I was wondering about this. Consider a formal power series

$$\sum_{n=1}^{\infty} a_n x^n$$.

We can find its formal exponential, given by

$$\exp\left(\sum_{n=1}^{\infty} a_n x^n\right) = \sum_{n=0}^{\infty} \frac{B_n(1! a_1, 2! a_2, \cdots, n! a_n)}{n!} x^n$$

where $B_n$ is a complete Bell polynomial.

So what I wonder is, is there some kind of generalization, some kind of "super Bell polynomials" if you will, for a multivariate series like

$$\sum_{\substack{(n,k) \in \mathbb{N}^2 \\ (n,k) \ne (0,0)}} a_{n,k} x^n y^k$$


EDIT 1: Actually, I'm interested more in the case where the indices are like

$$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} a_{n,k} x^n y^k$$.

EDIT 2: I'm looking for a way to isolate the $a_{n,k}$ (that is, have a formula for them like we can get via the Bell polynomials in the univariate case.).

  • $\begingroup$ How do we obtain/prove the second equality? $\endgroup$
    – lynx
    Aug 26 '21 at 14:41

Sure. Just allow the $a_n$ to be formal power series in $y$. (The identity you describe is valid for $a_n$ in any commutative ring, or equivalently for $a_n$ a sequence of formal variables.)

  • $\begingroup$ Thanks, but I'm looking for a way to isolate the $a_{n,k}$, like how we can isolate the coefficients in the univariate case and express them in terms of Bell polynomials. I just edited the question to reflect this. $\endgroup$ Sep 3 '12 at 6:14
  • $\begingroup$ You get one from substituting the corresponding formal power series in $y$ into the Bell polynomials. I don't think it's particularly helpful to write out what this is explicitly. $\endgroup$ Sep 3 '12 at 6:20

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