Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Sorry, I misread your equation as being a product:( –  plusepsilon.de Apr 2 '13 at 9:07
1  
I have now given an answer. In the case that the expression is in the argument, it will be clearly no! –  plusepsilon.de Apr 2 '13 at 9:14
    
Yes. You make things clearer. –  Shpigle Apr 2 '13 at 9:17

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
"Coincidentally", math.stackexchange.com/questions/351944/… –  Yemon Choi Apr 5 '13 at 10:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.