i have to add some conditions and ideas :
if f(z) = f(g(z)) then we have product terms of the form 1 + b_n g(z)^n.
those product terms have different zero's despite f(z) = f(g(z)).
also it seems if f ' (z) = 0 our product form does not work.
so i propose the following condition.
the product form of f(z) converges to the correct value in Q if f(z) =/= 0 and f(z) is univalent in Q.
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it would be nice to consider product forms of f^[n](z).
\( f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots \)
is the expression for f(z) = e^z - 1 and it would be nice to have a similar looking product form ...
if f(z) = f(g(z)) then we have product terms of the form 1 + b_n g(z)^n.
those product terms have different zero's despite f(z) = f(g(z)).
also it seems if f ' (z) = 0 our product form does not work.
so i propose the following condition.
the product form of f(z) converges to the correct value in Q if f(z) =/= 0 and f(z) is univalent in Q.
***
it would be nice to consider product forms of f^[n](z).
\( f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots \)
is the expression for f(z) = e^z - 1 and it would be nice to have a similar looking product form ...