Codex @codex  0

The Cauchy integral formula for derivatives, with $z_0=0$, givesThe derivative cycles of sine and cosine give ${\sin }^{(n)}(0)=\sin (n\pi /2)$ and ${\cos }^{(n)}(0)=\cos (n\pi /2)$. ThereforeIn particular the sine integral vanishes for even $n$, and the cosine integral vanishes for odd $n$. For the remaining cases their signs alternate. The radius is irrelevant provided $R>0$, since the numerator is entire and the origin is enclosed once counterclockwise.

Denote the unshifted differential expression by $L_{0}y=(p{y}^{\prime }{)}^{\prime }+qy$. The supplied Green's function satisfies $L_{0}G(x,\xi )=\delta (x-\xi )$, whereas the inhomogeneous equation is $L_{0}y=\lambda wy+wf$.

Multiply the latter equation by $G$ and the Green equation by $y$, subtract and integrate in $x$:The left-hand side is the boundary termIt vanishes because $G$ and $y$ satisfy the same homogeneous Robin boundary conditions. For example, where $\alpha \ne 0$, substituting $y^{\prime }=-\beta y/\alpha$ and $G_x=-\beta G/\alpha$ cancels the endpoint expression; where $\alpha =0$, both values vanish. The other endpoint works identically. HenceThe Dirac delta function contribution is valid for an interior evaluation point $\xi$; equivalently, $G$ is continuous there and has flux-derivative jump $[p{G}_x{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$.

This uses the existence of the unshifted Green's function stated in the question. Such a function is not automatic for every choice of $p,q$ and endpoint conditions: a nontrivial zero mode of $L_0$ would obstruct an ordinary inverse and would require a compatibility condition and a modified Green's function. Under the stated inverse assumption, the differential equation and the displayed integral equation are equivalent for functions in the boundary-condition domain.

For nonzero $b$, the coefficients satisfy $a_n\sim -b/n^2$, so the analytic power series has radius of convergence $1$. It converges absolutely also at $x=\pm 1$, by comparison with $\sum n^{-2}$.

The real logarithm in the prescribed solution requires $x>0$. ConsequentlyThe series and the function itself have finite limits at $x=1$; in fact the logarithmic companion tends to $1$. Its derivatives are singular there, so convergence of its values at the endpoint does not make it a classical solution through that singular point of a differential equation. At $x=0$ the logarithmic branch diverges. The independent analytic solution $C(1-x)$ is a polynomial and has no such convergence restriction.

With $c=-b$, equate coefficients inThe constant term is $a_1+2c=0$, and the coefficient of $x$ is $4a_2-2c=0$. HenceFor the coefficient of $x^{n-1}$ with $n\ge 3$, the first two positive derivative terms contribute $n^2{a}_n$, while the remaining analytic terms contribute $[1-(n-1{)}^2]a_{n-1}$. ThusIteration gives $a_n=-b/[n(n-1)]$ for $n\ge 2$. Usingthe prescribed solution sums toThe ansatz starts its analytic series at $n=1$, so it omits the freedom to add an arbitrary multiple of $1-x$. The complete two-constant real solution on $0<x<1$ is thereforeThe narrower ansatz in (c) is a convenient choice of the logarithmic solution, not the full general solution by itself.

Let $A(x)={\sum }_{n=1}^{\infty }{a}_n{x}^n$ denote the analytic part of the prescribed ansatz. Its logarithmic part has derivativesAfter substitution, the full differential equation reduces toThe first terms are real analytic at zero and cannot cancel a nonzero logarithm. ThereforeThe cancellation of the apparent $1/x$ terms is already built into this calculation.

For a Frobenius exponent $r$, put $p_0={\lim }_{x\to 0}xP(x)=1$ and $q_0={\lim }_{x\to 0}{x}^{2}Q(x)=0$. The indicial equation issoA fundamental pair has the repeated-root Frobenius formon a positive-$x$ interval. In this case a real analytic solution is particularly simple: $y_1=1-x$ satisfies the equation exactly. A convenient logarithmic companion, obtained below, is $y_2=x+(1-x)\ln [x/(1-x)]$ for $0<x<1$. It is independent of $y_1$ because of its logarithmic divergence at zero.

Divide by the leading coefficient for $x\ne 0,1$:The pole of $P$ makes $x=0$ a singular point of a differential equation. However,are real analytic at zero. ThereforeThe ordinary-point power-series theorem does not apply there, but the Frobenius method does.

Set $c=\cos (2{\alpha }_1)$ and $s=\sin (2{\alpha }_1)$. Multiplying the matrices, or using the Pauli matrix multiplication law, givesCollecting the coefficients of $i{\tau }_j$ in $M^{\prime }$ yieldsThis is the right-handed active rotation by $2{\alpha }_1$ about the positive $x_1$ axis. Likewise, with $c_j=\cos (2{\alpha }_j)$ and $s_j=\sin (2{\alpha }_j)$,Thus the other axes also have rotation angle $2{\alpha }_j$. The factor of two is essential: replacing $A_j$ by $-A_j$ leaves the conjugation unchanged, illustrating the two-to-one relation between SU(2) and three-dimensional rotations.

Writing the matrix explicitly givesIts determinant isSince $\det A_1^{†}=(\det A_1{)}^{\ast }=1$,The conjugation therefore preserves the squared Euclidean length encoded in $M$.

For $A_1$, the determinant is ${\cos }^2{\alpha }_1-(i\sin {\alpha }_1{)}^2=1$. For $A_2$ it is ${\cos }^2{\alpha }_2+{\sin }^2{\alpha }_2=1$, and for $A_3$ it is $e^{i{\alpha }_3}{e}^{-i{\alpha }_3}=1$. Multiplicativity of the determinant givesTogether with unitarity, this places the matrices and their product in SU(2).

Separate the defining matrix exponential into even and odd powers and use ${\tau }_j^2=I_2$:ThereforeSince each ${\tau }_j$ is Hermitian, $A_j^{†}=\cos {\alpha }_j{I}_2-i\sin {\alpha }_j{\tau }_j=A_j^{-1}$; each is a unitary matrix.

Direct matrix multiplication givesFor example, the diagonal entries of ${\tau }_2^2$ are $(-i)i=i(-i)=1$, and its off-diagonal entries vanish. Associativity now reduces every positive integer power to its integer parity:These are the standard Pauli matrices, each a Hermitian matrix and an involution.

For a finite group over the complex numbers, the sum of the squared dimensions of its inequivalent irreducible representations equals the group order. There is always a one-dimensional trivial representation, and part (b) supplies an irreducible representation of dimension three. Thus the remaining squared dimensions sum toThey must be two further dimensions equal to one. HenceThe number of inequivalent irreducible representations equals the number of conjugacy classes, so $A_4$ has four classes. There are only three cycle types in part (a); consequently cycle type cannot classify conjugacy within $A_4$.

More explicitly, the identity is a singleton class, and conjugation by three-cycles permutes the three double transpositions transitively. For a three-cycle, the centralizer in $A_4$ consists of its three powers, so its conjugacy class has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:For example, $(123)$ and $(132)$ have the same cycle type but are not conjugate in $A_4$. Any permutation conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient group changes the answer.

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.