Conider two power series is definded as follows :

$$ f(x) = \sum_{n=0}^\infty f_n x^n $$ $$ g(x) =\sum_{n=0}^\infty g_n x^n $$

Then , Hadamard product of $f$ and $g$:

$$ (f\odot g)(x) =\sum_{n=0}^\infty f_n g_n x^n $$ $$= (1/2π) \int_0^{2π}f(\sqrt{x}e^{it})g(\sqrt{x}e^{-it})dt $$

Questions :

  1. consider the situation where we know everything about $(f\odot g)$ and $g$ then how to find (constructively) $f$ ? (Equivalent "Hadamard Division")
  2. In general can we construct inter-relative "Hadamard Algebra" ?
  3. Can we get upper bound on $f$ if we know the upper bounds on $(f\odot g)$ and we know everything ( poles and zeroes ) about $g$

Note : I'm seeing in comment section there is an induction of trivial confusion due to my insufficiency of stating . By " we know everything about .." I mean we know the compact and series representation of them ( $(f\odot g)$ and $g$ ) . The problem is we only know infinite series representation of $f$ and due to nature of it's mysterious coefficients we can't get certain properties of $f$ . so I'm trying to get alternative representation (likely to be compact ) of $f$. Also all $f_n$ & $g_n$ are non zero .

  • $\begingroup$ If we "know everything" just divide $(f\odot g)_n$ by $g_n$. What is " inter-relative algebra"? There is a nice theory of Hadamard products for rational functions and algebraic functions. $\endgroup$
    – Conifold
    Jan 13, 2020 at 7:05
  • 1
    $\begingroup$ If there was such a formula you'd get an integral formula for "Hadamard reciprocal". But series with even a single zero coefficient have no reciprocals, so there is no regular representation of this sort. You have to recover the coefficients and divide them, if possible. $\endgroup$
    – Conifold
    Jan 13, 2020 at 10:55


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