andFor a mode with middle displacement zero, write $v=(X,0,Z{)}^T$. The middle row requires $X/b+Z/c=0$, so $Z=-cX/b$. A nonzero mode then has $X,Z\ne 0$, and the outer rows require ${\omega }^2/g=a^{-1}+b^{-1}=c^{-1}$. Therefore the necessary and sufficient condition isWhen it holds, the mode shape is $(1,0,-c/b)$ with $\omega =\sqrt{g/c}$. This is the force-cancellation mechanism of the stationary middle mass in a three-mass spring chain, applied to the equivalent quadratic stiffnesses here.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.

The residue theorem states that the contour integral of a meromorphic function around a positively oriented simple closed contour equals $2\pi i$ times the sum of its enclosed residues, provided there are no singularities on the contour. The roots of $1+z^n$ are $z_k=e^{(2k+1)i\pi /n}$, $k=0,\dots ,n-1$. They are simple poles, with residues $1/(n{z}_k^{n-1})=-z_k/n$.

Use the sector contour of angle $2\pi /n$, running outward along the positive axis and inward along the rotated ray. It encloses only $z_0=e^{i\pi /n}$. The large-arc contribution is $O(R^{1-n})$, hence vanishes. With $J={\int }_0^{\infty }dx/(1+x^n)$, the two ray contributions giveSince $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$,For the second integral, use $z^{n/2}/(1+z^n)$ on a branch continuous across the sector away from zero, with a small inner arc of radius $ϵ$. Its large arc is $O(R^{1-n/2})$ and its small arc is $O({ϵ}^{1+n/2})$, both vanishing for $n\ge 3$. On the rotated ray, $z^{n/2}$ changes by $e^{i\pi }=-1$, so the two radial contributions add rather than subtract. The enclosed residue is $z_0^{1-n/2}/n=-i{e}^{i\pi /n}/n$. ConsequentlyBoth answers are instances of the Mellin integral of a reciprocal binomial. The direct contour argument also tracks the branch needed when $n$ is odd.