Paper IV: Residue-section reciprocals and Appell systems

9 Residue-section reciprocals and Appell systems

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The section \({\mathsf E}_{N,j}\) has a forced zero of order \(j\) at the origin. Removing it leaves a unit series in \(z^N\).

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Definition 9.1

For \(0\leq j<N\), define

\[ \Phi_{N,j}(w) = j!\sum_{n\geq0}\frac{w^n}{(Nn+j)!}. \]

Then

\[ {\mathsf E}_{N,j}(z) = \frac{z^j}{j!}\Phi_{N,j}(z^N). \]

Define the residue-section reciprocal function

\[ {\mathsf C}_{N,j}(z) = \Phi_{N,j}(z^N)^{-1} = \frac{z^j/j!}{{\mathsf E}_{N,j}(z)}. \]

For \(j>0\), the last quotient at \(z=0\) is interpreted by its removable continuation, namely the first expression in the display. Write

\[ {\mathsf C}_{N,j}(z) = \sum_{n\geq0}C_{N,j}(n)\frac{z^{Nn}}{(Nn)!}. \]

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Theorem 9.2: Triangular inverse system

The coefficients are determined by \(C_{N,j}(0)=1\) and

\[ \sum_{q=0}^{n} \binom{Nn+j}{Nq} C_{N,j}(q) = \delta_{n0}. \]

For \(n\geq1\),

\[ C_{N,j}(n) = - \frac{ \displaystyle\sum_{q=0}^{n-1} \binom{Nn+j}{Nq}C_{N,j}(q) }{ \displaystyle\binom{Nn+j}{Nn} }. \]

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Proof of Theorem 9.2

Multiply the defining series for \({\mathsf C}_{N,j}\) and \({\mathsf E}_{N,j}\). Their product is \(z^j/j!\). At degree \(Nn+j\), multiplication by \((Nn+j)!\) gives the stated binomial coefficient. The diagonal coefficient is nonzero, so the system is triangular.

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Definition 9.3

For \(r\in{\mathbb C}\), define formal coefficients \(C_{N,j,n}(r)\) by

\[ {\mathsf C}_{N,j}(z)^r = \sum_{n\geq0}C_{N,j,n}(r)\frac{z^{Nn}}{(Nn)!}. \]

On a simply connected region containing \(0\) and avoiding the zeros of \(\Phi_{N,j}(z^N)\), interpret this power as \(e^{rL}\), where \(L\) is the holomorphic logarithm of \({\mathsf C}_{N,j}\) normalized by \(L(0)=0\).

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Theorem 9.4: Parameter-addition law for residue-section reciprocals

For each \(n\geq1\), the coefficient \(C_{N,j,n}(r)\) is a polynomial in \(r\) of exact degree \(n\). Its leading coefficient is

\[ \frac{(Nn)!}{n!} \left(-\frac{j!}{(N+j)!}\right)^n. \]

It satisfies

\[ C_{N,j,n}(r+s) = \sum_{q=0}^{n} \binom{Nn}{Nq} C_{N,j,q}(r)C_{N,j,n-q}(s). \]

In particular,

\[ C_{N,j,n}(1)=C_{N,j}(n). \]

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Proof of Theorem 9.4

The unit series \(\Phi_{N,j}(w)^{-1}\) has a formal logarithm with zero constant term. Its linear term is

\[ -\frac{j!}{(N+j)!}w. \]

Exponentiation after multiplication by \(r\) shows that the coefficient of \(w^n\) is a polynomial of degree at most \(n\). The \(n\)th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly \(n\). Multiplication of the powers with parameters \(r\) and \(s\) gives the coefficient identity. More explicitly, before the factorial normalization the coefficient of \(r^nw^n\) is

\[ \frac1{n!}\left(-\frac{j!}{(N+j)!}\right)^n; \]

thus multiplication by \((Nn)!\) gives the displayed leading coefficient.

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Definition 9.5

Define polynomials \(E_m^{(N,j,r)}(x)\) by

\[ {\mathsf C}_{N,j}(z)^re^{xz} = \sum_{m\geq0}E_m^{(N,j,r)}(x)\frac{z^m}{m!}. \]

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Theorem 9.6: Appell laws

For \(m\geq1\), the polynomials satisfy

\[ \frac{d}{dx}E_m^{(N,j,r)}(x) = mE_{m-1}^{(N,j,r)}(x) \]

and, for \(m\geq0\),

\[ E_m^{(N,j,r)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x)y^{m-q}. \]

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Proof of Theorem 9.6

Differentiate the generating function with respect to \(x\) for the first identity. Multiply it by \(e^{yz}\) and compare coefficients for the second.

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Proposition 9.7: Explicit Appell polynomial

For every \(m\geq0\),

\[ E_m^{(N,j,r)}(x) = \sum_{0\leq n\leq m/N} \binom{m}{Nn} C_{N,j,n}(r)x^{m-Nn}. \]

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Proof of Proposition 9.7

Multiply the block series for \({\mathsf C}_{N,j}^r\) by the exponential series and collect the coefficient of total degree \(m\).

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Theorem 9.8: Joint Appell convolution in the space and order parameters

For all \(r,s,x,y\in{\mathbb C}\) and \(m\geq0\),

\[ E_m^{(N,j,r+s)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x) E_{m-q}^{(N,j,s)}(y). \]

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Proof of Theorem 9.8

Multiply the generating functions

\[ {\mathsf C}_{N,j}(z)^re^{xz} \quad\hbox{and}\quad {\mathsf C}_{N,j}(z)^se^{yz}. \]

Their product is the generating function with parameters \(r+s\) and \(x+y\). Coefficient comparison gives the formula.

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Proposition 9.9: Formal differential-operator representation on polynomials

Let a formal series \(A(z)=\sum_{q\geq0}a_qz^q/q!\) act on polynomials by

\[ A(\partial_x)f(x) = \sum_{q\geq0}\frac{a_q}{q!}\partial_x^qf(x). \]

The power \(r\) is applied to the formal symbol before the substitution \(z=\partial_x\). Then

\[ E_m^{(N,j,r)}(x) = {\mathsf C}_{N,j}(\partial_x)^r x^m. \]

The inverse relation is

\[ x^m = \Phi_{N,j}(\partial_x^N)^r E_m^{(N,j,r)}(x). \]

Both sums are finite on every polynomial.

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Proof of Proposition 9.9

Applying \(A(\partial_x)\) to \(e^{xz}\) multiplies it by \(A(z)\). Taking \(A={\mathsf C}_{N,j}^r\) gives the first identity by the defining generating function. Constant coefficient operators on polynomials satisfy

\[ (AB)(\partial_x)=A(\partial_x)B(\partial_x), \]

because all derivative sums are finite and powers of \(\partial_x\) commute. The two formal symbols

\[ {\mathsf C}_{N,j}(z)^r=\Phi_{N,j}(z^N)^{-r} \quad\hbox{and}\quad \Phi_{N,j}(z^N)^r \]

multiply to one. Their substitutions at \(z=\partial_x\) are therefore inverse constant coefficient operators on polynomials, which proves the second identity.

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Corollary 9.10: Appell symmetry

Each polynomial \(E_m^{(N,j,r)}\) is monic of degree \(m\), and

\[ E_m^{(N,j,r)}(\omega x) = \omega^mE_m^{(N,j,r)}(x). \]

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Proof of Corollary 9.10

The term \(n=0\) in Proposition 9.7 is \(x^m\). Every other exponent is \(m-Nn\), so replacement of \(x\) by \(\omega x\) multiplies every term by \(\omega^m\).

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Example 9.11

For \((N,j)=(2,0)\),

\[ {\mathsf C}_{2,0}(z)={\operatorname{sech}} z. \]

For \((N,j)=(2,1)\),

\[ {\mathsf C}_{2,1}(z)=\frac z{\sinh z}. \]

The first case is a scaled and translated Euler family. In fact, if \(E_m(x)\) denotes the classical Euler polynomial, then

\[ E_m^{(2,0,1)}(x) = 2^mE_m\left(\frac{x+1}{2}\right). \]

If \(B_m(x)\) denotes the classical Bernoulli polynomial, the second case has the exact identification

\[ E_m^{(2,1,1)}(x) = 2^mB_m\left(\frac{x+1}{2}\right), \]

obtained by substituting \(t=2z\) in \(t e^{yt}/(e^t-1)\).

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