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

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

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

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.

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)

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

• May I ask what is OGF? Commented Jul 31 at 20:45
• Ordinary generating function. Commented Jul 31 at 21:16