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Has the following operation $*$ on formal power series $f,g$ been studied before?

$$[X^n](f*g) = n! \cdot [X^n]f \cdot [X^n]g$$

where $n$ is a nonnegative integer? This is the typical Hadamard product multiplied by $n!$.

Clearly the identity is $\exp$, our product is associative and commutative — this means infinitely differentiable functions without pole at $x=0$ form a monoid under our product.

Other straightforward identities (forgive my abuse of notation):

  • $e^{ax} * e^{bx} = e^{abx}$;
  • $(cf)*g = f*(cg) = c(f*g)$;
  • $\cosh x * e^{ix} = \cos x$; and
  • If $c_n = \frac{f^{(n+1)}(0)}{f^{(n)}(0)}$, then the exponential generating function $\mathrm{EG}(c_n; - ) * f = f'$.
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    $\begingroup$ note that in general this series is not convergent, even with analytic functions, e.g. $f=g=\frac{1}{1-x}$ gives $(f*g)(x)=\sum_{n=0}^\infty n!x^n$. Maybe you want to consider it as an operation for formal power series. $\endgroup$
    – Pietro Majer
    Commented Feb 6, 2020 at 19:50
  • $\begingroup$ I was thinking of that example, I guess I didn't know the name "formal power series". Will make the edit, thank you! $\endgroup$
    – Jake Lai
    Commented Feb 6, 2020 at 20:46
  • $\begingroup$ Dividing the coefficients of a series by $n!$ is the Borel transform $\mathcal{B}$. Your product is $f * g = \mathcal{B}(\mathcal{B}^{-1}(f) \cdot \mathcal{B}^{-1}(g))$ where I wrote the usual Hadamard product as $\cdot$. So all the properties of the Hadamard product translate to yours via $\mathcal{B}$ and $\mathcal{B}^{-1}$. $\endgroup$
    – Igor Khavkine
    Commented Feb 7, 2020 at 10:56

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