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Edit: Using asymptotic to and not equal to as it keeps the meaning intact and makes sense of the relation

Would it be correct to imply from Euler's prime equality function
$x is ~ (asymptotic) to \frac {(\log x)^{\pi(x)}}{\prod \log p}$ where $\pi(x)$ is prime counting function?

$(\sum_{i=0}^x 1/i)$ ~ $(1 + \frac {1}{2} + \frac {1}{2^2} + \frac {1}{2^3} + \dots)(1 + \frac {1}{3} + \frac {1}{3^2} + \frac {1}{3^3} + \dots)(1 + \frac {1}{5} + \frac {1}{5^2} + \frac {1}{5^3} + \dots)\dots$

In famous Euler's prime euqality equation

$\sum x^{-1} = \prod (1-p^{-1})^{-1}$

We have x terms of left hand side but can we write x as

$x$ ~ $(\text{no of terms of 2})(\text{no of terms of 3})(\text{no of terms of 5})\dots$

or as $x$ ~ $\prod (\log_p x)$

If we find the no. of terms it contributes to $x$ say for some power of each prime number say $p$ so for some $k$ we can say $p^k \le x$

i.e $k = \log_p x = \frac {\log x}{\log p}$

Therefore we can rewrite

$x$ ~ $\frac {\log x}{\log 2}$$\frac {\log x}{\log 3}$$\frac {\log x}{\log 5}\dots$

Simplifying

$x$ ~ $\frac {(\log x)^{\pi(x)}}{\log 2 \log 3 \log 5 \dots}$

We can then put upper & lower bounds on ${\log 2 \log 3 \log 5 \dots}$, for eg for lower bound we can say each term is less than equal to $\log 2$.

i.e $x$ ~ $\left(\frac {\log x}{\log 2}\right)^{\pi(x)}$

Taking log on both sides

$\log x$ ~ $\pi(x) \log \frac {\log x}{\log 2}$

$π(x)$ ~ $\frac {\log x}{\log (\frac {\log x}{\log 2})}$

Or over simplifying

$\pi(x)$ ~ $\frac {\log x}{\log \log x}$

Edit:

After reading throgh some old proofs of PNT this answer resonates my thought process https://math.stackexchange.com/a/1892114/1295152

ajay sharma
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    It's recommended to use MathJax to make your question more legible - editing this to be easier to read will attract more engagement and hopefully a quicker answer – George Feb 26 '24 at 18:11
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    Thank you @George for guiding me to MathJax, it was helpful – ajay sharma Feb 27 '24 at 12:09
  • I took the liberty yo improve the formatting on the post @ajaysharma – Sahaj Feb 27 '24 at 12:30
  • It's not at all clear what you are asking. You mention "Euler's function", but it's not clear which function you have in mind. $x={\log x^{\pi(x)}\over\prod\log p}$ is certainly false, e.g., for $x=2$ it gives $2=1$. – Gerry Myerson Feb 27 '24 at 12:31
  • Hi Gerry, I asume it is false for small x because the infinite series dont hold for small x. Nevertheless I was talking in the ballpark of x limiting to infinity and yes the Euler's famous prime equality leading to Riemann Zeta function, thanks for reading – ajay sharma Feb 27 '24 at 12:35
  • Hi @Gerry Myerson I was brosing some existing posts and this answer here resonates my thought process https://math.stackexchange.com/a/1892114/1295152 – ajay sharma Feb 27 '24 at 14:09
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    Can you explain your line $(\sum_{i=0}^x i^2) = (1 + \frac {1}{2} + \frac {1}{2^2} + \frac {1}{2^3} + \dots)(1 + \frac {1}{3} + \frac {1}{3^2} + \frac {1}{3^3} + \dots)(1 + \frac {1}{5} + \frac {1}{5^2} + \frac {1}{5^3} + \dots)\dots$? It seems to resemble the Proof of the Euler product formula for the Riemann zeta function but it doesn't quite make sense. – Benjamin Wang Feb 27 '24 at 14:39
  • @Benjamin Wang, here goes my thought process. On the left hand side (1, 2, 3, .... up to x) we have x terms, we can sya x is the product of have these infinite series of each prime extending up to power of each prime (p) less than x i.e $\log_p x$ – ajay sharma Feb 27 '24 at 14:44
  • We may basiclly write assuming x goes to infinity x = $\prod (\log_p x)$ – ajay sharma Feb 27 '24 at 14:51
  • I had the same question as Benjamin Wang. Why does the left hand side not read $\sum_{i = 1}^x 1/x$? If I multiply out the right hand side then, by Unique Factorization into Primes I get every $1/i$ for $i \leq 1$ exactly once, but I have no idea how to get to the huge number $i^2$ – Vincent Feb 27 '24 at 15:52
  • Sorry @Vincent, Benjamin Wang and others who gave it a go, I did a typo I intended to write the 1/i, somewhere I copied the latex and forgot to change it and even after being mentioned in comments read over it, sorry again. I hope nevertheless the question makes sense now – ajay sharma Feb 27 '24 at 16:04
  • I still don't know what you mean by "Euler's function", nor by "Euler's famous prime equality", I still think you shouldn't write an equals sign between two quantities that aren't equal, and even after you fixed the typo in your "equation" for $\sum(1/i)$ it still isn't right. You know, one of the big things about mathematics is that it is possible to say exactly what we mean, rather than settle for something sorta kinda like what we mean. Your post needs review and revision, top to bottom. – Gerry Myerson Feb 27 '24 at 22:22
  • I will try to explain in simplest terms possible I guess, the riemann zeta function originated from Euler's discovery of the relation between the harmonic series and the prime numbers, from there can we imply $x = \prod_2 (P_n <= x) P_i $ – ajay sharma Feb 28 '24 at 11:57
  • Can you not just state what you mean by "Euler's function"? That should be easy enough. I suppose by "Euler's famous prime equality" you mean $\prod(1-p^{-s})^{-1}=\sum n^{-s}$ for $s>1$, but why can't you just say that? And the formula in your most recent comment is so garbled that I have no idea what it is supposed to say. – Gerry Myerson Feb 29 '24 at 12:35
  • Thanks for your patience first of all. Yes I meant the prime equality Euler's function. From the equation $\prod (1-p^{-s})^{-1} = \sum n^{-s}$ we have n terms on right hand side and I attempt to write $n = \prod P_k$ where each $P_k$ implies the no of prime terms less than n – ajay sharma Feb 29 '24 at 19:16
  • @GerryMyerson if i ask in Euler's equation you emntioned and how you formulated it how muny terms of each prime number exists peoprtionally, surely no of 2s are > no of 3s > no of 5s > no of 7s & so on, even though each series goes to infinity, we can say no of 2's/no of 3's is = $\frac {\log 3}{\log 2}$ & so on. The no of prime numbers less than n are less than $\log_2 n$ because that's the greatest power of 2 less than equal to n. Can you see where this idea may lead... – ajay sharma Feb 29 '24 at 19:28
  • "...we have $n$ terms on the right hand side..." No, we have infinitely many terms, on both sides. "...$n=\prod P_k$ where each $P_k$ implies the no of prime terms less than $n$" Incomprehensible. Take $n=360$. Please tell me what $P_k$ is for $k=1,2,3,\dots$ so I can understand your notation. – Gerry Myerson Feb 29 '24 at 21:37
  • $P_2 = [360/2] = 180$, $P_3 = [360/3] = 120$, $P_5 = [360/5] = 72$, $P_7 = [360/7] = 51$... – ajay sharma Mar 01 '24 at 12:07
  • OK, so, $n=\prod P_k$ says $360=180\times120\times72\times51\times\dotsb$. Good. Now I'm absolutely convinced that you have nothing worth looking at. Bye! – Gerry Myerson Mar 02 '24 at 10:11
  • @Gerry Myerson, not sure why you dont treat x as infinite but think it like this simply by notation we can say if x tends to infinity it should be asymptotic to product of all primes. If not you would be actually saying there is some prime greater than infinity. – ajay sharma Mar 02 '24 at 16:53

1 Answers1

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It is true for any finite number of primes $p_1,\dots,p_k$ that, when $x$ is sufficiently large in terms of $\{p_1,\dots,p_k\}$, the number of integers up to $x$ made up of powers of those primes alone is asymptotic to $\prod_{j=1}^k \frac{\log x}{\log p_j}$.

However, there is an error term in this approximation, and if we tried to extend this idea to an unbounded number of primes, the error term would become far larger than the main term, making the approximation unreliable. Thus trying to examine all integers/all primes this way doesn't end up working.

Greg Martin
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  • The error term would be a product of errors in powers of primes i.e <|1| which in comparison to the powers itself would be negligible – ajay sharma Feb 29 '24 at 21:28
  • Try working it out :) Even for finitely many primes, the right formula is $\prod_{j=1}^k \bigl( \frac{\log x}{\log p} + O(1) \bigr)$; multiplying this product out yields one main term and $2^k-1$ error terms, each of which is definitely larger than $1$. – Greg Martin Feb 29 '24 at 22:31
  • Hi Greg after thinking more about it, I believe we agree on x is asymptotic to product $\prod_{p} \log_p x$ than (I am sorry I have been using equal to and my maths is not polished). Would it be possible for you to overthink from there that if we use the concept of Geometriuc Mean <= Arithmetic Mean, we can treat the product (right hand side) as Geometric Mean and sort of put an upper bound on it by replacing it with AM equivalent ($(\log_p x)^{\pi (x)}$) and basically saying it is asymptotic to x. So even if we say there is an error term it is reallt taken care of by that transformation – ajay sharma Mar 02 '24 at 16:44
  • I'm not sure what you're proposing, but $\pi(x)$ is asymptotic to $x/\log x$, so anything of the shape $(\log x)^{\pi(x)}$ will be way larger than $x$. – Greg Martin Mar 02 '24 at 21:00