6
$\begingroup$

First of all, I imagine it will not be correct, just because of its simplicity, but I would also want to know why, as I can't find any mistake on it.

The "proof" would be based on convining two main theorems/formulae. The first one, would be this one , due to Nicolas, where it is stated that RH would hold iff:

$$\frac{N_k}{\phi{(N_k)}} > e^{\gamma} \ln{(\ln{(N_k)})}$$

holded for every $k$ being $N_k$ the primorial of order $k$ and $\phi{(N_k)}$ its Euler's Totient function.

Then, my main aim here will be to prove that formula for every $k$. To do that, I will use this other theorem:

$$ \prod_{p \le n} \frac{p}{p-1} > e^{\gamma}(\ln{n})(1-\frac{1}{2(\ln{n})^{2}})$$

Taken from "Approximate Formulas for Some Functions of Prime Numbers" (link), Theorem 8 (3.28).

As, in this case, $\frac{N_k}{\phi{(N_k)}} = \prod_{p \le p_k} \frac{p}{p-1}$, we can try to see if this holds:

$$e^{\gamma}(\ln{p_k})(1-\frac{1}{2(\ln{p_k})^{2}})>e^{\gamma} \ln{\ln{N_k}}$$

Hence

$$\ln{p_k}-\frac{1}{2\ln{p_k}}>\ln{\ln{N_k}}$$

For it to be more clear, we can change $\ln{N_k}$ by $\theta{(k)}$ (Chebyshev's First Function) so that

$$\ln{p_k}-\frac{1}{2\ln{p_k}}>\ln{\theta{(k)}}$$

From there, we could easily get to

$$\frac{1}{2\ln{p_k}}<\ln{\frac{p_k}{\theta{(k)}}}$$

And, with the bounds of Theorems 3 (3.12) and 4 (3.15), we get

$$\frac{1}{2\ln{p_k}}<\ln{\frac{\ln{k}}{1+ \frac{1}{2\ln{k}}}}$$

What would be true for every big enought k, meaning that

$$\frac{N_k}{\phi{(N_k)}} > e^{\gamma} \ln{(\ln{(N_k)})}$$

holds, and, with so, RH.

Is this correct? Why would/would not it prove the RH?

Thank you!

Edit thanks to Jyrki Lahtonen

$\endgroup$
22
  • 1
    $\begingroup$ I don't understand your question, try to make it readable. And there is no elementary proof of the Lagarias/Robin criterion. $\endgroup$
    – reuns
    Commented Oct 31, 2016 at 17:47
  • 2
    $\begingroup$ Assuming the Riemann hypothesis and you get some tight approximations of the functions involved, for example $\sum_{p < x} \frac{1}{p} = \ln \ln x + B+\mathcal{O}(x^{-1/2+\epsilon})$, $\sum_{p < x} \ln p = x+\mathcal{O}(x^{1/2+\epsilon})$, $ \sum_{p < x} \ln(1-\frac{1}{p}) = C- \ln \ln x + \mathcal{O}(x^{-1/2+\epsilon})$ and so on. This way you can see if what you are writing is a nonsense, and if it is tautologic. And again, I don't understand what you wrote in your question, try to restate it clearly $\endgroup$
    – reuns
    Commented Oct 31, 2016 at 19:21
  • 1
    $\begingroup$ To be clear "with the bounds of Theorems 3 (3.13) and 4 (3.16)," means that you are using some theorems you don't understand. And I don't see how a bound for $1/\ln p_k$ would help for bounding $\prod_{p < x} (1-\frac{1}{p})$. So again, write only the necessary the steps clearly, with the main step made obvious (and separate what is comparatively trivial : $p_k \sim k \ln k$, $\sum_{p < x} \frac{1}{p} \sim \ln \ln x$, i.e. the Mertens theorems and the prime number theorem, from the less trivial things) $\endgroup$
    – reuns
    Commented Nov 1, 2016 at 0:02
  • 1
    $\begingroup$ and this is obviously not true : $$\frac{1}{2\ln{p_k}}<\ln{\frac{\ln{k}+\ln{\ln{k}}}{1+ \frac{1}{\ln{k}}}}$$ since the RHS $\sim -\ln(1+\frac{1}{\ln k})\sim \frac{-1}{\ln k}$, while the LHS $\sim \frac{1}{2\ln k}$ $\endgroup$
    – reuns
    Commented Nov 1, 2016 at 0:25
  • 1
    $\begingroup$ $\frac{1}{2\ln{p_k}}<\ln{\frac{\ln{k}+\ln{\ln{k}}}{1+ \frac{1}{\ln{k}}}}$ should be $\frac{1}{2\ln{p_k}}<\ln{\frac{\ln{k}+\ln{\ln{k}}}{1- \frac{1}{\ln{k}}}}$ by the equations you cited. $\endgroup$
    – user361424
    Commented Nov 1, 2016 at 8:51

1 Answer 1

Reset to default
7
$\begingroup$

With $p_1, p_2, \ldots$ being a list of primes in increasing order we have $N_k = p_1 p_2 \cdots p_k$ for the primorial. Therefore $$\frac{N_k}{\phi(N_k)}=\prod_{p\le p_k}\frac{p}{p-1}.$$

Hence the lower bound is only $$ \frac{N_k}{\phi(N_k)} > e^\gamma \log p_k \left(1 - \frac1{2 \log^2 p_k}\right), $$ which does not work for the remaining argument.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged .