Hadamard's product formula for the derivative

Question

Let $f$ be an entire function of order $ρ<\infty$. Assume that $f$ does not vanish identically on $\mathbb{C}$. Then, we know that $f$ has a Hadamard's product formula

$$ f(s) =e^{g(s)}s^{r}\prod _ {k=1}^{\infty}\frac{s _ {k}-s}{s _ {k}} e^{s/s _ k} $$

the integer $r$ is the order of vanishing of $f$ at $s=0$, the $s_{k}$ are the other zeros of $f$ listed with multiplicity, $g$ is a polynomial of degree at most $ρ$, and the product converges uniformly in bounded subsets of $ℂ$. My question is how I can deduce directely a Hadamard's product formula for the derivative $f^′$ from the one of the function $f$.

Answer 1

(1) It seems your formula for the Hadamard product is only correct for $\rho<2$; more generally the exponent of $e^{s/s_k}$ contains a power series in $s$ of order $q={\rm Int}\;\rho$; see for example Eq. 1 in these lecture notes.

(2) To find a similar expression for $f'$, just take the logarithmic derivative:

$$ f'(s)/f(s)=g'(s)+r/s+\sum_{k=1}^{\infty}\frac{(s/s_k)^q}{s-s_k}.$$

Answer 2

The statement of the question must be corrected. First, as Carlo pointed out, the Hadamard representation as in the question is not valid for all functions of finite order. The correct Hadamard representation is $$f(z)=z^me^{P(z)}\prod_{n=1}^\infty \left( 1-\frac{z}{z_n}\right) \exp\left(\frac{z}{z_n}+\ldots+\frac{1}{q}\left(\frac{z}{z_n}\right)^q\right).$$ Here $q$ is the genus of the function.

You have to specify whether you are talking about functions of finite order (and thus finite genus) or functions of genus $1$.

Second, what does it mean "to deduce" a representation for the derivative? The derivative of a function of finite order is of finite order, so there is a similar representation for the derivative. To "find" it means to find the zeros of derivative in terms of zeros of the function, and to find the number $q$ and polynomial $P$.

Can you "deduce" the zeros of derivative of a polynomial in terms of zeros of this polynomial?

Of course, by taking log and differentiating the Hadamard formula, you obtain a formula for $f'$ which Carlo wrote, but this is not the Hadamard representation of $f'$.

By the way, in the beginning of 20-s century, the question of whether the genus of $f$ is the same as that of $f'$ was intensively discussed. If I remember correctly, it can be different.