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( I've been thinking about asking this for a long time . Though this is not rigorous; It can be thought of as heuristic or extraction of information from different viewpoint.)

We know "super-regularized"( the term coined by authors of the paper : http://cds.cern.ch/record/630829/files/sis-2003-264.pdf) product of primes $4π²$ i.e.

$$\infty \# = \prod_{k=1}^\infty p_k = 4\pi^2$$

Now , by operating by $\log$ and converting product to summation we get :

$$\sum_{k=1}^\infty \log(p_k)$$ .

In their paper they considered (for strange reasons somehow) $1$ as prime.

So, $G(k)=p_{(k+1)}= p(k+1)$ can be considered our new $n$th prime function ( "most natural" $n$th prime function with all nice analytic properties desired for Abel Plana Summation)

Let's define $F(k)= \log(p_{(k+1)})$

Now ,we can use Abel -Plana or Ramanujan Summation as follows :

$$i\int_0^\infty\frac{F(ix)-F(-ix)}{e^{2\pi x}-1}\,dx = \frac{1}{2}F(0)+\sum_{n=1}^\infty F(n) $$

Now ,we can see that $F(0) = 0$

so , now

$$i\int_0^\infty\frac{F(ix)-F(-ix)}{e^{2\pi x}-1}\,dx = \log(4π²)$$

Question : Can we "guess" the function $F(x)$ ( as we desire ) from the above equation ?

Approach :

Consider we have some "suitable" infinite product like $4π² = \prod ^ \infty a_n $ ( lots of infinite product for numbers like $π$ and $\sqrt2$ exist and lots of we can successively construct.)

Hence , we can get an infinite sum by taking $\log$ on both sides .

Now , equation (t):

$$i\int_0^\infty\frac{\log(\frac{p(1+ix)}{p(1-ix)})}{e^{2\pi x}-1} dx = \sum \log(a_n)$$

Now , Assume $$i\log(\frac{p(1+ix)}{p(1-ix)})$$ is continuous and $\infty$ - differentiable ( should be ?).

So ,it has a Taylor series around $0$ : $$\sum \frac{c_n x^n}{\Gamma(n+1)}$$

Now ,we know the integral $$\int_0^\infty \frac{x^{(a-1)}}{e^{b x} - 1} = \frac{\Gamma(a)\zeta(a)}{b^a}$$

So ,the equation $(t)$ becomes $$\sum _{n=1}^\infty \frac{c_n\zeta(n+1)}{(2π)^{(n+1)}}$$

Now only thing remaining is correlation between $\log(a_n)$ and $\frac{c_n\zeta(n+1)}{(2π)^{(n+1)}}$.

Can we guess the better $p_n$ using above analysis (by using some known facts like approximately $p_n$ is $n\ln(n)$ or using bounds on $p_n$) ?

( Motivation : I calculated various regularized products by using this method and it worked. )

(Note : I know this is an $\infty$-ly underdetermined system . So don't consider this a trivial or anything I'm just asking for suitable "guess" using the known facts. )

(Apologies in advance if there are any partial mistakes )

(Edit 1: consider $F(k)=\ln(k\ln(k)+ E(k))$ for initiation. ( Use of calculation of variation /?) .Here $E$ is sort of an error function. You can replace the $k\ln(k)$ by better known bounds on $p_n$ for example by P. Dusart but it'll become mess)

Edit 2: the general solution to the integrand in equation (t) is :

$$p(x) = e^{\frac{f(x)}{2}}E(x)$$ Where, $E(x)$ is any even function

$$f(x) =\sum_{n=0}^\infty \frac{a_n(2π)^{2n+2}(-1)^{n+1}(x-1)^{2n+1}}{\zeta(2n+2)\Gamma(2n+2)}$$

with $a_n$ being $\log$ of terms in the infinite product for $4π²$

Now ,the question is for $E(x)$.

(Can we use calculus of variations to get more insights? I haven't tried it.)

Edit 3:

Had an email exchange with Sir Roger Heath-Brown about the validity of argument . He confirms that argument is valid but also suggested that to uniquely find out the function for nth prime I have to use some additional known unconditional information of $n$th prime like growth condition of $n$th prime and then carefully selecting the coefficients to construct the function explicitly.

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    $\begingroup$ Looks like Salvador Dalí doing math. $\endgroup$ Nov 15, 2019 at 22:17
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    $\begingroup$ Surrealism, yes $\endgroup$ Nov 16, 2019 at 9:34
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    $\begingroup$ Me too, even though I don't fully grab the machinery used in this. $\endgroup$ Apr 13, 2020 at 23:07
  • 3
    $\begingroup$ Sockpuppet accounts attempting to generate a conversation will not be tolerated. $\endgroup$
    – Todd Trimble
    Apr 16, 2020 at 11:48
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    $\begingroup$ I think $p(1+ix)$ makes more sense when written as $p_n\approx nln(n)$ as appears in Edit 1, maybe OP could clarify how he got the integrand in edit 2 as well as maybe improve the explanation of the discussion mentioned in edit 3 (I don't see the need to have the text in edit 3 repeated as a comment), more importantly as his motivation is that he was able to calculate various regularized product with this method maybe he could share an example, sockpuppetry isn't necesarry the question is interesting I'm not an expert but I think more detail wouldn't hurt. $\endgroup$
    – Dabed
    Apr 17, 2020 at 5:56

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