# Math Help - Help with Riemann's Hypothesis

1. ## Help with Riemann's Hypothesis

Hi,.

My name is Robin Randall and I live in Silicon Valley. I have a Math Sciences degree from Stanford University and have been dabbling in the Clay Millennium Problems for
several years.

I picked your group because it seemed serious and open to new ideas and members. What would be best is if I could find a sort of Math pen pal who knows
more math than I do. That's my main problem - Finding someone who knows Number Theory and won't ridicule me for making a simple mistake.

Attached is my work so far on proving Riemann's Hypothesis. I think I understand the basics of the zeta zeros. What have I proved? And what is left to prove? (that the Real part of s = 1 / 2 for all k.in the "critical strip")

Thanks,
Robin

2. ## Re: Help with Riemann's Hypothesis

This proof is 7 years old, but anyway... I don't have a great background in number theory but I think the mistake is when you take $k^{1-s}=k^s$. I think a previous step stated that $\Sigma_{k=0}^{\infty}\frac{1}{k^{1-s}}=\Sigma_{k=0}^{\infty}\frac{1}{k^{s}}$ which is different altogether

3. ## Re: Help with Riemann's Hypothesis

Thanks, Shakarri,

The point is that $\displaystyle 0 = \sum_{k=1}^{\infty}\frac{1}{k^{1-s}}=\sum_{k=1}^{\infty}\frac{1}{k^{s}}$ and I proved that $\displaystyle 0 = \sum_{k=1}^{\infty}\frac{k^{1-s} - k^{s}}{k}$
So if we can find an s such that the numerator equals 0, then every element is 0 and the series collapses to 0
If $0 = k^{1-s} - k^{s}$ Then $k^{1-s} = k^{s}$ And we get $1-s = s$ Or $s = \frac{1}{2}$ for all $k$.

This is probably the heart of the proof, but it has to be stated just right to get credit. That is where I could use some help.
What more must I prove? If you are convinced $s=1/2$ already, it seems to fall in line.

4. ## Re: Help with Riemann's Hypothesis

Originally Posted by robinlrandall
Thanks, Shakarri,

The point is that 0 = \Sigma_{k=1}^{\infty}\frac{1}{k^{1-s}}=\Sigma_{k=1}^{\infty}\frac{1}{k^{s}}
and I proved that 0 = \Sigma_{k=1}^{\infty}\frac{k^{1-s} - k^{s}}{k}
So if we can find an s such that the numerator equals 0, then every element is 0 and the series collapses to 0
If 0 = k^{1-s} - k^{s} Then k^{1-s} = k^{s} And we get 1-s = s Or s = \frac{1}{2} for all k.

This is probably the heart of the proof, but it has to be stated just right to get credit. That is where I could use some help.
What more must I prove? If you are convinced s=1/2 already, it seems to fall in line.
Sorry this is not rendering the way I hoped, but I think you all can still read the meaning.
Robin

5. ## Re: Help with Riemann's Hypothesis

Originally Posted by robinlrandall
Sorry this is not rendering the way I hoped, but I think you all can still read the meaning.
Robin
Wrap the LaTeX in dollar signs: $\$\ or in TEX tags: 

Anyway, I, too, am not well-versed in this type of number theory. Question: Is $\text{Re}\left(\xi(s)\right) = \zeta\left(\text{Re}(s)\right)$? It appears that you are making that assumption, however, $\sum_{k=1}^\infty \dfrac{1}{k^{1/2}}$ is not defined over the reals (it diverges faster than the harmonic series), so it certainly would not converge to zero. That seems to indicate that $\text{Re}\left(\xi(s)\right) \neq \zeta\left(\text{Re}(s)\right)$... I could be wrong.

6. ## Re: Help with Riemann's Hypothesis

Thanks SlipEternal,

I think I have to understand better how Hardy move zeta(s) into the Complex Plane. Could <TEX> \sum_{k=1}^\infty \dfrac{1}{k^{1/2 +it}} </TEX> converge to 0?
Possibly, if we are in the Complex Plane. I need to do a bit more research. But thanks again. Oh, on <TEX>\sum_{k=1}^\infty \dfrac{1}{k^{1/2}}</TEX> I hadn't really thought of it that way, but perhaps I concluded similarly. Thanks again.

Better understanding of below should help:
The Riemann hypothesis discusses zeros outside the region of convergence of this series, so it must be analytically continued to all complex s. This can be done by expressing it in terms of the Dirichlet eta function as follows. If the real part of s is greater than one, then the zeta function satisfies

<TEX>\left(1-\frac{2}{2^s}\right)\zeta(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \cdots.
However, the series on the right converges not just when s is greater than one, but more generally whenever s has positive real part. Thus, this alternative series extends the zeta function from Re(s) > 1 to the larger domain Re(s) > 0, excluding the zeros s = 1 + 2\pi in/\ln(2) of 1-2/2^s (see Dirichlet eta function). The zeta function can be extended to these values, as well, by taking limits, giving a finite value for all values of s with positive real part except for a simple pole at s = 1.

In the strip 0 < Re(s) < 1 the zeta function satisfies the functional equation

\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s).
One may then define ζ(s) for all remaining nonzero complex numbers s by assuming that this equation holds outside the strip as well, and letting ζ(s) equal the right-hand side of the equation whenever s has non-positive real part. If s is a negative even integer then ζ(s) = 0 because the factor sin(πs/2) vanishes; these are the trivial zeros of the zeta function. (If s is a positive even integer this argument does not apply because the zeros of sin are cancelled by the poles of the gamma function as it takes negative integer arguments.) The value ζ(0) = −1/2 is not determined by the functional equation, but is the limiting value of ζ(s) as s approaches zero. The functional equation also implies that the zeta function has no zeros with negative real part other than the trivial zeros, so all non-trivial zeros lie in the critical strip where s has real part between 0 and 1. </TEX>
Robin

7. ## Re: Help with Riemann's Hypothesis

I am just cleaning up the LaTeX. Forum tags require square brackets instead of angle brackets.

Originally Posted by robinlrandall
Thanks SlipEternal,

I think I have to understand better how Hardy move zeta(s) into the Complex Plane. Could $\sum_{k=1}^\infty \dfrac{1}{k^{1/2 +it}}$ converge to 0?
Possibly, if we are in the Complex Plane. I need to do a bit more research. But thanks again. Oh, on $\sum_{k=1}^\infty \dfrac{1}{k^{1/2}}$ I hadn't really thought of it that way, but perhaps I concluded similarly. Thanks again.

Better understanding of below should help:
The Riemann hypothesis discusses zeros outside the region of convergence of this series, so it must be analytically continued to all complex s. This can be done by expressing it in terms of the Dirichlet eta function as follows. If the real part of s is greater than one, then the zeta function satisfies

$\left(1-\frac{2}{2^s}\right)\zeta(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \cdots.$
However, the series on the right converges not just when s is greater than one, but more generally whenever s has positive real part. Thus, this alternative series extends the zeta function from $\text{Re}(s) > 1$ to the larger domain $\text{Re}(s) > 0$, excluding the zeros $s = 1 + 2\pi in/\ln(2)$ of $1-2/2^s$ (see Dirichlet eta function). The zeta function can be extended to these values, as well, by taking limits, giving a finite value for all values of s with positive real part except for a simple pole at $s = 1$.

In the strip $0 < \text{Re}(s) < 1$ the zeta function satisfies the functional equation

$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$.
One may then define ζ(s) for all remaining nonzero complex numbers s by assuming that this equation holds outside the strip as well, and letting ζ(s) equal the right-hand side of the equation whenever s has non-positive real part. If s is a negative even integer then ζ(s) = 0 because the factor sin(πs/2) vanishes; these are the trivial zeros of the zeta function. (If s is a positive even integer this argument does not apply because the zeros of sin are cancelled by the poles of the gamma function as it takes negative integer arguments.) The value ζ(0) = −1/2 is not determined by the functional equation, but is the limiting value of ζ(s) as s approaches zero. The functional equation also implies that the zeta function has no zeros with negative real part other than the trivial zeros, so all non-trivial zeros lie in the critical strip where s has real part between 0 and 1.
Robin

8. ## Re: Help with Riemann's Hypothesis

Thanks so much!

9. ## Re: Help with Riemann's Hypothesis

With this statement from Wolfram's there seems to be very little to prove:

The Riemann hypothesis is equivalent to the statement that all the zeros of the Dirichlet eta function (a.k.a. the alternating zeta function)

$\eta(s)=\sum_{k=1}^\infty_{\frac{{-1}^{k-1}}{k^s}}=$${1-2^{1-s}}$$\zeta$${s}$$ = \frac{{1}}{1^s} - \frac{{1}}{2^s} + \frac{{1}}{3^s} - \frac{{1}}{4^s} ...$
(1)
falling in the critical strip 0<R[s]<1 lie on the critical line R[s]=1/2.

Wiener showed that the prime number theorem is literally equivalent to the assertion that the Riemann zeta function zeta(s) has no zeros on sigma=1 (Hardy 1999, pp. 34 and 58-60; Havil 2003, p. 195).

In 1914, Hardy proved that an infinite number of values for s can be found for which zeta(s)=0 and R[s]=1/2 (Havil 2003, p. 213). However, it is not known if all nontrivial roots s satisfy R[s]=1/2. Selberg (1942) showed that a positive proportion of the nontrivial zeros lie on the critical line, and Conrey (1989) this to at least 40% (Havil 2003, p. 213).

If I can show that ALL nontrivial roots s satisfy R[s]=1/2 that would be something. I don't understand why no one has proved that part. Does anyone see something else which must be proved? It seems Hardy should get a lot of credit in solving this problem.
Robin

10. ## Re: Help with Riemann's Hypothesis

The reason why no one has solved it yet is because it is incredibly difficult to solve. It is possible that there is no solution.

11. ## Re: Help with Riemann's Hypothesis

Since the analytic continuation function η(s) has the exact same zeroes as ϛ(s) then we expect η(s) = η(1-s) (Hardy.)

$0=η(s)≡(1- 2/2^s )ϛ(s)=∑_(n=1)^∞▒〖(-1)〗^(n+1)/n^s =∑_(n=1)^∞▒〖(-1)〗^(n+1)/n^((1-s)) =η(1-s) R>0
= 1/1^s - 1/2^s + 1/3^s - 1/4^s … = 1/1^((1-s)) - 1/2^((1-s)) + 1/3^((1-s)) - 1/4^((1-s)) …
Subtracting …
0 = ∑_(n=1)^∞▒〖〖(-1)〗^(n+1)/n^((1-s)) -〗 〖(-1)〗^(n+1)/n^s = ∑_(n=1)^∞▒〖(〖(-1)〗^(n+1) n^s)/(n^((1-s) ) n^s ) - 〗 (〖(-1)〗^(n+1) n^((1-s)))/(n^((1-s) ) n^s )
0 = ∑_(n=1)^∞▒〖(〖(-1)〗^(n+1) 〖(n〗^s- n^((1-s))))/n 〗
Now this is true for all n so if 0 = n^s- n^((1-s)) => n^s= n^((1-s))$

and this is possible only if s = (1 – s) => s = 1/2 or R[s] = 1/2
There are no other zeroes to be found in the equation, so I believe we have them all. Otherwise we disprove Riemann’s Hypothesis.

12. ## Re: Help with Riemann's Hypothesis

Since the analytic continuation function $\eta{(s)}$ has the exact same zeroes as $\zeta{(s)}$then we expect : $
\eta{(s)} = \eta{(1-s)} (Hardy.)
0=\eta_{s}={1- \frac{2}{2^s}\zeta_{s}
=\sum_{n=1}^{\infty}\frac{{(-1)}^{n+1}}{n^s}}
=\sum_{n=1}^{\infty}\frac{{(-1)}^{n+1}}{n^{1-s}} =\eta_{1-s} R >0
=\frac{1}{1^s} - \frac{1}{2^s}+\frac{1}{3^s} - \frac{1}{4^s}\dots
=\frac{1}{1^{1-s}} - \frac{1}{2^{1-s}} + \frac{1}{3^{1-s}} - \frac{1}{4^{1-s}}\dots
Subtracting \dots
0 =\sum_{n=1}^{\infty}{\frac{{-1}^{n+1}}{n^{{1-s}}} - \frac{{-1}^{n+1}}{n^s}
=\sum_{n=1}^{\infty}\frac{{-1}^{n+1} n^s}{n^{{1-s}} n^s } -
\frac{(-1)^{n+1}n^{1-s}}{n^{1-s}n^s}
0 =\sum_{n=1}^{\infty}\frac{{-1}^{n+1} ({n}^s - n^{1-s})}{n}$

Now this is true for all n so if $0 = n^s- n^{(1-s)} => n^s= n^{(1-s)}$

and this is possible only if s = (1 – s) => s = 1/2 or R[s] = 1/2
There are no other zeroes to be found in the equation, so I believe we have them all. Otherwise we disprove Riemann’s Hypothesis.
Robin
(Having trouble with line breaks and formatting.)

13. ## Re: Help with Riemann's Hypothesis

When dealing with infinite sums, you have no reason to suspect $n^s=n^{1-s}$ for all $n$

14. ## Re: Help with Riemann's Hypothesis

OK, Then show me a zero from the equation where this is NOT true. You have just disproved
R.H. since it says ALL zeroes have R[s] = 1/2.
Robin
(I'll see if I can frame it so there is a possibilty that it does NOT apply for ALL n) Sometimes we
have blinders on, right?)

15. ## Re: Help with Riemann's Hypothesis

Maybe I should say for all positive n since the sum is from n=1 to infinity. Does that make a difference?

Also can you tell how to do a line break in LaTEX? or I guess I can google it.
Robin

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