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Here $\zeta(s)$ is the usual Riemann zeta function, defined as $\sum_{n=1}^\infty n^{-s}$ for $\Re(s)>1$.

Let $A_n=${$s\;:\;\zeta(s)=n$}. The behaviour of $A_0$ is basically just the Riemann hypothesis; my question concerns $A_n$ for $n\neq0$.

1) Is determining this just as hard as the Riemann hypothesis?

2) If we know the behaviour of some $A_n$, does it help in deducing the behaviour of other $A_m$?

3) For which $n$ is $A_n$ non-empty?

Question 3 has now been answered for all strictly positive $n$ - it is non-empty, and has points on the real line to the left of $s=1$. For $n=0$, it is known to be non-empty. Any idea for negative $n$? (the same answer won't work, since $\zeta(s)$ is strictly positive on the real line to left of $s=1$. Big Picard gives it non-empty for all but at most one $n$. How can we remove the 'at most one'?

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For (1) answer is probably that it is harder problem (there is less structure). The (2) is probably a wishful thinking, but hard to refute. (3) is almost solved by Big Picard. –  Boris Bukh Dec 2 '09 at 16:11
    
I'd also tag this as analytic-number-theory. –  Ben Weiss Dec 2 '09 at 16:36
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Question (1) is interesting to me. I have the impression that computational techniques are such that it should be possible to plot a substantial portion of the inverse image of 1 (or 2, or 3....) under the zeta function: it would be interesting collect and look at the data and see if there are any visible patterns. –  Jonah Sinick Dec 2 '09 at 22:46
    
You say that "$\zeta(s)$ is strictly positive on the real line to left of $s=1$". In fact $\zeta(s)$ is strictly negative on $(0,1)$. This follows from $(1-2^{1-s})\zeta(s)=1-2^{-s}+3^{-s}-4^{-s}+\cdots$. In addition, $\lim_{s\to 1-}\zeta(s)=-\infty$ because $\zeta(s)=\frac{1}{s-1}+O(1)$ for $|s-1|<1$. –  GH from MO May 1 '11 at 1:54

9 Answers 9

up vote 25 down vote accepted

Regarding 3), this "Big Picard" stuff is serious overkill.

Think like an undergraduate real analysis student:

The p-series $\zeta(p)$ converges for real $p > 1$, whereas $\zeta(1)$ = sum of the harmonic series = oo.

An easy argument using (e.g.) the integral test shows that

$$lim_{p \rightarrow \infty} zeta(p) = 1$$

The function $\zeta(p)$ is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value $n \ge 2$ at least once -- and, since it is a decreasing function of p, exactly once -- on the real line.

Thus $A_n$ is nonempty for all $n > 1$.

EDIT: Let me show that zeta(s) takes on all real values infinitely many times on the negative real axis.

For this, note that for all $n > 0$,

$$\zeta(-(2n-1)) = - \frac{B_{2n}}{(2n)}$$,

where $B_{2n}$ is the $(2n)$th Bernoulli number. It is known that the $B_{2n}$'s alternate in sign and grow rapidly in absolute value:

$$|B_{2n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{(\pi e)^{2n}}\right)$$

The claim follows from this and the Intermediate Value Theorem.

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Isn't the the limit of $\zeta(p)$ as $p\to\infty$ 1? I mean, that certainly what it looks like from the Dirchlet series. Also, you get all negative integers on the real line below $p=1$. –  Ben Webster Dec 2 '09 at 22:04
    
Yes, of course it is -- I'll fix it. (This doesn't affect my argument.) –  Pete L. Clark Dec 2 '09 at 22:54
    
Oops, yes it does, a bit: it shows that zeta(s) does not take on the value 1 in the half plane Re(s) > 1. I'll fix this by looking at negative integer values, as you suggest. –  Pete L. Clark Dec 2 '09 at 23:02

There are infinitely many roots of $\zeta(s)-a=0$ for every complex number $a$. When $a\ne 0$, these are called "$a$-values" and there is a whole chapter discussing their distribution in Titchmarsh's book on the zeta-function. Selberg also discusses $a$-values in his (now famous) paper "Old and new conjectures and results about a class of Dirichlet series" where he defines the Selberg class.

Here is an overview of some of the important results:

1) There are $\frac{T}{2\pi}\log T + O(T)$ $a$-values of $\zeta(s)$ in the strip $0<\Im s\leq T$.

2) Like the zeros of $\zeta(s)$, Levinson proved that $a$-values cluster near the half-line. That is to say, almost all $a$-values are arbitrarily close to the half-line.

3) Unlike the zeros of $\zeta(s)$, there are provably a lot of $a$-values away from the half-line (though not a positive proportion). Namely, there are $\gg T$ roots of $\zeta(s)=a$ for $a\neq 0$ in any region $A\leq \Re s \leq B$ and $0<\Im s\leq T$ where $A\in (1/2,1)$ and $A$ strictly less than $B$. This is proved in Titchmarsh's book. On the other hand, standard zero-density estimates for the zeta-function tell us that there are $o(T)$ zeros in such a region. Some have suggested that this is evidence for the Riemann Hypothesis.

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Here is a link to Levinson's paper: pnas.org/content/72/4/… –  Micah Milinovich Aug 2 '10 at 16:10

About (1) I agree with Boris Bukh. There is no conjecture about the location of the a-points of the Riemann zeta function ($\zeta(s) = a$), in the way that we have for the 0-points. And by my next remark, there may well be no reasonable description of the m-points for $m \neq 0$.

About (2) I also agree with Boris Bukh, for a particular reason. There is a universality result for the Riemann zeta function, to the effect that as you move the disk $|s - 3/4| \leq 1/4$ vertically upwards, the Riemann zeta function can be made to approximate arbitrarily well in the sup-norm an arbitrary continuous function on the closed disk that is holomorphic in the open disk. It is just a matter of moving the disk far enough up. Since for an arbitrary holomorphic function there is no connection between the a-points and the b-points for a pair of values $a \neq b$, neither would you expect such a relationship for the Riemann zeta function.

The universality result is due to M. Voronin, see page 308 of the second edition of The Theory of the Riemann Zeta-function by E. C. Titchmarsh. It is crucial to get the second edition, with the end-of-chapter notes by D. R. Heath-Brown. This is the standard reference on the Riemann zeta function, though there is also a very useful book by Aleksandar Ivic.

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This is by no means my area of expertise, but doesn't $|\zeta(s)|$ get arbitrarily large along the negative real line between the trivial zeros? And also alternate in sign at the odd negative integers? This would imply (median value theorem!) that it hits every integer, with none of this "at most one exception" stuff.

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Yes, I modified my response above in this way before I noticed yours. (I'm upvoting it.) –  Pete L. Clark Dec 2 '09 at 23:11

There are in fact rather precise conjectures as to the location of roots $s$ of the equation $\zeta(s) = a$ for any $a$ complex (this includes $a = n$ of course).

For $a = 0$ this is the Riemann Hypothesis.

For $a \neq 0$ it is conjectured that half of the points cluster slightly to the left of the half-line (precisely $\sqrt{\log\log t}/\log t$ to the left, if $t$ is the imaginary part) while the other half clusters very closely to the half-line (possibly $o(1/\log t)$). Around each of these lines the zeros spread uniformly to the left and right. In particular this led Selberg to conjecture that $3/4$ of the $a$-points lie to the right of the half-line while $1/4$ lies to the right. Selberg complemented this with the conjecture that on any given line, there are at most two $a$-points. Selberg has in addition obtained a central limit theorem for the distribution of the $a$-points around $1/2 - c \sqrt{\log\log t}/\log t$ with $c$ varying. A good reference for these questions are the paper of Selberg mentionned above, the thesis of Kai-Man Tsang (written under Selberg), which can be found on the internet, and finally for some more recent developments I would refer you to the papers:

http://arxiv.org/abs/1402.0169

http://arxiv.org/abs/1402.6682

http://www.ams.org/journals/proc/2012-140-12/S0002-9939-2012-11275-4/home.html

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Let's look at $\zeta(s)$ for some large $\sigma$ (the real part of $s$). We can bound the function by $\int_1^\infty \frac{dx}{x^\sigma} + 1$, which is $\frac{\sigma}{\sigma-1}.$ So all these values in any $A_m$ ($m > 1$) can be bounded by a half-plane. This definitely doesn't pin down where they are, but does give a nice bound on where they are not.

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Well, for 3, it'd be all of them (well, at most one of them is empty, by Big Picard). The zeta function is meromorphic, but not rational, and so has an essential singularity, at infinity.

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Surely you mean "all of them (well, all but at most one of them)"? I.e. A_n is empty for at most one value of n. –  Tom Leinster Dec 2 '09 at 16:53
    
Ahh, yes, that'd be a typo as I dashed off to class. Correcting it now. –  Charles Siegel Dec 2 '09 at 18:07
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I don't understand other people's objection to using Big Picard. It's a big theorem, sure, but we know it's true and it enables (3) to be answered without calculation. –  Tom Leinster Dec 3 '09 at 0:40

Regarding Jonah's remark about plotting the inverse images of n. A quick way to visualize the zeros of a meromorphic function $f(s)$ is to plot (by color-coding with four colors) the quadrant of the value of $f(s)$. Points at the junction of four differently colored regions are either zero or poles. For $f(s) = \zeta(s) - n$ there is only one pole, so all the other 4-color junctions are inverse images of $n$. The plots I've looked at show some behavior that looks related to many other plots one has seen in connection with zeta. But the plot of the inverse image of the set of all (Gaussian) integers obtained from $f(s) = Mod[\zeta(s),1]$ (using $Mathematica$ notation) seems particularly interesting, especially (for example) in the three by three square with center 1, or in much smaller regions in the left half plane, for example, the square centered at $-25 + \frac 12 i$ with side length $10^{-5}$.

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There is entire chapter of a book written on this very subject (Titchmarsh, The Theory of The Riemann zeta-function, chapter XI)

In the rectangle 0 \leq Re(s) \leq 1,0< Im(s)>T zeros of \zeta(s)-a in the box 3/4< Re(s) < 4/5, 0

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The book recommendation already appears in Micah Milinovich's answer. –  S. Carnahan Jan 10 '12 at 8:43

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