Paper IV: Scaled reciprocal coupling and theta polynomials

10 Scaled reciprocal coupling and theta polynomials

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Definition 10.1

Set

\[ H_N(z)={\mathsf E}_{N,0}(z) = \sum_{n\geq0}\frac{z^{Nn}}{(Nn)!}. \]

For \(u\in{\mathbb C}\), define \({\mathrm{\Theta}}_{N,n}(u)\) by

\[ H_N(uz)^{-1}H_N(z)^{-u} = \sum_{n\geq0}{\mathrm{\Theta}}_{N,n}(u)\frac{z^{Nn}}{(Nn)!}. \]

This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of \(\log H_N\) normalized by \(\log H_N(0)=0\).

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Theorem 10.2: Theta reconstruction

For every \(n\geq0\),

\[ {\mathrm{\Theta}}_{N,n}(u) = (-1)^n \sum_{q=0}^{n} \binom{Nn}{Nq} P_{N,q}(1)u^{Nq}P_{N,n-q}(u). \]

Thus \({\mathrm{\Theta}}_{N,n}\) is a polynomial of degree at most \(Nn\).

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Proof of Theorem 10.2

Choose \(\rho\) with \(\rho^N=-1\). Then

\[ H_N(z)={\mathsf D}_N(\rho z). \]

The coefficient of \(z^{Nn}\) in \(H_N(z)^{-r}\) is therefore \((-1)^nP_{N,n}(r)/(Nn)!\). Apply this to \(H_N(uz)^{-1}\) and \(H_N(z)^{-u}\), then multiply the two block series.

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Proposition 10.3: Trivial parameter factors

For \(n\geq1\), the polynomial \({\mathrm{\Theta}}_{N,n}(u)\) is divisible by \(u\). If \(N\) is even, it is also divisible by \(u+1\).

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Proof of Proposition 10.3

At \(u=0\), the defining product is one. Hence every positive-index coefficient vanishes. If \(N\) is even, then \(H_N(-z)=H_N(z)\). At \(u=-1\), the product is again one, so every positive-index coefficient vanishes there as well.

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Proposition 10.4: Theta Newton recurrence

Put

\[ \theta_{N,n}(u)=\frac{{\mathrm{\Theta}}_{N,n}(u)}{(Nn)!}. \]

Then \(\theta_{N,0}(u)=1\) and

\[ n\theta_{N,n}(u) = \sum_{m=1}^{n} (-1)^m\kappa_{N,m} \bigl(u^{Nm}+u\bigr) \theta_{N,n-m}(u). \]

In particular,

\[ {\mathrm{\Theta}}_{N,1}(u)=-(u^N+u). \]

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Proof of Proposition 10.4

Write \(w=z^N\). Since \(H_N(z)=\Delta_N(-w)\),

\[ -\log H_N(z) = \sum_{m\geq1}(-1)^m\kappa_{N,m}\frac{w^m}{m}. \]

The logarithm of the theta generating function is therefore

\[ \sum_{m\geq1} (-1)^m\kappa_{N,m} \bigl(u^{Nm}+u\bigr) \frac{w^m}{m}. \]

Logarithmic differentiation and coefficient comparison give the recurrence. The case \(n=1\) uses \(\kappa_{N,1}=1/N!\).

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Example 10.5

For \(N=2\),

\[ {\mathrm{\Theta}}_{2,1}(u)=-u(u+1) \]

and

\[ {\mathrm{\Theta}}_{2,2}(u) = u(u+1)(5u^2+u+2). \]

The two factors predicted by Proposition 10.3 are visible.

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