4
$\begingroup$

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.).

$\endgroup$
0
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – The_Sympathizer 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. $\endgroup$ – Qiaochu Yuan Sep 3 '12 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.