How do I find coefficients of a product expansion

Question

Any power series $f(t) = 1 + t \mathbb{Z}[[t]]$ can be uniquely expanded in the following two ways:

$$1 + \sum_{i=1}^\infty f_i t^i = \prod_{i=1}^\infty (1-t^i)^{-n_i}$$

Here, the $f_i$ and $n_i$ are integers.

Is there an explicit formula for the $n_i$ in terms of the $f_i$?

Similarly a power series $f(t, L) = 1 + t \mathbb{Z}[[t, L]]$ can be uniquely expanded

$$1 + \sum_{i=1}^\infty \sum_{j=0}^\infty f_{i,j} t^i L^j = \prod_{i=1}^\infty \prod_{j=0}^\infty (1-t^i L^j)^{-n_{i,j}}$$

How do I find the $n_{i,j}$ in terms of the $f_{i,j}$?

More generally, in an arbitrary $\lambda$-ring $\Lambda$, a power series $f(t) \in 1 + t \Lambda[[t]]$ can be uniquely expanded in the following two ways:

$$1 + \sum_{i=1}^\infty f_i t^i = \prod_{i=1}^\infty (1-t^i)^{-n_i}$$

where the $f_i, n_i \in \Lambda$, and the exponent signifies $(1-t^i)^{-M} := \sum_{k=0}^\infty t^{ik} \mathrm{Sym}^k(M)$

Is there an explicit formula for the $n_i$ in terms of the $f_i$?

Answer 1

I'll use the notation $[t^i]$ to stand for the coefficient of $t^i$ in a Maclaurin series. Let

$$ C(i) = [t^i] \log(1+t F(t)) = \sum_{j=1}^\infty (-1)^{j+1} [t^i] \dfrac{t^j F(t)^j}{j} = \sum_{j=1}^i (-1)^{j+1} [t^{i-j}] \dfrac{F(t)^j}{j} $$ Then $$ C(i) = [t^i] \log \prod_{j=1}^\infty (1-t^j)^{-n_j} = - \sum_{j | i} n_j [t^i] \log(1 - t^j) = \sum_{j | i} \dfrac{j n_j}{i}$$ and then $$ n_j = \sum_{i | j} \mu(j/i) \dfrac{i}{j} C(i)$$

Answer 2

Robert Israel's answer works if you have access to the coefficients of $\log f(t)$, which isn't always the case (one example that comes to mind is when the $f_i$'s are classes of punctual Hilbert schemes, and you're trying to get invariants as in here.)

To answer your question more specifically, here is my attempt at spelling out the strategy you mention. Let's start with the identity $$\frac{1}{1-at}=\prod_{k\geq 1}\frac{1}{(1-t^k)^{M_k(a)}},$$ where $M_k(a)=\frac{1}{k}\sum_{d|k}a^d\mu(k/d)$ are the necklace polynomials.

Next, let $h_i\in \Lambda[x_1,x_2,\dots]$ be the complete homogeneous symmetric polynomials. We have the following fundamental generating function $$1+h_1t+h_2t^2+\cdots=\prod_{r\geq 1}\frac{1}{(1-x_rt)}.$$ So combining with our previous observation we have $$1+h_1t+h_2t^2+\cdots=\prod_{k\geq 1}\frac{1}{(1-t^k)^{M_k(x_1)+M_{k}(x_2)+\cdots}}.$$ Now, $n_k=\sum_r M_k(x_r)=\frac{1}{k}\sum_{d|k}p_d(x_1,x_2,\dots) \mu(k/d)$, so to finish off we need to express the power sum symmetric functions, $p_k$, in terms of the $h_k$'s. This is given by a variation of the Newton identities $$p_n(h_1,h_2,\dots)=(-1)^{n-1}\begin{vmatrix}h_1 & 1 & 0 & \cdots & \\ 2h_2 & h_1 & 1 & 0 & \cdots \\ 3h_3 & h_2 & h_1 & 1 & \\ \vdots & & \\ nh_n & h_{n-1} & \cdots & & h_1\end{vmatrix}.$$

So the polynomials you're looking for are $$n_k(f_1,f_2,\dots)=\frac{1}{k}\sum_{d|k}\mu(k/d)p_d(f_1,f_2,\dots).$$

Hope this helps.