# Math Help - Thoughts needed for this "proof" of riemann s hypothesis

1. ## Thoughts needed for this "proof" of riemann s hypothesis

Using the norm

$| \frac {1}{1-1/p^{s}} |^2 =
\left( \frac {1}{1-1/p^{2a}} \right)
\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (1)$

valid for any $s = a + ib \in \field{C}$, $p \in \field{N}$, consider the case of $a=Re\left( s \right) > \frac {1}{2}$, and multiplying (1) over all primes p, define a real function of two variables:

$
| M \left( a, b \right) |^2 \equiv \prod^{\infty}_{p=primes} | \frac {1}{1-1/p^{s}} |^2
=\zeta \left( 2a \right) \prod^{\infty}_{p=primes}
\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (2)$

where $\zeta$ is the Riemann Zeta function.

For $a=Re\left( s \right) > 1$, the right-hand product of (2) converges absolutely and in that range one can equate:
$
\ln \left( \frac {|M \left( a, b \right) |^2 } { \zeta \left( 2a \right) } \right) =
\ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)
=\sum^{\infty}_{p=primes} \ln \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (3)$

Apply the Laplacian $\nabla^2=\frac{\partial^2 }{\partial a^2} + \frac{\partial^2 }{\partial b^2}$ on each side of (3):

$\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)
=\sum^{\infty}_{p=primes} \nabla^2 \ln \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)
= -4 \left( \sum^{\infty}_{p=primes} \left( \frac{ p^{2a} \ln ^2(p)} { (p^{2a}-1)^2 } \right) \right) (4)$

The right-hand side of (4) converges absolutely for $a=Re\left( s \right) > \frac {1}{2}$ without claims as to the convergence of (3) in the same range. Thus the left-hand side of (4) exists, is finite and always negative for all
$s = a + ib \in \field{C}, a=Re\left( s \right) > \frac {1}{2}$ which precludes the existence of zeroes of Zeta to the right of the critical line(?)

Can anyone spot an obvious flaw? I have been twisting it in my head for quite a while before posting but can't seem to see what is wrong except it is too simple...The last step in (4) was checked with Mathematica online and is pretty straightforward. Thanks.

2. From what I can see the left hand side is the laplacian of the logarithm term and you have shown that it is never zero. I don't see what that would imply for the logarithm let alone for the zeta function.

Here is what I think. You have shown that the laplacian is never zero. Since it is the sum of two second order partial derivatives we have a few cases.

If $\displaystyle\frac{\partial^2 f}{\partial a^2} \ge 0$, then $\displaystyle\frac{\partial^2 f}{\partial b^2} < 0$ and is greater in magnitude than the first.

If $\displaystyle\frac{\partial^2 f}{\partial a^2} <0$, then $\displaystyle\frac{\partial^2 f}{\partial b^2}$ can be any real number as long as it has smaller absolute value than the other partial.

So the partials are either both negative or one is negative and the other is non-negative. I don't know how this would imply that the zeta is non-zero.

3. Thank you for your reply. To avoid confusion in the following, is it ok for me to assume that the text of your answer is correct ("second order derivative" is what you meant) and that the Latex expression $\displaystyle\frac{\partial f}{\partial a} < 0$ was actually meant to be written $\displaystyle\frac{\partial^2 f}{\partial a^2} < 0$ ?

My argument is as follows:

Since it is known that the real-valued function $\zeta(2a)$ is never 0 for $a> 1/2$ and
$| \zeta \left( a, ib \right) | ^2$ can only be 0 or positive by definition (except at its unique singularity a=1, b=0)

then if there are zeroes of $\zeta (s)$ to the right of the critical line, we must have at those zeroes, and only at those zeroes:

$\ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right) \rightarrow -\infty$

and at those zeroes both second-order derivatives should be positive (I am contradicting your argument here that one or the other second order partial derivative can be negative) yielding

$\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right) > 0$

However equation (4)

$\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)
=\sum^{\infty}_{p=primes} \nabla^2 \ln \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)
= -4 \left( \sum^{\infty}_{p=primes} \left( \frac{ p^{2a} \ln ^2(p)} { (p^{2a}-1)^2 } \right) \right) (4)$

is negative (and finite) everywhere to the right of the critical line and thus shows there cannot be zeroes to the right of the critical line, no?

4. Yea you are right. I meant 2nd order derivatives! I'll change it now.

I think argument would be pretty compelling if you can justify why both 2nd order partials need to be positive.

5. Originally Posted by jlb
$\ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right) \rightarrow -\infty$

and at those zeroes both second-order derivatives should be positive (I am contradicting your argument here that one or the other second order partial derivative can be negative) yielding

$\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right) > 0$
Why is that?

Unless I'm mistaken in general I don't think that's necessarily true.

Let $\displaystyle f(x,y)=\log(x+y) \implies \nabla^2 f(x,y) = -2(x+y)^{-2}\leq0$

Edit: Oh, but we have a local minimum in your case though... Now we need to look out for cusps.

6. Originally Posted by jlb
$\displaystyle \frac {1}{\left|1-1/p^{s} \right|^2} =
\left( \frac {1}{1-1/p^{2a}} \right)
\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)$
Could you go a little in depth in saying why this is true?

Edit: Never mind, I made a dumb mistake.

7. I checked the identity for some complex numbers and it is correct up to the precision of the computer so it's unlikely to be false. You can probably derive the right hand side from the left hand side via the rules of complex numbers and using the principal branch Log.

Edit: With a few steps I get

[LaTeX ERROR: Convert failed]

EDIT: here

[LaTeX ERROR: Convert failed]

The identity follows immediately and the absolute value brackets fall because it can never be <0 in the range we are discussing.

8. Originally Posted by jlb
Consider the case of $a=Re\left( s \right) > \frac {1}{2}$, and multiplying (1) over all primes p, define a real function of two variables:

$
| M \left( a, b \right) |^2 \equiv \prod^{\infty}_{p=primes} | \frac {1}{1-1/p^{s}} |^2
=\zeta \left( 2a \right) \prod^{\infty}_{p=primes}
\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) (2)$
When $\displaystyle \frac12<\text{Re}(s)<1,\; \prod_p \frac1{\left|1-1/p^{s}\right|^2}$ diverges...

9. Apologies for the delay in replying, your questions thankfully made me re-think and I will forgo the unnecessary argument that if there are zeroes of Zeta to the right of the critical line then $\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)$ should be positive at those zeroes.

This is (probably-possibly) wrong but also not essential, since all that is needed to prove (and easier) is instead that, if there are zeroes of Zeta $(a_n, b_n)$ to the right of the critical line, then at those zeroes $(a_n, b_n)$, and only at those zeroes:

$\ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right) \rightarrow -\infty$, and $\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)$ also diverges at those points.

(keep in mind that the function

$f(a, b) = \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right)}$ has absolute zeroes at $(a_n, b_n)$
and $f(a_n, b_n) = f_a' (a_n, b_n)=f_b' (a_n, b_n)=0$ and $f_a_a'' (a_n, b_n) > 0, f_b_b'' (a_n, b_n) > 0$).

However, if the sum over primes for the Laplacian shown in equation (4) at the beginning of this post is right, it shows instead that the Laplacian is actually finite everywhere to the right of the critical line.

I need a bit more time to write this out in Latex, sorry.

With regards to the comment that equation (2) diverges for $0.5< \Re(s)< 1$, that would actually be hard to prove or disprove (even numerically) since it is an infinite product of Poisson kernels and I am not sure any one knows what that leads to. However, it is not necessary to know this in this case: when you use the Laplacian on the logarithm of (2), the infinite sum has the terms in $1/p^a$ vanish, and only the terms in $1/p^2^a$ remain which extends the range of convergence for the Laplacian (and not equation (2)!) inside the critical strip at $0.5< \Re(s)< 1$.

10. Originally Posted by jlb
$f(a, b) = \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right)}$ has absolute zeroes at $(a_n, b_n)$
and $f(a_n, b_n) = f_a' (a_n, b_n)=f_b' (a_n, b_n)=0$ and $f_a_a'' (a_n, b_n) > 0, f_b_b'' (a_n, b_n) > 0$).
Why do $f_a' (a_n, b_n)=f_b' (a_n, b_n)=0?$ Couldn't it be that a cusp forms?

If not, then as stated before: this is a compelling argument.

11. So I've modified this proof a bit to see what I'd get:

I considered $\displaystyle \frac1{\left|1-p^{-s}\right|^4} = \left( \frac {1}{1-1/p^{4a}} \right)
\left(\frac {p^{4a}-1}{\left(p^{2a}-2 p^{a}\cos(b\ln{p})+1\right)^2} \right)$

Next define $\displaystyle |M\left( a, b \right)|^4 \equiv \prod_p \frac1{\left|1-1/p^{s}}\right|^4}
=\zeta \left( 4a \right) \prod_p
\left(\frac {p^{4a}-1}{\left(p^{2a}-2 p^{a}\cos(b\ln{p})+1\right)^2} \right)$

I follow your proof analogously and then get to your Laplacian (boy what a mess!). This time the Laplacian of the LHS will converge absolutely for $\displaystyle a>\tfrac14$. Going along with what you claim, the Laplacian should be positive within a neighborhood of a zero of $\displaystyle \zeta(s)$ e.g. $\displaystyle (a,b)=\left(\tfrac12,14.137425\ldots\right)$.

Unfortunately the laplacian is negative everywhere. Here is an image of this, where $(a,b)\in[0,1]\times[14,15]$ and the blue dot is the previously mentioned zero:

Assuming I've done things correctly this should show there's a major hole in your argument. What is everyone else's thoughts?

12. I'm not quite sure of the following steps but, if for every positive number N, we consider the product over all primes smaller than N, we get that

[LaTeX ERROR: Convert failed]

Since they agree for every number N they must agree in the limiting case, i.e.

[LaTeX ERROR: Convert failed]

Here we have that [LaTeX ERROR: Convert failed] diverges for s with real part between 1/2 and 1. Hence the absolute value squared must also diverge and thus [LaTeX ERROR: Convert failed] must also diverge.

13. Originally Posted by Vlasev
I'm not quite sure of the following steps but, if for every positive number N, we consider the product over all primes smaller than N, we get that

[LaTeX ERROR: Convert failed]

Since they agree for every number N they must agree in the limiting case, i.e.

[LaTeX ERROR: Convert failed]

Here we have that [LaTeX ERROR: Convert failed] diverges for s with real part between 1/2 and 1. Hence the absolute value squared must also diverge and thus [LaTeX ERROR: Convert failed] must also diverge.
You're absolutely correct, but the OP is claiming this doesn't matter. IMO he makes a plausible argument and could be correct, I'm a little too distracted to think about it long enough.

So my above post shows there is a hole though, I'm thinking it's either the convergence issue or the zeros form a cusp.

14. I think I spotted a hole. In (2) the OP considers real part greater than 1/2 and defines the function M. Then to define (3) the OP says the sum is absolutely(EDIT: uniformly) convergent when s has real part greater than 1. Then he takes the laplacian of the expression and interchanges differentiation (the laplacian) with the infinite sum which only works on absolutely (EDIT: uniformly) convergent series. This is fine though, whenever the real part is greater than 1. However after doing the interchange, the OP makes comments in the strip 1/2 < Re(s) <1 but in that part he couldn't have done the interchange since the sum is not absolutely(EDIT: uniformly) convergent.

15. Originally Posted by Vlasev
I think I spotted a hole. In (2) the OP considers real part greater than 1/2 and defines the function M. Then to define (3) the OP says the sum is absolutely convergent when s has real part greater than 1. Then he takes the laplacian of the expression and interchanges differentiation (the laplacian) with the infinite sum which only works on absolutely convergent series. This is fine though, whenever the real part is greater than 1. However after doing the interchange, the OP makes comments in the strip 1/2 < Re(s) <1 but in that part he couldn't have done the interchange since the sum is not absolutely convergent.
I am not sure that this argument implies a hole since I never make use of the range of convergence of the series in (3) (as stated before, it obviously converges for Re(s)> 1 and diverges for Re(s)< 1/2 but I am not certain what the series does for 1/2 < Re(s) < 1). As far as I understand, there can be several series representations for any function with different ranges of convergence for each. Perhaps as a couterexample, I could refer to the Dirichlet series of Zeta (with semi-convergence for Re(s) > 0) which is obtained by subtracting two series that diverge for Re(s) < 1.

In the case of equation (4), I believe the Laplacian ends up performing a similar "trick" by adding two infinite series for which the terms in $1/p^a$ cancel out due to the fact that the Poisson kernels have vanishing Laplacian.

That is also why I am not sure, as stated in one of the replies, that the Laplacian of the log of:

$\displaystyle |M\left( a, b \right)|^4 \equiv \prod_p \frac1{\left|1-1/p^{s}}\right|^4}
=\zeta \left( 4a \right) \prod_p
\left(\frac {p^{4a}-1}{\left(p^{2a}-2 p^{a}\cos(b\ln{p})+1\right)^2} \right)$

converges for Re(s) > 1/4 since the RHS terms in the infinite product do not obey Laplace's equation and should not cancel out when adding the second order derivatives. I need to verify this claim though because that would indeed be a major hole. Thank you again for your input.

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