6
$\begingroup$

Let us assume there exists some infinite order differential equation whose solution is:

$$ y= \sum_{n=1}^\infty A_n \exp(p_n^sx) $$

Where $p_n$ is the $n$'th prime. Substituting $ y=\exp(\lambda x)$ as a trail solution and factorizing. The differential equation must be simplified and factorized to:

$$ \prod_{j=1}^\infty (\lambda-p_n^s) = 0$$ $$ \implies \prod_{j=1}^\infty (1-\frac{p_n^s}{\lambda}) = 0$$

Expanding the above equation:

$$\implies 1 - (\sum_{i} p_i^s)\frac{1}{\lambda} + \sum_{i < j} (p_i p_j)^s \frac{1}{\lambda^2} - \sum_{i<j<k} (p_i p_j p_k)^s \frac{1}{\lambda^3} + \dots + \lim_{n \to \infty} \sum_{i<j<\dots<n}(-1)^n (p_i p_j \dots p_n)^s \frac{1}{\lambda^n} = 0 $$

Writing more compactly:

$$ 1+ \sum_{n=1}^\infty \sum_{x_i<x_j<\dots<x_n} (p_{x_i}p_{x_j} \dots p_{x_n})^s = 0 $$

We conjecture that $\sum_{x_i<x_j<\dots<x_n} (p_{x_i}p_{x_j}\dots p_{x_n})^s $ can be analytically continued as some function of $ P(s)$ where $P(s)$ is the prime zeta function

Hence, for $f_1(P(s)) = P(s)$

And $ f_2(P(s)) = P(s)^2 - P(2s)$

And so on ...

Where $f_n(P(s))$ is given by comparing the below equation to the equation before the analytic continuation:

$$ 1 + \frac{f_1(P(s))}{\lambda} + \frac{f_2(P(s))}{\lambda^2} + \dots = 0 $$

Writing it compactly:

$$ 1 + \sum_{i}^\infty \frac{f_i(P(s))}{\lambda^s} =0 $$

Note this can be thought as of the solution of a integral:

$$ y + f_1(P(s))\int_{-\infty}^x y(x) dx + f_2(P(s))\int_{-\infty}^x (\int_{-\infty}^k y(z) dz) dk + \dots$$

(To verify this subsitute $y=Ae^{2x}$) Writing it compactly:

$$ y + \sum_{i=1}^\infty f_i(P(s)) \int_{- \infty}^x (\int_{- \infty}^x (\dots( \int_{- \infty}^x y dx )dx)\dots \text{$i$ times})dx) = 0 $$

Strategy from Here

My idea was to take some asymptotic limit as $s \nearrow 1$ and rewrite this as a integral equation for $y^2$.

Since,

$$ y=\sum_{i=1}^\infty A_i \exp{p_i x} $$

Setting $A_1=0$ and squaring:

$$ y^2 = \sum_{i,j}A_{ij}' \exp{(p_i + p_j)x} = \sum_i^\infty B'_i e^{2ix} $$

By Goldbach's conjecture the rightmost side must contain all the even powers.

Questions

Is this a viable strategy or useful reformulation of Goldbach's conjecture? (Is it easier to show $y^2 =\sum_i^\infty B'_i e^{2ix} $as a solution)? Is the analytic continuation valid?

$\endgroup$
  • 1
    $\begingroup$ It seems unlikely that the desired analytic continuation exists. $\endgroup$ – Kyle May 9 '17 at 14:27
  • 2
    $\begingroup$ Your 'ansatz' is both wrong (you have the term corresponding to $p=2$ in your $y$ which means that you'll get plenty of terms in $k$ corresponding to $e^{nx}$ for odd $n$; e.g., you have a term proportional to $e^{5x}$ coming from terms of the form $e^{2x}$ and $e^{3x}$ in $y$) and not nearly enough (you need to show that all of the even coefficients of $k$ are nonzero). I don't want to rule anything out, but this looks more like complication for complication's sake than a viable path. $\endgroup$ – Steven Stadnicki May 11 '17 at 21:50
  • 1
    $\begingroup$ Not sure that offering a bounty will help counterbalance the fact that there is no real question here. $\endgroup$ – Did May 14 '17 at 11:36
  • 1
    $\begingroup$ @Servaes This is very honestly explained... Knowing better "this kind of math" would not change your "inclination", (un)fortunately. (As an aside, I would be curious to hear from the 4 upvoters why they think there is a question suitable to the site here, and even, a good one...) $\endgroup$ – Did May 14 '17 at 12:02
  • 2
    $\begingroup$ If ever you embarked on the task of clarifying the $$\cdots$$ in this equation, to make precise the bounds of integration and to get rid of some annoying repeated variables ($dxdx$, oh my...), then and only then might all this reach the status of being an acceptable question. I know that this is hard, this is called doing mathematics... (Unrelated: Please use @.) $\endgroup$ – Did May 14 '17 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.