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I have seen a proof outline that $\zeta$ has infinitely many zeros on the critical line here but what I really want is:

Simplest possible (least "magic") argument that explains why zeta has infinitely many zeros in the critical strip (including the edges) (note: I have already shown all the nontrivial zeros lies in there) as well as being able to fill in the rigorous details myself.

I have also studied the prime number theorem in which it is shown no zeros lie on the edges of the strip: I think there should be a proof that there are no zeros in the strip including the edges which comes logically before this and is much simpler.

I don't feel like I understand such a deep difficult result as Hardy's, so I really want something that I can break every step down. Thanks for your help.

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A simple intuitive thing might be: If there where only finitely many (notrivial) zeroes, then this number of zeroes would be a very special natural number - there are no special natural numbers. –  Hagen von Eitzen Feb 9 '13 at 13:22
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+1 for "intuitive explanation with rigorous details" :-) –  joriki Feb 9 '13 at 15:05
    
To the contrary, every natural number is special. For, suppose that some natural numbers are not special. Then there is a smallest natural number that is not special. But this number is special! Q.E.D. –  marty cohen Feb 10 '13 at 16:43

2 Answers 2

up vote 3 down vote accepted

One way to think about intuitively why there aren't a fininte number of non-trivial zeroes of $\zeta$ is that, if there were, then it would grow in a way that we know it doesn't. To show this, we use the completed zeta fucntion, which is defined as: $$\xi = \begin{cases} s \left({s - 1}\right) \pi^{-s/2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) & : \Re \left({s}\right) > 0 \\ \xi \left({1 - s}\right) & : \Re \left({s}\right) \le 0 \end{cases}$$

Some authors use a constant factor of $\frac12$ in this definition and others don't. It doesn't really matter. This is incredibly helpful, because we can ignore all of the trivial zeros, of $\zeta$, leaving only the non-trivial ones, because we know $\zeta(s) > 0$ for $\Re(s) > 1$ and the same obviously applies for the other factors of this function. The second part of this definition is incredibly useful as well, because we have by the same argument that $\xi >0$ for all $s$ with real part less than $0$. We therefore see that all trivial zeroes of the zeta function are not zeros of our new completed function, while all of the non-trivial zeroes are. Since other than poles at $0,1$ $\xi$ is non-vanishing where $\zeta$ is, we have that the statement: "there are an infinite number of nontrivial zeros to $\zeta$" is equivalent to "there are an infinite number of zeros for $\xi$ of any type."This already may seem overly abstract and distant from the problem, but we really are just looking at a completion of the same function, that isn't $0$ where we don't want it to be.

It can be shown, with a little difficulty concerning the order of $\xi$, by the Hadamard factorisation theorem that $$\xi(s) = \prod_\rho\left(1-\frac s\rho\right)$$ for its zeros $\rho$. This is a little trickly analytically, but the intuition is basically the same as that for the Weierstass factorization theorem. Plus, it's on Wikipedia. Now, if this function had a finite number of zeros, it would be a polynomial of degree, say $n$. But we know that asymptotically for a polynomial $P(x)$ that $\log(P(x)) \sim n\log(s)$ on the real line. This goes against what we know about $\zeta$, which gives that on $\Bbb{R}$ we have that $\log(\xi(s))\sim s\log (s).$ Therefore $\xi$ has an infinite number of zeros and $\zeta$ has an infinite number of non-trivial zeros.

I glossed over a really important point, which is that you can apply the factorisation theorem, and that you can arrive at the formula I gave. This wouldn't really assist much with intuition, however, as those parts are a little technical. You would be well off learning them though.

In the end, functions with finite order and with a finite number of zeros behave in a very particular way, and we know for certain that $\xi$ doesn't behave in this way. This $\xi$ contains information about the non-trivial zeros of $\zeta$, and the fact that it has an infinite number of zeros is enough to prove our claim.

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This is really nice argument and I get the ideas but I don't know how to do the bits you skip. For one thing, can we get the Hadamard factorization without knowing anything about the zeros? I've seen one way to get it but it required an asymptotic upper bound on the number of zeros –  user58512 Feb 9 '13 at 15:39
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Are you familiar with Hadamard's theorem about arbitrary entire functions of finite order? Try applying that to $(s)(s-1)\xi$. You can say the same thing, that $\xi = e^{cs}P(x)$ for some polynomial P(x), and the functional equation gives that c = 0. –  Sam DeHority Feb 9 '13 at 15:54
    
ok! I think the proof I saw uses asymptotics of the zeros to skip having to calculate the order of the function! thanks again –  user58512 Feb 9 '13 at 16:18
    
@user58512 I could write that out, but it's written out in full here: proofwiki.org/wiki/…. I'm glad I could help. –  Sam DeHority Feb 9 '13 at 16:36
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And truth, I'm glad to say, is in the mind of the beholder (to misquote the distinguished Professor Lehrer). When correctly viewed, everything is skewed: I could tell you things about Terence Tao, who's a wizard of math, but I don't know how. –  marty cohen Feb 10 '13 at 16:50

This is all from The Riemann Zeta-Function - Karatsuba and Voronin except the last bit where I show $\xi$ grows fast

Let $G$ be an analytic function of order 1.

Then by a theorem in complex analysis we can factorize it as $G(s) = s^{\nu(0)} e^{As+B}f(s)$ with $$f(s) = \prod_{\rho}\left(\left(1 - \frac{s}{\rho}\right)e^{s/\rho}\right)^{\nu(\rho)}$$ where $\rho$ are the zeros of $f$ and $\nu$ counts their multiplicity.


If it had finitely many zeros then

$$\begin{array}? \max_{|s|=R}|f(s)| &\le& \max_{|s| = R} \prod_{\rho}\left(1 + \frac{|s|}{|\rho|}\right)e^{|s|/|\rho|} \\ &\le& \prod_{\rho}\exp\left(2\frac{R}{\rho}\right) \\ &=& \exp\left(2R\sum_{\rho} \frac{1}{|\rho|}\right) \end{array}$$

so $$\log \max_{|s| = R}|G(s)| \le c (R+1)$$ for some constant $c > 0$.


$\xi$ is an entire function of order at most 1 since:

$$\begin{array}{rcl} |\zeta(s)s(s-1)| &=& O(|s|^3) \\ |\Gamma(s)| &\le& \exp(c|s| \log(|s|)) \\ |\pi^{-s/2}| &\le& \exp(c |s|) \end{array}$$

and $\xi(s) = \frac{1}{2}s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$

and by Stirlings approximation $\log \Gamma(s) \sim s \log s$ we find $\xi$ has order exactly 1.

This means we have such a factorization for $\xi$.


As real $s \to +\infty$ $$|\xi(s)| \ge e^{cs}$$ for any fixed $c > 0$ by stirlings approximation and lower bounding the zeta function by $\int_1^\infty \frac{dn}{n^s} = \frac{1}{s-1}$ which cancels out so you get something like $$\xi(s) = \tfrac{1}{2} s \sqrt{\frac{2\pi}{s}} e^{-s \log(\pi/2)} e^{s \log(s/e)} (1 + O(1/s))$$ which gives what we want due to the $e^{s \log(s)}$

This means that it cannot have finitely many zeros, and we have already seen that they are all in the critical strip.

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This depends on $\xi(0) \not = 0$ which Eric Naslund kindly showed how to compute here –  user58512 Feb 11 '13 at 16:25

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