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With $x=\cos \theta$, the weight becomes ${\sin }^2\theta d\theta$. Sincethe sine orthogonality relations show that $P_n$ is a scalar multiple of the Chebyshev polynomial of the second kind $U_n$. Its roots areThe printed normalization $P_n(0)=1$ is possible only for even $n$; then $U_n(0)=(-1{)}^{n/2}$ andFor odd $n$, every such orthogonal polynomial is odd and vanishes at zero, so that normalization is inconsistent.

Write $f=Q_{2n-1}+E_n$, where $Q_{2n-1}$ contains the terms through degree $2n-1$. On $[-1,1]$,The quadrature error of $Q_{2n-1}$ is zero. Bounding the integral and quadrature errors of $E_n$ separately, using positive weights with total $\int r$, gives exactly

Given $ϵ>0$, the Weierstrass approximation theorem supplies a polynomial $Q$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For all sufficiently large $n$, the quadrature is exact on $Q$. Positivity of the weights and ${\sum }_j{A}_j=\int r$ then giveLetting $ϵ\downarrow 0$ proves convergence.

Let ${\ell }_j$ be the Lagrange interpolation polynomial for the roots ${\omega }_j$, and defineFor $\mathrm{deg}Q\le 2n-1$, polynomial division gives $Q=S{P}_n+R$ with $\mathrm{deg}S,\mathrm{deg}R<n$. Orthogonality kills the first term, while interpolation gives $R={\sum }_{j}R({\omega }_j){\ell }_j$. ThereforeTaking $Q={\ell }_j$ proves uniqueness. Taking $Q={\ell }_j^2$ also showsThis is Gaussian quadrature.

Let $x_1,\dots ,x_k$ be the distinct zeros in $(-1,1)$ at which $P_n$ changes sign, and put $Q(x)={\prod }_{j=1}^k(x-x_j)$. Then $P_{n}Q$ has constant sign and is nonzero except at finitely many points, soIf $k<n$, this contradicts the orthogonality of $P_n$ to the lower-degree polynomial $Q$. Hence $k\ge n$; degree $n$ forces exactly $n$ distinct, simple roots, all in $(-1,1)$.

Apply the Gram-Schmidt process to $1,x,\dots ,x^n$. The last resulting vector $P_n$ is nonzero, has degree exactly $n$, and is orthogonal to every polynomial in $\mathrm{span}(1,x,\dots ,x^{n-1})$.

Bilinearity and symmetry are immediate. Since $r\ge 0$, $⟨u,u⟩\ge 0$. If this is zero, continuity of $u^{2}r$ makes it identically zero. The points where $r>0$ are dense because only finitely many are excluded, so $u$ vanishes on a dense set and hence everywhere. Thus the form is an inner product.

Insertinto the continuity equation. Its zeroth-order part is the homogeneous cosmological continuity equationSince $\bar{\rho }$ is spatially homogeneous, the terms of first order in $ϵ$ giveOn the other hand, the quotient rule and the background equation implySubstitution of the first-order equation yields the linearized cosmological continuity equationThe ratio $\delta \rho /\bar{\rho }$ is the density contrast.

The elementary Newtonian derivation of the Friedmann equation follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the Euler equation and Poisson equation must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same Hubble flow, without selecting a physical centre. This is the improvement captured by the Newtonian fluid derivation of the Raychaudhuri equation.

In the homogeneous background, $\mathbf{v}=0$ and the physical velocity is the Hubble flowThe comoving coordinate $\mathbf{x}$ is held fixed by the time derivative in the fluid equation, soMoreover, spatial homogeneity gives $\nabla \bar{P}=0$. The Euler equation therefore reduces toTaking the divergence in three spatial dimensions givesConsistency with the Poisson equation,then requiresThis is the pressureless, zero-cosmological-constant Raychaudhuri equation, obtained by the Newtonian fluid derivation of the Raychaudhuri equation.

The Euler--Lagrange equations for the quadratic Lagrangian areFor a normal mode $x(t)=a{e}^{i\omega t}$, they become the generalized eigenvalue problem for small oscillationsPut $\lambda =m{\omega }^2/k$. With the stated spring constants,The final matrix has characteristic polynomialso its eigenvalue are ${\nu }_{\pm }=\pm \sqrt{1+{\delta }^2}$. HenceThe two angular frequencies therefore satisfyThis is the spectrum recorded by normal modes of two equal masses between three springs.

The kinetic energy of the two equal masses isThe extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their potential energy is thereforeThus the mechanical Lagrangian $L=T-U$ has the requested quadratic form with the symmetric small-oscillation mass and stiffness matrices

False under the hypotheses as printed. Nonzero coefficients do not ensure the two-column dominant-subspace condition.

For a counterexample, take $n=3$, eigenvaluesand an orthonormal eigenbasis $w_1,w_2,w_3$ for whichThe two displayed coefficient vectors are orthonormal and can be completed to an orthogonal matrix, so such a real symmetric matrix $A$ exists. Every $b_i$ and $c_i$ is nonzero, butIndeed, with $t=1/\sqrt{2}$,is a nonzero multiple of $w_1$, while the remaining direction in
$\mathrm{span}(A^k{e}_1,A^k{e}_2)$ isThusBy the simultaneous iteration interpretation of the QR algorithm, the leading two columns of the accumulated $Q$ factor span this same space. The leading block $B_k$ is therefore an orthogonal-coordinate representation of the compression of $A$ to this space. Sincewe havefor every relevant $k$, rather than $\{-2,2\}$. Hence the requested Hausdorff convergence does not hold.

For completeness, the intended statement becomes true if one addsLet $Z_k$ consist of the first two columns of ${\overline{Q}}_{k-1}$. The Accumulated QR factorization identity givesAfter division by $|{\lambda }_n{|}^k$, every component along
$w_1,\dots ,w_{n-2}$ tends to zero because of the strict spectral gap, while $\Delta \ne 0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the dominant invariant subspaceThere are two-by-two orthogonal matrices $U_k$ such thatBecauseup to the harmless index convention at $k=0$, we obtainin the matrix 2-norm. Orthogonal similarity preserves the spectrum, and the stated spectral perturbation bound now gives

Each QR decomposition $A_k=Q_k{R}_k$ givesInduction therefore yields the first Accumulated QR factorization identity:

We next prove the corresponding factorization of a matrix power. The case $k=0$ is just $A=Q_0{R}_0$. Ifthen the similarity formula above impliesConsequently,The product ${\overline{Q}}_k$ is orthogonal, and the product ${\overline{R}}_k$ is upper triangular. Henceis a QR decomposition.

At larger frequencies the denominatorcan vanish. Its zeros satisfyand are the normal-mode frequencies for a pressure node at the outer sphere and zero radial velocity at the inner sphere. Near one of these radial acoustic resonances, the ideal undamped harmonic response becomes very large; exactly at resonance the linear lossless initial-value problem develops secular growth rather than a bounded periodic solution.

In a physical system, fluid viscosity, thermal losses, acoustic radiation, and damping in the elastic shell keep the amplitude finite and shift its phase. A nonzero pressure-velocity phase component then supplies positive mean power. At sufficiently large amplitude the linear acoustic approximation can also fail, leading to nonlinear frequency shifts or shock formation.