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Writing $Y=GZ$ with an invertible matrix-valued function gives $Z^{\prime }=\Lambda Z$ exactly when $\Lambda =G^{-1}AG-G^{-1}{G}^{\prime }$. For a formal power series transformation at an irregular singular point, coefficient comparison recursively removes off-diagonal terms when leading diagonal eigenvalues are distinct. This is a differential-system change of frame, not specifically an electromagnetic gauge transformation.

This equation has a regular singular point at zero and an irregular singular point at infinity. For a nonnegative integer $\nu =n$, the solution analytic at zero with value one is the Laguerre polynomial $L_n$. Its weighted self-adjoint form is $(x{e}^{-x}{y}^{\prime }{)}^{\prime }+\nu e^{-x}y=0$, connecting it to the Laguerre differential operator on polynomials.

Divide by $2x$ away from zero to obtain $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, with $P(x)=x/2-1/(2x)$ and $Q(x)=\lambda /2$. At zero, the quantities $xP(x)=(x^2-1)/2$ and $x^{2}Q(x)=\lambda x^2/2$ are analytic. The pole in $P$ makes zero singular, so $x=0$ is a regular singular point. Its indicial equation isAt $x=1$, both normalized coefficients are analytic functions, so $x=1$ is an ordinary point. These are the standard normalized-coefficient and indicial tests.^[ref]

For infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The chain rule gives $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, so the transformed normalized equation isThe product of the first-derivative coefficient with $t$ still has a second-order pole at zero. Hence the criterion for a regular singular point fails and $x=\infty$ is an irregular singular point, including when $\lambda =0$. The primes on $Y$ here refer to differentiation with respect to $t$.

Put $N=n(n+1)$. After division by $1-x^2$, the Legendre differential equation has coefficientsBoth are analytic functions at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm 1$, $p$ has a genuine simple pole; hence these are singular points even when $n=0$ makes $q$ vanish. However,and both products extend analytically across $x_0$. By the regular singular point criterion for a second-order equation,None of the three specified points is an irregular singular point.

Normalize the Laguerre differential equation as $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, where $P(x)=1/x-1$ and $Q(x)=n/x$. At $x=0$, the products $xP=1-x$ and $x^{2}Q=nx$ are analytic. Thus zero is a regular singular point, with indicial equationAt $x=1$ the coefficients $P,Q$ are analytic, so it is an ordinary point. No singular-point indicial equation is needed there.

For infinity put $z=1/x$ and $Y(z)=y(1/z)$. Then the equation becomesAt $z=0$, the coefficient $z(z^{-1}+z^{-2})=1+1/z$ is not analytic. Therefore infinity is an irregular singular point, including when $n=0$. A terminating polynomial solution does not change the classification of the differential equation.

Use the local coordinate $t=1/\lambda$. The differential system becomesFor the nonzero quadratic leading term, the coefficient of the local linear ordinary differential equation system has a pole of order four, so infinity is an irregular singular point of Poincaré rank three in this gauge. The matrix-valued function $A(\lambda )$ itself has a pole of order two; the extra factor $d\lambda /dt=-t^{-2}$ is essential when classifying the equation. If $A_0$ vanishes the degree and these orders decrease. The specified distinct eigenvalues $1,-1$ give the nondegenerate unramified case.

The leading exponential function factors have actions $q_{\pm }=\pm {\lambda }^3/3$, so their difference is $2{\lambda }^3/3$. Equal exponential magnitudes occur onThese are commonly called Stokes rays in the sectorial ODE convention. With the action-phase convention used by the wiki's Stokes line article they are anti-Stokes lines; the same-phase rays instead haveGiving both defining equations removes the nomenclature ambiguity; both sets contain six directions.

The standard construction with distinct leading eigenvalues uses six sectorial fundamental solutions per turn, with basic angular spacing $\pi /3$. They are fundamental matrices, each containing two independent column solutions. On the logarithm cover choose six consecutive overlapping sectors of width $\pi /3+2\delta$, with $\delta >\epsilon /2$ small. Their union has angular opening $2\pi +2\delta$, covering the requested $2\pi +\epsilon$. One often also writes a seventh index to compare the closing sector with the first after a full turn; that is a closing copy, not a seventh independent sector in the six-sector construction. The number is a generic sectorial fundamental solution normalization count, not a universal minimum for exceptional systems with vanishing Stokes matrices: the exactly diagonal system already has one global elementary fundamental matrix.

There is a missing qualification in the requested product identity. Polynomial coefficients give an entire invertible fundamental matrix, hence trivial actual monodromy about a large circle: the circle contracts through ordinary points. But the formal matrix generally has the form $\hat{G}{\lambda }^{\Theta }{e}^Q$, and a positive turn changes its logarithm branch by the formal monodromy $M_f=e^{2\pi i\Theta }$. This factor cannot be discarded.

To specify the printed product order, let $S_j$ act on column vectors of solution coefficients across the $j$th sector boundary, so adjacent normalized matrices satisfy $Y_{j+1}=Y_j{S}_j^{-1}$. Six transitions give $Y_7=Y_1(S_6\cdots S_1{)}^{-1}$. The normalized formal branch after the turn is ${\hat{Y}}_1{M}_f$, so sectorial uniqueness gives $Y_7=Y_1{M}_f$. ConsequentlyUsing Stokes matrices for right changes of fundamental matrices instead reverses/inverts the corresponding convention; the formal monodromy factor remains necessary. The formal monodromy correction to an entire-system Stokes product shows that the printed identity holds when $M_f=I$, including the corrected Lax specialization discussed below.

It does not follow from polynomiality alone. For example takeThe coefficient matching in question 2 gives $\Theta =\frac{1}{2}\mathrm{diag}(1,-1)$, hence $M_f=-I$ and the ordered Stokes matrix product is $-I$, not $I$, despite having no finite singularity. This is an explicit counterexample to the unqualified final clause.

In the normalized second-order linear differential equation, an ordinary point $z_0$ is one at which both $p$ and $q$ are holomorphic functions in a neighbourhood of $z_0$. A singular point of a differential equation is a point that is not ordinary. A singular point $z_0$ is a regular singular point if $(z-z_0)p(z)$ and $(z-z_0{)}^{2}q(z)$ extend analytically to $z_0$. Equivalently, $p$ has at most a simple pole and $q$ at most a double pole there. If a singular point fails this test, it is an irregular singular point. The normalized form matters: zeros of a leading coefficient must first be assessed after division by that coefficient.

After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ are analytic functions, so $0$ is a regular singular point. Its indicial equation is $r(r-1)+2r=r(r+1)=0$. There are no other finite singular points of a differential equation. At infinity the regular singular point criterion for a second-order equation would require $P=O(z^{-1})$, but $P\to -1$; hence infinity is an irregular singular point.

Substituting $y=f/z$ reduces the equation to $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two characteristic exponents at a regular singular point, $0,-1$. The integer difference does not require a logarithm. At infinity $e^z/z$ has an essential singularity, consistent with the irregular singular point; the existence of the simpler solution $1/z$ does not make infinity regular.

Fora finite point $z_0$ is a regular singular point when $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ extend holomorphically to $z_0$. A singularity at infinity is classified by applying this criterion after the change of variable $w=1/z$.

For the displayed Bessel differential equation,At $z=0$, the functions $zP(z)=1$ and $z^{2}Q(z)=z^2-1/4$ are holomorphic, so zero is regular singular. To inspect infinity, write $Y(w)=y(1/w)$. The equation becomesHere $w^{2}Q(w)=w^{-2}-1/4$ is not holomorphic at zero, so infinity is an irregular singular point. Thus

Set $y=z^{-1/2}f$. Direct differentiation makes every lower-order term cancel:Hence $f^{\prime \prime }+f=0$, and on any domain carrying a branch of $\sqrt{z}$ two linearly independent solutions areNear the regular singular point, these behave as $z^{-1/2}$ and $z^{1/2}$, the two Frobenius exponents. At infinity their combinations are $z^{-1/2}{e}^{\pm iz}$; after $w=1/z$ these contain $e^{\pm i/w}$, the exponential behavior characteristic of an irregular singular point.

For an equation $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, infinity is a regular singular point only when $p(x)=O(x^{-1})$ and $q(x)=O(x^{-2})$. Here $p=-1$ and $q=-2/x-2/x^2$, so infinity is an irregular singular point.

A singular point which is not regular singular is an irregular singular point: at least one of $(z-z_0)p(z)$ and $(z-z_0{)}^{2}q(z)$ fails to extend holomorphically to $z_0$.

After setting $z=1/x$, the transformed coefficients have a singularity beyond the regular-singular growth bounds at $z=0$. Thus infinity is an irregular singular point.

Write a second-order linear ordinary differential equation in monic form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$. At a finite point $x_0$, an ordinary point has analytic $P,Q$. A singular point is a regular singular point if $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically there. A singular point failing this condition is an irregular singular point. The point at infinity is classified by the reciprocal coordinate $t=1/x$.

For the Laguerre differential equation, $P=1/x-1$ and $Q=\nu /x$. Thus $xP=1-x$ and $x^{2}Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, the transformed equation isHere $t{P}_t=1+1/t$ is not analytic at zero, so infinity is an irregular singular point, for every real $\nu$.

For the Frobenius solution, the lowest power $x^{\sigma -1}$ has coefficient $a_0{\sigma }^2$. Hence the indicial equation is ${\sigma }^2=0$, and equating the remaining powers givesWith the repeated root $\sigma =0$, the Frobenius series for the Laguerre differential equation obeysIf the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the ratio test gives infinite radius of convergence. It is an entire function, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.

For $\nu =n\ge 0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})/k!$, giving the Laguerre polynomialsThe same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.

Singular points. Normalizing the equation gives $p=2/x$ and $q=1-\ell (\ell +1)/x^2$. The only finite singular point is $x=0$, and it is regular because $xp$ and $x^{2}q$ are analytic there. At infinity put $t=1/x$ and $Y(t)=y(1/t)$; the equation becomesThe fourth-order pole makes $t=0$ an irregular singular point, so infinity is irregular.

Exponents and recurrence. Substitution of a series for a Frobenius solution gives the indicial equationWith negative-index coefficients set to zero, the coefficient equation isFor $\sigma =\ell$, it gives $a_1=0$ andFor $\sigma =-\ell -1$, it givesFor $\ell \ge 1$, $a_1=0$. At the exceptional odd index $j=2\ell +1$, the recurrence becomes $a_{2\ell -1}=0$, which is already satisfied. Thus $a_{2\ell +1}$ is free; choosing it to be zero gives an independent pure-even series times $x^{-\ell -1}$. A nonzero choice adds a multiple of the regular solution. For $\ell =0$, this freedom occurs at $j=1$ itself. This Frobenius resonance in the spherical Bessel equation is compatible, so no logarithmic term is needed despite the integer separation of exponents.

Normalized regular solution. Set $\sigma =\ell$ and $a_0=2^{\ell }\ell !/(2\ell +1)!$. The proposed even coefficientssatisfy $a_{2s}/a_{2s-2}=-1/[2s(2s+2\ell +1)]$, precisely the recurrence above. ThereforeThis is the regular Spherical Bessel function. The coefficient ratio tends to zero, so its series converges for every finite $x$; the other even-series factor also has infinite radius, although the prefactor has a pole at zero.

For $\ell =0$, the coefficients reduce to $(-1{)}^s/(2s+1)!$, givingDirectly, $x^2{j}_0^{\prime }=x\cos x-\sin x$, whose derivative is $-x\sin x$. Thus $(x^2{j}_0^{\prime }{)}^{\prime }+x^2{j}_0=0$, verifying the equation. The independent singular branch at $\ell =0$ can be chosen as $\cos x/x$.

Expanding the derivative and dividing by $1-x$ givesThe only finite singular point is $\boxed{x=1}$. There $(x-1)p(x)=1$ and $(x-1{)}^{2}q(x)=c^2(x-1{)}^2-\lambda (x-1)$ are analytic, so it is a regular singular point. At infinity the coefficient of $y$ tends to $c^2\ne 0$, giving an irregular singular point after reciprocal substitution; the finite classification is independent of $\lambda$.

Away from zeros of the leading coefficient, the normal-form differential equation isThe only finite singular points are $z=0,i,-i$. At zero, $zP(z)=3/(1+z^2{)}^2$ and $z^{2}Q(z)=1/(1+z^2{)}^2$ are analytic, so zero is a regular singular point. At $z=\pm i$, the coefficient $P$ has a genuine double pole, so $(z\mp i)P$ is not analytic; these are irregular singular points. ThusIf infinity is included, set $t=1/z$ and $Y(t)=y(1/t)$. The transformed coefficients are $\tilde{P}=2/t-3t^3/(1+t^2{)}^2$ and $\tilde{Q}=t^2/(1+t^2{)}^2$. They satisfy the regular singular point at infinity criterion, so $\boxed{\infty \text{ is regular singular}}$ as well.

At an unramified irregular singular point with distinct leading eigenvalues, a sectorial fundamental solution is a fundamental matrix whose entries are holomorphic functions asymptotic to a prescribed formal matrix on a sector of the logarithm cover. For competing cubic exponentials, the usual construction uses six overlapping sectors with basic angle $\pi /3$. A formal power series need not converge, and two sectorial realizations may differ by exponentially small terms across a sector boundary.