Taylor expansion of $\prod_{k=0}^\infty (1-z \mu_k)^{-1}$

Question

In this paper, the author claimed that the following identity is easy to see:

$$\frac{1}{n!} \sum_\pi \prod_{i=1}^n \left(\sum_{k=0}^\infty \mu_k^{i}\right)^{N_i} = \text{coefficient of $z^n$ in the Taylor expansion of $\prod_{k=0}^\infty (1-z \mu_k)^{-1}$}.$$

In the left hand side, the summation is over all the permutation $\pi$ in symmetric group $S_n$ and $N_i$ is the number of $i$-cycles in $\pi$. The right hand side is the coefficient of $z^n$ in $\prod_{k=0}^\infty \sum_{i=0}^\infty \mu_k^i z^i$ where $\mu_k \in (0,1)$.

For a simple example, consider $n = 2$. Then we have

$$LHS = \frac{\sum_{k=0}^\infty \mu_k^2}{2} + \frac{(\sum_{k=0}^\infty \mu_k)^2}{2} = \sum_{k=0}^\infty \mu_k^2 + \sum_{k<k'} \mu_k \mu_{k'} = RHS.$$

But for general $n$, I cannot directly see that this is true. Are there any good ways to view it?

Answer 1

This is a variation of Newton's identities, in this case the ones expressing the complete homogeneous symmetric polynomials $h_n$ in terms of the power sum symmetric polynomials $p_n = \sum \mu_k^n$. There is a short proof using the logarithmic derivative. Write $P(z)$ for the RHS, which is equal to

$$P(z) = \sum h_n z^n.$$

Taking the logarithm gives

$$\log P(z) = \sum \log \frac{1}{1 - z \mu_k} = - \sum \log (1 - z \mu_k)$$

and taking the (formal) derivative gives

$$\frac{d}{dz} \log P(z) = \sum \frac{\mu_k}{1 - z \mu_k} = \sum p_{n+1} z^n.$$

Integrating gives

$$\log P(z) = \sum_{n \ge 1} \frac{p_n}{n} z^n$$

and exponentiating gives

$$P(z) = \boxed{ \sum h_n z^n = \exp \left( \sum_{n \ge 1} \frac{p_n}{n} z^n \right) }.$$

Expanding out this exponential gives the LHS; this is the "permutation form of the exponential formula," see e.g. this old blog post of mine. The same argument but starting with the product $\prod (1 + z \mu_k) = \sum e_n z^n$ describing the elementary symmetric polynomials gives the corresponding identity for $e_n$, which has signs inserted.

Answer 2

The LHS is the cycle index of the symmetric group $S_n$ evaluated at $a_\ell = \sum_{k\ge 0} \mu_k^\ell.$ This cycle index has OGF $$Z(S_n) = [z^n] \exp \left(\sum_{\ell \ge 1} a_\ell \frac{z^\ell}{\ell}\right).$$ Doing the substitution, $$[z^n] \exp \left(\sum_{\ell \ge 1} \sum_{k\ge 0} \mu_k^\ell \frac{z^\ell}{\ell}\right) = [z^n] \exp \left(\sum_{k\ge 0} \sum_{\ell \ge 1} \mu_k^\ell \frac{z^\ell}{\ell}\right) \\ = [z^n] \exp \left(\sum_{k\ge 0} \log\frac{1}{1-\mu_k z}\right) = [z^n] \prod_{k\ge 0} \frac{1}{1-\mu_k z}.$$ This is the claim.

Remark. The ordinary generating function cited here is given by the combinatorial class (notation from Analytic Combinatorics)

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}( \mathcal{A}_1 \times \textsc{CYC}_{=1}(\mathcal{Z}) + \mathcal{A}_2\times \textsc{CYC}_{=2}(\mathcal{Z}) + \mathcal{A}_3\times \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$