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For a symmetric positive-definite matrix $H$, the Conjugate gradient method minimizes the error in the induced weighted norm over its affine Krylov subspace. Writing the error as $p(H)e_0$ givesFor ${\lambda }_{\min }<{\lambda }_{\max }$, choose the scaled Chebyshev polynomial $p(\lambda )=T_k(({\lambda }_{\max }+{\lambda }_{\min }-2\lambda )/({\lambda }_{\max }-{\lambda }_{\min }))/T_k(({\lambda }_{\max }+{\lambda }_{\min })/({\lambda }_{\max }-{\lambda }_{\min }))$. On the spectral interval its numerator has absolute value at most one. With $\rho =(\sqrt{\kappa }+1)/(\sqrt{\kappa }-1)$, the denominator equals $({\rho }^k+{\rho }^{-k})/2\ge {\rho }^k/2$. This proves the displayed bound. If the two endpoints coincide, $H$ is a positive scalar multiple of the identity and one iteration solves the system exactly.

With zero initial guess, the generalized minimal residual method minimizes the Euclidean norm of the linear-system residual over each Krylov subspace. The Arnoldi decomposition reduces this to ${\min }_y\Vert {\tilde{H}}_{k}y-\Vert b{\Vert }_2{e}_1{\Vert }_2$, followed by $x_k=Q_{k}y$. Givens rotations solve the small problem in $O(k^2)$ operations once the decomposition is available. For an invertible matrix $A$, the same iterate minimizes the error in the weighted norm induced by $A^{T}A$, because $\Vert Ax-b{\Vert }_2=\Vert x-x_{\ast }{\Vert }_{A^{T}A}$.

Multiplication by an integrating factor givessince $p^{\prime }=ap$ and hence $(p{w}^{\prime }{)}^{\prime }=p(w^{\prime \prime }+a{w}^{\prime })$.

For the singular-endpoint equation, take $p=x^{-1/2}$, $r=x^{1/2}/2$, $q=0$. The indicial equation is $\alpha (2\alpha -3)=0$. The constant branch violates the zero endpoint value unless its coefficient vanishes; the nontrivial admissible branch therefore has $\boxed{\alpha =3/2}$. More precisely, an eigenfunction has $w=C{x}^{3/2}+O(x^{9/2})$ and $w^{\prime }=(3C/2)x^{1/2}+O(x^{7/2})$ near zero.

Integrate the self-adjoint equation after multiplying by $w$, initially on $[a,1]$ with $a>0$:At zero the boundary expression is $O(x^{3/2})$, the first integrand is $O(x^{1/2})$, and the second is $O(x^{7/2})$. Both integrals converge and the boundary term vanishes. Division by the positive weighted norm gives $\boxed{\lambda =I(w)}$.

The change of coordinate $z=x^{3/2}$, $w(x)=W(z)$ gives exactlyThus ${\lambda }_n=9n^2{\pi }^2/2$ and $w_n=\sin (n\pi x^{3/2})$. To prove a minimum over trial functions, rather than just identify the least eigenvalue, transform the quotient itself:The Dirichlet Poincaré inequality gives the last step, with equality precisely for a nonzero multiple of $W=\sin \pi z$. The trial class is the nonzero finite-energy class with zero Sobolev traces; endpoint values alone without an integrability requirement do not define a finite quotient. This is the Power-coordinate reduction of a singular Sturm-Liouville quotient.

Assume the regular real Sturm-Liouville problem, with $p,w>0$, ordered eigenvalues ${\lambda }_0<{\lambda }_1<\cdots$, and real orthonormal eigenfunctions in the weighted inner productThe Rayleigh quotient for an admissible trial function isThe second expression follows by integration by parts. It is the quantity used in the Rayleigh-Ritz method.

For $\tilde{y}=y_0+{\sum }_{n\ge 1}{a}_n{y}_n$, orthonormality eliminates every cross-term in the denominator:Applying $\mathcal{L}$ to the eigenfunction expansion and using the same orthogonality in the numerator givesHence, whenever the trial function has finite energy,Put $S=\sum a_n^2$ and $\Delta =\sum a_n^2({\lambda }_n-{\lambda }_0)$. The exact error and its expansion areIn particular, under a common small scaling $a_n=\epsilon b_n$ with fixed finite sums $\sum b_n^2$ and $\sum b_n^2({\lambda }_n-{\lambda }_0)$,The fourth-order remainder includes mixed products $a_n^2{a}_j^2$, not just individual fourth powers.

Error comparison and sign. Since $\Vert y_0{\Vert }_w=1$, the relative weighted norm error of the supplied trial function isFor ${\lambda }_0\ne 0$, the relative eigenvalue error is $\Delta /(|{\lambda }_0|(1+S))=O({\epsilon }^2)$. Thus the Rayleigh quotient has a second-order error for a first-order error in the trial eigenfunction, provided the energy-weighted coefficient sum is controlled. Small weighted norm error alone does not control arbitrarily high-frequency contamination, because ${\lambda }_n$ can grow without bound. If ${\lambda }_0=0$, an absolute error must replace the undefined relative one.

Finally every summand of $\Delta$ is nonnegative. Thuswith strict inequality whenever some $a_n\ne 0$. This is the upper-bound property of the Rayleigh-Ritz variational principle for the lowest eigenvalue.

Multiplying by $y$ and integrating with the zero endpoint values first gives $\lambda {\int }_1^b{y}^{2}dx/x={\int }_1^{b}x(y^{\prime }{)}^{2}dx>0$ for a nonzero real eigenfunction; hence $\lambda >0$. Using the printed change of variables $\xi =\lambda \ln x$, write $y(x)=Y(\xi )$. Then $(x{y}^{\prime }{)}^{\prime }={\lambda }^2{Y}_{\xi \xi }/x$, so $Y_{\xi \xi }+Y/\lambda =0$. The first endpoint gives $Y=A\sin (\xi /\sqrt{\lambda })$, and the other gives $\sin (\sqrt{\lambda }\ln b)=0$. ThusThis is the logarithmic change of variable in a Sturm-Liouville problem. The measure $dx/x=d(\ln x)$ gives squared weighted norm $A_n^2\ln b/2$, so an orthonormal choice is $A_n=\sqrt{2/\ln b}$.

Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

Only two coefficients in the Sturm-Liouville eigenfunction expansion are nonzero: $a_0=2$ and $a_1=2-\alpha$. Since $\alpha \ne 2$, the result of part (i) simplifies toUsing Parseval's identity, the prescribed weighted norm becomesThus $1-\alpha =\pm 2$, so $\alpha =-1$ or $3$. Positivity selectsThe standing nonresonance assumption still applies: this answer requires that $3$ is not one of the unspecified higher eigenvalues. Their actual values are not supplied. If a higher eigenvalue were $3$, the stated exclusion would fail and the data would not have an admissible $\alpha$ under that exclusion.

For a nonzero admissible radial profile define the radial diffusion Rayleigh quotientMinimizing the numerator subject to ${\int }_0^{1}r{R}^{2}dr=1$ introduces a multiplier $\nu$. The Euler-Lagrange equation for $r\kappa R^{\prime 2}-\nu r{R}^2$ isThe allowed variations vanish at $r=1$, and the natural condition at the centre is zero radial flux. Thus the minimum, when attained, is the smallest decay eigenvalue ${\nu }_0$. Equivalently, expansion in weighted-orthogonal eigenfunctions makes $\mathcal{R}[R]$ a weighted average of their decay eigenvalues. The Rayleigh-Ritz variational principle therefore givesAdmissible trial functions must obey the outer boundary condition, be nonzero and have finite weighted norm and energy; they need not solve the differential equation. If the diffusion operator has no lowest eigenfunction, the same variational argument bounds the infimum of its spectrum. Existence of a fundamental diffusion mode is an additional assumption for strongly vanishing diffusivity, not a consequence of the identity alone.

For the proposed trial profile, $R^{\prime }=-q{r}^{q-1}$. Its weighted energy isThe squared weighted norm isBoth are finite for $q>0$ and $n\ge 0$, and $R(1)=0$. Therefore the Rayleigh-Ritz variational principle givesWhen a fundamental diffusion mode exists, its decay rate is the infimum and is bounded above by this expression.

The existence qualification matters for the full range of $n$ printed in the question. For $n>2$, trial functions concentrated within radius $\epsilon$ have quotient proportional to ${\epsilon }^{n-2}$, proving the loss of a spectral gap under strongly degenerate radial diffusion. The infimum is zero and is not attained, because zero energy together with $R(1)=0$ forces the zero function. Thus there is no positive lowest-mode decay rate in that case. Even at $n=2$, the mode equation is $r^2{R}^{\prime \prime }+3r{R}^{\prime }+\nu R=0$: its characteristic powers are $r^{-1\pm \sqrt{1-\nu }}$, and for $\nu >0$ every nonzero solution is unbounded at the centre. Hence the asserted bounded fundamental mode cannot be assumed for all $n\ge 0$. The trial quotient remains valid, and the next part's $n=0$ case has the usual fundamental mode.

For $n=0$, a constant solution and its normalization give $y_0=1$. For $n=1$, substitute $y=ax+b$: the equation gives $b=0$, and the normalization gives $a=1$. For $n=2$, put $y=a{x}^2+bx+c$. Comparing coefficients gives $b=0$ and $c=-a/2$; then $y(1)=1$ fixes $a=2$. ThusThese are the first three Chebyshev polynomials. For their squared weighted norms, use $x=\cos \theta$, which converts $dx/\sqrt{1-x^2}$ into $-d\theta$:More generally, $y_n(\cos \theta )=\cos (n\theta )$ makes their orthogonality the usual cosine orthogonality.

The coefficient formula from part (b) and the Leibniz rule yield Rodrigues' formula directly:The leading coefficient is $(-1{)}^n/n!$, so $p_n^{(n)}=(-1{)}^n$. Substitute this differentiated expression into one factor in the squared weighted norm and apply integration by parts $n$ times:Every endpoint term contains a derivative of $e^{-x}{x}^n$ of order less than $n$: it vanishes at zero, and its exponential decay beats every polynomial at infinity. This proves the normalization of the ordinary Laguerre polynomials for all $n$, including $n=0$ without any integrations by parts.

Write ${\mathcal{P}}_n$ for the vector space of polynomials of polynomial degree at most $n$, and put $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$. As the question calls the weighted integral an inner product, its finiteness is implicit; for example, a positive integrable weight $w$ suffices for all the continuous functions involved here.

For every $q\in {\mathcal{P}}_n$, differentiating the squared weighted norm $\Vert f-p_n-tq{\Vert }_w^2$ at its minimum $t=0$ gives the orthogonality equationsIn particular, the requested linear system isIts Gram matrix $G_{kj}={\int }_{-1}^{1}w(x)x^{k+j}dx$ is a symmetric positive-definite matrix: for any nonzero vector $a$,Indeed a nonzero polynomial is nonzero on some open interval, and the weight is positive. Therefore $G$ is an invertible matrix, proving uniqueness. Conversely, if these equations hold, the Pythagorean theorem in an inner-product space gives, for every $q\in {\mathcal{P}}_n$,so the solution really is the unique best approximation.

The claim is true, withIt is a symmetric positive-definite matrix: for every nonzero vector $z$, the fact that $A$ is an invertible matrix gives $z^{T}Mz=\Vert Az{\Vert }_2^2>0$. Since $A{x}_{\ast }=b$, every candidate $x$ satisfiesThe generalized minimal residual method therefore minimizes precisely this weighted norm of the error over the linear span of $Q_k$. The same matrix $M$ works for every $k$ and does not depend on $b$.

Under the permitted singular value bounds, $m=4n$ givesFor the symmetric positive-definite matrix $H=G^{T}G$ in part (iii), its eigenvalues are the squared singular values of $G$, soApply the supplied conjugate gradient convergence bound to $H$. For $e_k=x_k-x_{\ast }$ and nonzero $e_0$, the function $(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$ increases with $\kappa \ge 1$, givingThus the error in this weighted norm contracts at a guaranteed exponential factor of at most $1/2$ per iteration, up to the prefactor $2$. Here $\Vert e{\Vert }_H=\Vert Ge{\Vert }_2$. The Pythagorean identity in part (iii) also identifies $\Vert e_k{\Vert }_H^2$ with the excess squared linear least-squares problem objective, which consequently decreases by a bound proportional to $4^{-k}$. For the Euclidean norm, the same eigenvalue bounds give $\Vert e_k{\Vert }_2/\Vert e_0{\Vert }_2\le 62^{-k}$.

For relative accuracy $0<ϵ<1$ in the $H$-weighted norm, it is sufficient to takeCompute $c=G^{T}b$ once in $O(mn)$ operations. Each iteration applies $p\mapsto G^T(Gp)$ with two dense matrix multiplications and uses $O(n)$ further arithmetic operations; when $m=4n$ this costs $O(n^2)$. Hence the total cost isUsing finite termination of the conjugate gradient method in exact arithmetic refines this to $O(n^2\min \{n,\lceil {\log }_2(2/ϵ)\rceil \})$. Explicitly constructing $G^{T}G$ first would add $O(m{n}^2)=O(n^3)$ setup work and is unnecessary. These accuracy estimates concern relative error; an absolute tolerance replaces $2/ϵ$ by $2\Vert e_0{\Vert }_H/ϵ$ when $\Vert e_0{\Vert }_H>ϵ$. If $e_0=0$, the initial guess already solves the problem.

As in part 2(b)(v), the Gaussian assumption alone does not make the printed singular value interval a certainty. For example, the event that every entry of $G$ has absolute value below a sufficiently small positive number has positive probability and forces $\Vert G{\Vert }_F<\sqrt{n}$, contradicting the asserted lower endpoint. The displayed rate is therefore conditional on the singular-value bounds authorized by the paper. The cost and convergence derivation are exact under that condition; an unconditional probabilistic statement would need an additional failure-probability estimate.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.