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Question about Riemann Hypothesis

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#1 h May 26, 2021, 5:09 AM • 2 Y

Y by centslordm, etvat

I am just learning about the Riemann Hypothesis and a question that has been nagging me is why can't we just find the zeroes of the functional equation and see if it holds against the hypothesis? It seems to me that finding the zeroes of that equations would easily prove/disprove the hypothesis. But obviously, it's more complex (no pun intended) than that. Can someone provide a glimpse into the difficulty of the problem?

Thanks again

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#2 h May 26, 2021, 6:02 AM • 5 Y

Y by MathNinja7, hashtagmath, centslordm, Mango247, Mango247

I suppose that by the functional equation you mean Riemann's Functional Equation for the Zeta Function:
$\pi^{-s/2} \Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-(1-s)/2} \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s).$ If so, I don't understand what you mean by "finding the zeroes of that equation".
If we ignore for the moment the powers of $\pi$ and the Gamma Function (which we can assume to be well-understood, in particular regardings its zeroes and poles), the functional equation relates the values of $\zeta(s)$ and $\zeta(1-s)$ .
In particular, it tells us that a zero of $\zeta$ at $s$ gives us a zero at $1-s$ (unless we are at a pole of the Gamma Function which would cancel the zero, this happens exactly at the even negative integers, giving the so-called "trivial zeroes").
So we do know that the non-trivial zeroes of $\zeta$ possess some structure: For $s$ a zero, also $1-s$ is a zero.
We also know that since $\overline{\zeta(s)}=\zeta(\overline{s})$ , for $s$ a zero, also $\overline{s}$ is a zero.
So the zeroes are symmetrical around $s=\frac{1}{2}$ in several ways.
This explains naturally why something interesting could happen at the line $\text{Re}(s)=\frac{1}{2}$ but of course it does not tell us at all whether the zeroes are on that line or not.

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#3 h May 26, 2021, 4:17 PM • 1 Y

Y by centslordm

I see, thank you for your thorough explanation.

I was actually thinking about the equation $\zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$ (which I assume is a manipulation of the equation in your post). The trivial zeroes are completely obvious in the $\sin$ part of the equation. But I now I understand the difficult part about proving the zeroes lie on one line is the part such that $\zeta(s) = \zeta(1-s)?$ .

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#4 h May 26, 2021, 8:03 PM • 2 Y

Y by hashtagmath, centslordm

Indeed your version is just a rearrangement of mine (the other one is more natural, for instance because it generalizes more nicely to Dirichlet L-functions). However I don't understand what you mean by "the part such that $\zeta(s)=\zeta(1-s)$ ".
Of course, when $\zeta(s)=0$ , the equation $\zeta(s)=\zeta(1-s)$ is true, but I don't see why you would study exactly this equation.

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#5 h May 27, 2021, 12:21 AM • 2 Y

Y by centslordm, Mango247

It may be a misunderstanding, but wouldn't finding the zeroes of this functional equation directly prove/disprove if the hypothesis is true?

Also, I meant to say, wouldn't studying the behavior of $\zeta(1-s)$ help in finding when the equation is equal to zero? Intuitively, I would expect that part is the most nontrivial as it's almost like a recursive function

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#6 h May 27, 2021, 5:45 AM • 2 Y

Y by hashtagmath, centslordm

Well, as you maybe can see better from my version of the functional equation, this is really not a recurrence, but a symmetry:
You can relate $\zeta(s)$ to $\zeta(1-s)$ and vice versa, but applying it twice really does not give you anything interesting, but a tautology!
It doesn't simplify anything to work with $1-s$ instead of $s$ , this is just a change of variables.
Of course there are regions where $\zeta$ is well-understood. For instance, when $\text{Re}(s)>1$ we have the Euler product from which one can read off that there are no zeroes. One can then use the functional equation to "translate" this to the region $\text{Re}(s)<0$ where there will only be the trivial zeroes.
This leaves you with the so-called "critical strip" $0<\text{Re}(s)<1$ . Again you could translate from $\frac{1}{2}<\text{Re}(s)<1$ to $0<\text{Re}(s)<\frac{1}{2}$ and vice versa. Indeed, if you can prove that one of them does not contain any zeroes, then the other won't as well. The point is that we understand neither of the two regions and neither of them is a priori simpler than the other!

Maybe I should also mention that it would already be a major advance (most likely winning you a Fields medal, unless you miss the age restriction) to prove that there are no zeroes with $0.999999<\text{Re}(s)<1$ .

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#7 h May 27, 2021, 4:30 PM • 4 Y

Y by hashtagmath, Mango247, Mango247, Mango247

Not sure if I'm misunderstanding your notion, but just finding the zeros to $\zeta(s)$ is not sufficient. Computers have been able to detect over a quadrillion non-trivial zeros, all of which have a real component of $\frac{1}{2}$ . Just one counterexample would disprove the hypothesis, but until we can mathematically prove that all of the non-trivial zeros have this property, the hypothesis will remain hypothetical.

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#8 h May 27, 2021, 4:36 PM • 2 Y

Y by centslordm, Mango247

Tintarn wrote:

Well, as you maybe can see better from my version of the functional equation, this is really not a recurrence, but a symmetry:
You can relate $\zeta(s)$ to $\zeta(1-s)$ and vice versa, but applying it twice really does not give you anything interesting, but a tautology!
It doesn't simplify anything to work with $1-s$ instead of $s$ , this is just a change of variables.
Of course there are regions where $\zeta$ is well-understood. For instance, when $\text{Re}(s)>1$ we have the Euler product from which one can read off that there are no zeroes. One can then use the functional equation to "translate" this to the region $\text{Re}(s)<0$ where there will only be the trivial zeroes.
This leaves you with the so-called "critical strip" $0<\text{Re}(s)<1$ . Again you could translate from $\frac{1}{2}<\text{Re}(s)<1$ to $0<\text{Re}(s)<\frac{1}{2}$ and vice versa. Indeed, if you can prove that one of them does not contain any zeroes, then the other won't as well. The point is that we understand neither of the two regions and neither of them is a priori simpler than the other!

Maybe I should also mention that it would already be a major advance (most likely winning you a Fields medal, unless you miss the age restriction) to prove that there are no zeroes with $0.999999<\text{Re}(s)<1$ .

Thanks for the explanation!

naenaendr wrote:

Not sure if I'm misunderstanding your notion, but just finding the zeros to $\zeta(s)$ is not sufficient. Computers have been able to detect over a quadrillion non-trivial zeros, all of which have a real component of $\frac{1}{2}$ . Just one counterexample would disprove the hypothesis, but until we can mathematically prove that all of the non-trivial zeros have this property, the hypothesis will remain hypothetical.

I was originally thinking to find ALL zeroes to the Riemann zeta function, it may be sufficient to set each individual part of the function equation (i.e. $2^s$ , $\pi^{s-1}$ , etc.) equal to zero and find the values of $s$ such that it is true. But it seems like the $\zeta(1-s)$ creates a symmetry which makes it more difficult to solve. Is my understanding correct?

And yes, you are correct about the large amount of non-trivial zeroes; in fact, Hardy proved that there are infinitely many zeroes on the critical line.

A follow-up question that I may ask about this is, would all the zeroes of the Riemann zeta function arise from the functional equation? In other words, is every zero check/verified/found through finding when the functional equation equals zero?

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#9 h May 27, 2021, 5:10 PM • 2 Y

Y by centslordm, hashtagmath

That would be an extremely challenging way to calculate zeros. Instead, computers input complex numbers into the function and notice if they are trending closer towards $0$ . It's just like how computers calculate limits. So they are fund randomly, not from actual calculation. The zeta function is much more irregular than a polynomial like $f(x)=x^3-2x^2+1$ . We currently do not have a way of calculating zeros of zeta directly. They are just detected

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#10 h May 27, 2021, 6:26 PM

Y by

Interesting, is there a known pattern for the distribution of zeroes on the critical line? Or are they irregularly distributed?

Also, separately, is the Riemann functional equation the only possible analytic continuation of the Riemann zeta function $\left(\zeta(s) = \sum_{n = 1}^{\infty}\frac{1}{n^s}\right)?$ In other words, does there exist a possibility for a different equation to extend the domain of the Riemann Zeta function?

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#11 h May 27, 2021, 6:49 PM • 1 Y

Y by hashtagmath

Yes, any meromorphic function defined on a non-empty open set has at most one analytic continuation (Proof: The difference of two such analytic continuations vanishes on a non-empty open set hence is identically zero.) Nothing special about $\zeta$ here.
Regarding the zeroes: I think there is some fundamental misunderstanding. We don't (and we cannot!) "find" the zeroes from the functional equation. Again, regarding the zeroes inside the critical strip, the functional equation tells us that if $\zeta(s)=0$ then $\zeta(1-s)=0$ . No more, no less. The other factors (the power of $\pi$ and the sine) are completely irrelevant for that purpose.
Instead there are clever methods that actually allow you to check for any finite number of zeroes whether they are on the critical line or not. But as already mentioned several times in this thread now, there are infinitely many zeroes so we will never prove the Hypothesis in this way.
Regarding the distribution: There is no obvious pattern, but there is some regularity. One can estimate how many zeroes roughly are there up to a certain imaginary part. Also, there are very deep conjectures about the "fine-scale statistics" i.e. how the distances between successive zeroes should behave, but this is all very speculative and certainly at least as hard to prove as the Riemann Hypothesis. (At this point we cannot even decide whether there could be a double zero, or whether the imaginary part of some of the zeroes is rational etc... we certainly expect that all of this does not happen and that these are essentially "random" numbers on the critical line.)

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#12 h May 28, 2021, 4:42 AM • 5 Y

Y by hashtagmath, PhysKid11, centslordm, Mango247, Mango247

Just to elaborate on the above re analytic continuation, there is a theorem in complex analysis that I like to call "rigidity" ~~this post became longer than i thought it would be, so treat this as an advertisement for learning complex analysis because it's SO cool~~:

Definition (Accumulation Point): Let $S$ be a set of complex numbers. Then $z_0 \in \mathbb{C}$ is an accumulation point of $S$ if for every $r > 0$ , I can find $z \in S$ with $z \ne z_0$ such that $|z-z_0| < r$ .

For example, the set $\{1,1/2,1/4,1/8,\cdots\}$ has an accumulation point, namely $0$ . (The definition of accumulation point generalizes to arbitrary metric spaces, which you'll encounter very often in real analysis.)

Definition (Holomorphic): Let $\Omega \subseteq \mathbb{C}$ be an open region. Then $f:\Omega \to \mathbb{C}$ is holomorphic if it is "differentiable"; that is the limit $\lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}$ exists for all $z_0 \in \Omega$ .

This may sound boring for such a fancy word, but the fact that the limit holds over complex numbers actually spawns a crazy amount of properties that makes the word "holomorphic" a hundred time fancier than "differentiable". For example:
- If you're holomorphic, you're also analytic, i.e. have a power series representation. (Proof left as a very hard exercise, pirate a textbook for elaboration)
- If you're analytic, you're also holomorphic... (Left as an exercise)
- Shocking Conclusion: Any complex function that can be differentiated (i.e. is holomorphic) can be differentiated infinitely many times.

Note that the shocking conclusion above does not hold in real analysis, e.g. the function $f(x) = \begin{cases}x^2, & x \geq 0 \\ -x^2, x < 0\end{cases}$ can only be differentiated once. This shows that complex analysis is super lit.

These facts also show that being holomorphic and being analytic is the same thing in the complex world, so these words are used interchangeably when we're talking about complex analysis.

Theorem (Rigidity): Let $\Omega$ be an open region, $f:\Omega \to \mathbb{C}$ be holomorphic. Suppose the zeroes of $f$ (i.e. the set $\{z \in \Omega : f(z) = 0\}$ ) has an accumulation point. Then $f$ is the zero function.

Proof. Pirate a textbook. $\blacksquare$

For example, suppose $f:\mathbb{C} \to \mathbb{C}$ is holomorphic.
- If I know that $f$ is zero everywhere in the unit disk $\mathbb{D} := \{z \in \mathbb{C} : |z| < 1\}$ , then $f$ has to be zero everywhere else because $\mathbb{D}$ has an accumulation point (in fact any point in the closure $\overline{\mathbb{D}} = \{z \in \mathbb{C} : |z| \leq 1\}$ is an accumulation point).
- If I know that $f(i/n) = 0$ for all $n \in \mathbb{N}$ , then $f$ is zero everywhere else too.

But why do we care?

Corollary: Let $\Omega$ be an open region, $f,g:\Omega \to \mathbb{C}$ be holomorphic. Suppose the set for which $f = g$ has an accumulation point. Then $f \equiv g$ everywhere.

Proof. Apply the rigidity theorem on the holomorphic function $f-g$ . $\blacksquare$

This lets us show that analytic continuations are unique.

Definition (Analytic Continuation): Let $\Omega_1,\Omega_2$ be open regions with $\Omega_1 \subset \Omega_2$ . Let $f_1:\Omega_1 \to \mathbb{C}$ and $f_2,\Omega_2 \to \mathbb{C}$ be holomorphic. We say that $f_2$ is the analytic continuation of $f_1$ if $f_1$ and $f_2$ agree on $\Omega_1$ , i.e. $f_1(z) = f_2(z) \quad \forall z \in \Omega_1$ .

Theorem: Analytic continuations are unique.

Proof. Unless you're doing something incredibly dumb, $\Omega_1$ is non-empty, and by virtue of being open (look this up if you're unfamiliar) it definitely has an accumulation point. So if $f_2,f_3$ are both continuations of $f_1$ to $\Omega_2$ , then surely $f_2,f_3$ must agree on $\Omega_1$ . By the corollary it follows that $f_2$ and $f_3$ agree everywhere, i.e. they're the same analytic continuation. $\blacksquare$

With regards to the Riemann Zeta function, we have the "original" definition:
$\zeta(s) := \sum_{n=1}^\infty \frac{1}{n^s}$ Where does this sum actually converge? Write $s = \sigma + it$ . Then if $\sigma > 1$ then:
$\left|\sum_{n=1}^N \frac{1}{n^s}\right| \leq \sum_{n=1}^N \frac{1}{|n^s|} = \sum_{n=1}^N \frac{1}{|e^{\log(n)\sigma + i\log(n)t}|} = \sum_{n=1}^N \frac{1}{e^{\log(n)\sigma}} = \sum_{n=1}^N \frac{1}{n^\sigma}$ Any I think by sending $N \to \infty$ the RHS converges via an integral test or something. Then absolute convergence implies convergence, so it follows that $\zeta(s)$ converges (and is thus well-defined) for all $s$ with real part greater than $1$ .

But these silly mathematicians somehow defined $\zeta(-1)$ and stuff, how could they possibly do that? The short version is that using black magic, you can prove that
$\zeta(s) = \frac{\pi^{s/2}\xi(s)}{\Gamma(s/2)} \qquad \forall \text{Re}(s) > 1$ and that using additional black magic you can show that $1/\Gamma$ extends everywhere (this proof is actually pretty cool) and $\xi$ extends everywhere except at $s=0,1$ or something, idk I don't remember the proof and this is above my paygrade of zero dollars an hour. Anyways what's important is that an analytic continuation of $\zeta(s)$ exists over $\mathbb{C} \setminus \{1\}$ . And by the previous theorem, it's the only possible analytic continuation!

tl;dr pls study complex analysis its fun i promise ~~might want to learn some real analysis first though~~

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