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Let $f$ be an analytic function verfifying

$f(s)=\epsilon f(2-s)$

where $\epsilon=\pm 1$. The expression of Hasse-Weil L-function $f$ is

$$f(s)=N^{s/2}(2\pi)^{-s}\Gamma(s)\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$$

where $N$ is an integer and $\Gamma(s)$ is the gamma function.

Let $r$ be an integer. I have a set of equations of the form

$$f^{(k)}\left(1-2\prod_{j=1}^{k}s_{j}\right)=f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)$$

for all $k=1,...,r$. Here $f^{(k)}$ is the $k$-th derivative of $f$.

Can I deduce that

$$(t_1,t_2,\ldots,t_{r})=(s_1,s_2,\ldots,s_{r})$$

under some conditions on the derivatives of $f$?

The injectivity is not possible for this case.

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  • $\begingroup$ Sorry, I misread your equation as being a product:( $\endgroup$
    – Marc Palm
    Apr 2, 2013 at 9:07
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    $\begingroup$ I have now given an answer. In the case that the expression is in the argument, it will be clearly no! $\endgroup$
    – Marc Palm
    Apr 2, 2013 at 9:14
  • $\begingroup$ Yes. You make things clearer. $\endgroup$ Apr 2, 2013 at 9:17

1 Answer 1

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No, that is impossible. The $k$-th derivative of a L function has necessarily infinitely many zeros. So you can choose $s_j$ and $t_j$ inductively such that the products give distinct zeros of $f^j$. Moreover, if one of $s_j$ is zero you can't say anything clever either, but I assume that you simply have forgotten that condition in your question.

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