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Hi everyone.

Could you plz tell me where the zeros of $f(s)$ in the strip $\{0 < \Re s < 1 \}$ are ?

Do they all have $\Re s= 1/2$ ?

$$f(s) = 1 - 2^{-s} - 3^{-s} + 4^{-s} - 5^{-s} + ...$$

$$= \sum a(n)/n^s$$

with $a(n) = -1\\ $ if $n = -1 \mod 3\\ $ or $n = -1 \mod 4,\\ $ $a(n) = 1\\ $ otherwise.

$$f(s) = \zeta(s) - 2 \big[ 3^{-s} \zeta(s,-1/3) + 4^{-s} \zeta(s,-1/4) - 12^{-s} \zeta(s,-1/12) \big]$$

Could you plz give me some zeros in the strip $\{0 < \Re z < 1 \}$?



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closed as too localized by Robin Chapman, Mark Sapir, Loop Space, Harald Hanche-Olsen, Harry Gindi Nov 14 '10 at 22:48

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Ouch, I thought at first that this reduced to RH but now I'm not so sure. It would also help if it were translated into English: my dictionary lacks "plz" for instance. – Robin Chapman Nov 13 '10 at 12:15
I don't think there will be any good description of the zeroes of this function. If you made $a(n)$ be $-1$ if $n$ is $-1$ modulo one of $3$ and $4$, but $1$ if $n$ is $-1$ mod both of them, then this would be some simple modification of the $L$-function for a Dirichlet character modulo $12$. But, as is, this is some linear combination of $L$ functions, so I don't see any way to understand its zeroes. – David Speyer Nov 13 '10 at 12:44
There should be some motivation. Why do you want information about this combination of Dirichlet L-functions? – Mark Sapir Nov 13 '10 at 13:03
I think the general point that Scott and I am making is that, if you just choose some random $a(n)$ and look at the zeroes of $\sum a(n)/n^s$, there is no reason to expect there to be any good control over the zeroes you get. The examples that work do so because they come from interesting cohomological or number theoretic constructions. Your function isn't even multiplicative, so it doesn't have an Euler product! (Note that $a(5)=-1$, $a(7)=-1$ and $a(35)=-1$.) I'm not one of the people voting to close, but I think that, without more motivation, you're not likely to get a better answer. – David Speyer Nov 13 '10 at 13:53
I should also add that phrasing such as "can you plz give me some zeros" is a bit off-putting to some of us. It makes it sound like you view us as your teachers or as your tech support – Yemon Choi Nov 13 '10 at 18:25

The question for the case of a linear combination of Dirichlet L-series is actually easier than the case of a single L-function (Since RH is not known). In fact in each strip $1/2 \leq \sigma_1 <\Re(s)<\sigma_2 \leq 1$ there exists $\gg T$ zeroes for $-T < \Im(s) < T$. This follows by e.g. the Joint Voronin universality theorem for Dirichlet L-functions of Bagchi (A good reference for these results is Jörn Steuding's SLN 1877 "Value Distribution of L-functions").

Update Nov 14. I found the recent paper of Saias and Weingartner "Zeros of Dirichlet series with periodic coefficients", Acta Arithmetica 2009 where they get the same results that I indicated above, but also that there exists zeros to the right of the critical strip. Namely there exists some $\eta>0$ such that there are $\gg T$ zeros in any strip $1 \leq \sigma_0 <\Re(s) < \sigma_1 \leq 1+\eta$. This is actually simpler to prove since the Dirichlet series is absolutely convergent and the joint universality result is not needed, and more classical results of Bohr can be used instead.

Regarding zeros on the left of the critical line. The same result should hold in that in any vertical strip there exists $\gg T$ zeros. While this is not done in Saias-Weingartner as far as I can see it follows from the functional equation and using joint universality for the L-series in $1/2<\Re(s)<1$. Now we have two different functional equations depending on whether the L-series is odd or even it differs slightly in the Gamma-factors (this is the reason why the argument in my first answer is not applicable. See below). However Stirlings formula should imply that they do not differ sufficiently for this argument not to hold.

Further results we can get unconditionally is that there are about $T \log T$ with imaginary part less than $T$. It is not too difficult to prove that if we have a closed vertical strip that does not include the critical line, that the right order of magnitude actually is $T$, from which it would follow that for any open vertical strip including the critical line would have $T\log T$, i.e. the majority of the zeros, so the zeros should cluster around the critical line. Explicit results in this direction are included in the paper of Jörn Steuding "On Dirichlet series with periodic coefficients", Ramanujan Journal 2002 where he proves these results, i.e. clustering around $\Re(s)=1/2$, as well as other estimates (Another related paper is Garunkstis-Steuding "On the zero distribution of the Lerch zeta-function" where they prove corresponding results for the Lerch zeta function).

However to prove that they lie exactly on the critical line I believe that they must satisfy the same functional equation (see below) so an analogue of the Hardy function can be found and worked with. Therefore it is not clear (I am not sure about this though) that there should be any zeros exactly on the critical line (at least in order for there to be zeros on any particular line there should be a reason for it, since the zeros are countable, but the reals in an intervals are uncountable). Numerical experiments are welcome (I am not doing them though.).

Edit after comment of John below: I had originally thought that Bombieri and Hejhal's, and Hejhal's and Selberg's later results on linear combinations of L-functions would have applied on this problem, but as John pointed out below, this should not be the case, since the L-functions have to have the same Functional equation. Selberg's latest (unpublished) result would have yielded a positive proportion (order of $T \log T$ of zeros on the critical line), and Bombieri-Hejhal's (conditional on RH and weak Montgomery pair correlation conjecture) would have yielded the true asymptotics, if this would have been the case.

I checked one of Hejhal's papers on this subject, and John is right in his comment below. The condition to apply this method is that the Gamma-factors in the functional equation and the modulus are the same. When we consider a linear combination of Dirichlet L-functions we have to have a combination of only odd or even Dirichlet characters and the same modulus. For the Hurwitz zeta-function of rational parameters all Dirichlet characters will appear and thus this example is not of this type.

Thus I do not have an answer regarding zeros on the critical line. However the argument that shows that we have at least the order of $T$ zeros in any vertical strip $1/2 \leq \sigma_1<\Re(s)<\sigma_2 \leq 1+\eta$ with imaginary part less than $T$ still holds. Thus at least we know that Riemann hypothesis is not true for this function.

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Well, so much for Scott and my pessimism. Nice answer! – David Speyer Nov 13 '10 at 15:43
@Johan: The function $f(s)$ in the question can be written as a linear combination of Dirichlet L-functions, but it cannot be written as a linear combination of Dirichlet L-functions which all have the same functional equation (i.e. all of the same modulus and are all either even or odd). I believe that if you look at the results of Hejahl and of Bombieri & Hejhal, you will see that they are always assuming that the linear combination of L-functions satisfy to same functional equation. In particular, I do not believe that their results can be used to study the zeros of Hurwitz zeta-functions. – Micah Milinovich Nov 13 '10 at 16:44
(continued) I am currently without access to MathSciNet, but I believe Gonek has conjectured that $\zeta(s,a)$ has at $\ll T$ on the line Re $s=1/2$ is $a\neq 1/2$ is a rational number with $0<a<1$. – Micah Milinovich Nov 13 '10 at 16:47
John, Thanks for your comment. I think you are right. I have listen to seminars of Dennis Hejhal on the subject, but not really worked on the problem myself. The use of Voronin universality to prove that there are $\gg T$ zeros for strips to the right on the critical line still holds though, so that RH is not true. I will change my answer to reflect this – Johan Andersson Nov 13 '10 at 16:51
eta: in some cultures it is considered impolite to get other people to do your work for you, especially if you do not consider that they may have to spend effort on it. – Yemon Choi Nov 14 '10 at 21:37

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