I've encountered the following problem while doing research (trying to invert a Taylor expansion). Let $$ c_{\alpha} = \begin{cases} \frac{1}{k!}, &\mbox{if } \alpha = k e_k = (0,\dots,0,k,0,\dots,0), \\ 0, &\mbox{otherwise}, \end{cases}$$ where $e_k$ is the $k$-th standard basis vector. For any given $M\in \mathbb{N}$, the goal is to find coefficients $d_{\beta}$ such that $$\sum_{\alpha,\beta\in \mathbb{N}_0^{\times M} : \, \alpha + \beta = \gamma} c_{\alpha} d_{\beta} = \begin{cases} 1, &\mbox{if } \gamma = (0,0,\dots,0), \\ 0, &\mbox{otherwise}. \end{cases}$$ If $\gamma$ has exactly one nonzero component , then the choice $d_{\beta} = \frac{(-1)^{\sum_{i=1}^M \beta_i}}{\prod_{i=1}^M \beta_i !}$ works because of the binomial identity $\sum_{j=0}^n (-1)^j \binom{n}{j} = 0$, but it doesn't work in general. I can't figure out the general solution. Any ideas ?

  • $\begingroup$ If you use multi-indices, it means that you deal with Taylor expansions with n variables. Wouldn't it be simpler to begin by $n=2$ ? $\endgroup$
    – Jean Marie
    May 2 at 15:21
  • $\begingroup$ Possibly connected: hindawi.com/journals/aaa/2013/310679 $\endgroup$
    – Jean Marie
    May 2 at 15:27
  • $\begingroup$ @JeanMarie Sure, you can assume $M = 2$ if you want, but I still don't see the general solution in that case. $\endgroup$ May 2 at 15:35
  • $\begingroup$ Possibly connected as well: math.stackexchange.com/q/3492355 $\endgroup$
    – Jean Marie
    May 2 at 15:47
  • 1
    $\begingroup$ @JeanMarie This is the original problem: math.stackexchange.com/questions/3908571/… The multivariate Lagrange theorem looks very complicated to apply. In the specific case here, it looks more like a combinatorics problem than anything else. $\endgroup$ May 2 at 15:56

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