At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.

Define the Wronskian by $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. Differentiating and using the normal-form differential equation givesIntegrating this first-order linear differential equation proves Abel's identity:Thus, on any interval avoiding zero and $m$,Absolute-value signs from real logarithms can be absorbed into the constant on each such interval. For linearly independent solutions, $C\ne 0$.

Set $t=x-m$ and use the Frobenius solution $y=t^{\nu }(a_0+a_{1}t+\cdots )$ with $a_0\ne 0$. From part (a), the leading coefficients are $P\sim 1/t$ and $Q\sim m^2{\omega }^2/t^2$. The coefficient of the lowest power $t^{\nu -2}$ is thereforeThe indicial equation reduces to ${\nu }^2+m^2{\omega }^2=0$, givingSince $m\omega >0$, the two indicial roots differ by a nonzero imaginary number, not an integer. Thus the two local Frobenius solutions exist independently on a chosen branch of $\log (x-m)$.

First simplify the coefficient of $y$ usingAfter dividing by $x(x-m)$, the second-order linear differential equation isAt zero, $xP\to 1$ and $x^{2}Q\to 0$, and both products are analytic there. At $m$, $(x-m)P\to 1$ and $(x-m{)}^{2}Q\to m^2{\omega }^2$, again with analytic extensions. Thus $\boxed{x=0\text{ and }x=m\text{ are regular singular points}}$, and there are no other finite singular points.

Infinity must also be considered because the question asks for all singular points. Put $t=1/x$ and $Y(t)=y(1/t)$. The coefficient of $Y$ in its normal-form equation is $Q(1/t)/t^4\sim {\omega }^2/t^4$. Multiplication by $t^2$ does not make it analytic at zero. Hence $\boxed{x=\infty \text{ is an irregular singular point}}$.

The orthogonal change of coordinates preserves distance to the origin. On the required quadric surface,Thus the squared distance is at least $1/4$. Equality holds precisely when $u=v=0$, and the constraint then gives $w=\pm 1/2$. Using $e_w=(1,0,-3)/\sqrt{10}$ from part (b), the nearest points in the original coordinates areSubstitution checks the constraint directly: $5x^2-3z^2+6xz=(5-27-18)/40=-1$. The inequality proves these are global nearest points, rather than merely stationary candidates.

In the orthonormal basis of part (b), the three quadric surfaces have the following forms. For $\alpha =1$,a two-sheeted hyperboloid with axis $u$, vertices $(\pm 1/\sqrt{6},0,0)$, and one sheet on each side of the plane $u=0$. For $\alpha =0$,an elliptic cone with vertex at the origin and the same axis. For $\alpha =-1$,a one-sheet hyperboloid with axis $u$ and a waist ellipse at $u=0$, whose semiaxes are $1/\sqrt{3}$ along $v$ and $1/2$ along $w$. The sections at fixed $u$ enlarge away from the waist.

The quadratic form is ${\mathbf{x}}^{T}Q\mathbf{x}$ for the real symmetric matrixThe $y$ axis has eigenvalue $-3$. On the $xz$ plane, the characteristic polynomial isThus the orthogonal diagonalization from part (a) yieldsFor later use, one possible orthonormal basis isThen $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the Euclidean norm is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.

Choose a real unit eigenvector $e_1$ of the real symmetric matrix $Q$, with eigenvalue ${\lambda }_1$. Its orthogonal complement is an invariant subspace: for any $v$ with $e_1^{T}v=0$,The restriction of $Q$ to this $(n-1)$-dimensional space is again real and symmetric. Applying the same argument inductively produces an orthonormal basis $e_1,\dots ,e_n$ of real eigenvectors. The base case $n=1$ is immediate. This proves the spectral theorem for real symmetric matrices without assuming that distinct eigenvalues are available.

One can also check that eigenvectors with different eigenvalues are automatically orthogonal: ${\lambda }_i{e}_i^T{e}_j=(Q{e}_i{)}^T{e}_j=e_i^{T}Q{e}_j={\lambda }_j{e}_i^T{e}_j$. Within a repeated eigenspace, the Gram-Schmidt process preserves the eigenvector property. Thus repeated eigenvalues cause no obstruction, and $\boxed{Q=OD{O}^T\text{ for an orthogonal }O\text{ and real diagonal }D}$.

Here $A^2=-I$. Splitting the matrix exponential into even and odd powers therefore givesFor the second matrix, $(I+A)(I-A)=I-A^2=2I$, so $(I+A{)}^{-1}=(I-A)/2$. HenceBoth are real orthogonal matrices, and hence also unitary matrices.

If $Av=\lambda v$, then $A^{n}v={\lambda }^{n}v$, and the matrix exponential series gives $Uv=e^{\lambda }v$. Also $(I+A{)}^{-1}v=(1+\lambda {)}^{-1}v$, since $1+\lambda \ne 0$. ThusThe assumed diagonalization ensures that these account for all eigenvalues, including multiplicities. In particular, for $\lambda =i\theta$,both of modulus one, in agreement with the unitary matrix property.

Taking the conjugate transpose term by term in the convergent matrix exponential givesThe matrices $A$ and $-A$ commute, so the matrix exponential product law gives $U^{†}U=e^{-A}{e}^A=I$ and likewise $U{U}^{†}=I$. Hence $\boxed{U\text{ is unitary}}$.

For $V$, use the inverse and product rules for the conjugate transpose:The factors $I\pm A$ commute, as do their inverses, since they are expressions in the same matrix. Consequentlyand similarly $V{V}^{†}=I$. Therefore $\boxed{V\text{ is unitary}}$. This rational construction is the skew-Hermitian form of the Cayley transform of a Hermitian matrix, obtained by writing $A=iH$ with $H$ a Hermitian matrix.

If $I+A$ were a singular matrix, a nonzero vector $v$ would satisfy $(I+A)v=0$, so $Av=-v$. This would make $-1$ an eigenvalue of the anti-Hermitian matrix, contrary to part (a). Thus $\boxed{I+A\text{ is invertible}}$. The same argument shows $I-A$ is an invertible matrix, which will also be used below.

Let $Av=\lambda v$ with $v\ne 0$. Taking the Hermitian inner product with $v$ gives $v^{†}Av=\lambda \Vert v{\Vert }^2$. For an anti-Hermitian matrix,Consequently $\overline{\lambda }\Vert v{\Vert }^2=-\lambda \Vert v{\Vert }^2$. As $\Vert v{\Vert }^2>0$, the eigenvalue satisfies $\boxed{\overline{\lambda }=-\lambda }$, or $\boxed{\lambda =i\theta \text{ with }\theta \in \mathbb{R}}$. Zero is included as a purely imaginary eigenvalue.

The squared norm of the original function isApplying the Parseval identity to part (c), using the evenness of the squared transform, givesThus the last integral equals ${\pi }^2/8$. Now set $u=\pi k/2$. Sincewe obtainThe apparent singularity at $u=\pi /2$ is removable: the continuous value of the integrand there is $1/4$. No principal-value interpretation is needed.

The real odd function has a purely imaginary odd Fourier transform. Setting $A=\pi /2$ and using a product-to-sum identity,The apparent singularities are removable. Direct evaluation or limits giveIn particular, the transform stays finite where its numerator and denominator both vanish.