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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.

Seek a positive weight such that the differential operator becomes $-w^{-1}(p{y}^{\prime }{)}^{\prime }$. Matching the second derivative gives $p=w(1-x^2)$, while matching the first derivative requires $p^{\prime }=-xw$. ConsequentlyChoosing the multiplicative constant to be one givesThis is Sturm-Liouville form with zero potential term. Multiply the equations for $y_n,y_m$ by $y_m,y_n$, subtract, and integrate. The Lagrange identity for a Sturm-Liouville operator givesThe boundary expression vanishes because $p\to 0$ and the functions and their derivatives are finite. Since $n,m\ge 0$ are distinct, their eigenvalues $n^2,m^2$ are distinct, provingThe endpoint singularities of the weight are integrable, so these weighted inner products exist.

For fixed endpoint values and a smooth variation vanishing at the endpoints, the first variation isThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationIf $f$ is independent of $y$, then $y$ is a cyclic coordinate, and integration givesThis is conservation of the corresponding generalized momentum.

For either exponent, the recurrence from part (c) impliesfor every finite complex $x$. The ratio test therefore givesThe factor multiplying $x^{\sigma }$ is an entire function in each case. For $\sigma =1/2$, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_n{x}^n$, not to a Taylor expansion of the branched full solution.

For $n\ge 1$, the coefficient of $x^{n+\sigma }$ isFactoring the bracket gives the recurrence relationFor $\sigma =1/2$ this is $a_n=-a_{n-1}/[n(2n-1)]$. Iteration and $(2n)!=2^{n}n!(2n-1)!!$ yieldFor $\sigma =1$, the recurrence is $a_n=-a_{n-1}/[n(2n+1)]$, givingHere the double factorial products are one at $n=0$, consistent with the chosen normalization. Recognizing the cosine and sine power series, for positive $x$ one may also write the two Frobenius solutions asThe constants multiplying the two independent solutions may be chosen separately.

Apply the Frobenius method. For $y=x^{\sigma }{\sum }_{n=0}^{\infty }{a}_n{x}^n$, the coefficient of the lowest power $x^{\sigma }$ comes only from $2x^2{y}^{\prime \prime }-x{y}^{\prime }+y$; the remaining term $xy$ starts one power higher. As $a_0\ne 0$, this lowest coefficient gives the indicial equationTherefore the two possible characteristic exponents at a regular singular point areTheir difference is not an integer, so the two corresponding Frobenius solutions do not require a logarithmic correction.

Divide by $2x^2$ to put the ordinary differential equation into normalized form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$:The coefficients are singular at zero, butare both analytic there. This is precisely the regular singular point criterion for a second-order equation, so $x=0$ is a regular singular point.

Expand the exponential in inverse powers of $z$:This is a Laurent series, not a Taylor series, and converges throughout $0<|z|<\infty$. Every displayed coefficient of a negative power is nonzero, so its principal part has infinitely many terms. ThusThe residue is the coefficient of $z^{-1}$:

Put $w=z-4i$. Factor out the nonzero constant in the denominator and use the geometric series:This Taylor series converges when $|z-4i|<|3-4i|$. The simple pole at $z=3$ is the only finite obstruction to analyticity, whose distance from the expansion centre is five. Hence the radius of convergence is

The derivative of an analytic function must be independent of the direction of approach. Taking a real increment gives $f^{\prime }(z)=u_x+i{v}_x$; taking an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:A real harmonic function is a twice continuously differentiable function whose Laplacian vanishes. For the given logarithmic function and $x^2+y^2>0$,It is therefore harmonic everywhere on the punctured complex plane.

On any domain admitting a branch of the complex logarithm, chooseIts derivatives are $v_x=-2y/(x^2+y^2)$ and $v_y=2x/(x^2+y^2)$, as required. For example, the principal complex logarithm gives such a harmonic conjugate on the plane cut along the negative real axis.

There is no single-valued harmonic conjugate on the whole punctured plane. Indeed its putative differential would satisfywhereas a globally defined function has zero integral of its differential around a closed curve. This is the log modulus has no global harmonic conjugate on the punctured plane obstruction. Thus the conjugate requested in the question must be interpreted locally or on a cut plane.

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.

Every row sums to six, so $\mathbf{e}=(1,1,1{)}^T$ is an eigenvector with eigenvalue six. Since $M$ is a real symmetric matrix, its other two eigenvalues, say ${\lambda }_2,{\lambda }_3$, are real. The trace givesAlso, for a real symmetric matrix, $\mathrm{tr}(M^2)$ is the sum of the squares of all its entries. HereThus ${\lambda }_2^2+{\lambda }_3^2=42-36=6$. Together with their zero sum, this implies

A square matrix $A$ is a diagonalizable matrix over a specified field if some invertible matrix $P$ makes $P^{-1}AP$ diagonal. Equivalently, the underlying vector space has a basis of eigenvectors of $A$.

For example,The first is already diagonal. The second has only the eigenvalue one, of algebraic multiplicity two, but its eigenspace consists of $(x,0{)}^T$ and has dimension one. It cannot supply a two-vector eigenbasis.

The finite-dimensional spectral theorem gives a spectral decomposition of the Hermitian matrix:The matrix exponential respects similarity, soThe diagonal entries are real, so the conjugate transpose of this expression equals itself. Its eigenvalues are $e^{{\lambda }_j}>0$, provingIn particular, for any nonzero $v$, $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$. Thus the exponential of a Hermitian matrix is a Hermitian positive-definite matrix, even when $A$ itself has negative eigenvalues.

If $A^{†}=-A$, then $H=-iA$ is a Hermitian matrix, because $H^{†}=i{A}^{†}=-iA=H$. By the finite-dimensional spectral theorem, there is a unitary matrix $Q$ withThus $A=Q\mathrm{diag}(i{\theta }_j)Q^{†}$. Since the matrix exponential respects similarity,Each diagonal entry has modulus one, so $C^{†}C=I$. Using $Q^{†}Q=Q{Q}^{†}=I$,Hence the exponential of a skew-Hermitian matrix is unitary.