Prime Number Theorem: Proving Equivalence of $P(x)$ and $Li(x)$

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Discussion Overview

The discussion revolves around the equivalence of two functions, \( P(x) \) and \( Li(x) \), in the context of the Prime Number Theorem. Participants are exploring the mathematical relationships and proofs necessary to establish that \( P(x) \sim Li(x) \) is equivalent to the Prime Number Theorem, which states that \( \pi(n) \sim \frac{n}{\log n} \).

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant defines \( P(x) \) and \( Li(x) \) and states the Prime Number Theorem, seeking assistance in proving the equivalence \( P(x) \sim Li(x) \).
  • Another participant notes that since \( \sim \) is transitive, it is sufficient to show \( Li(x) \sim \frac{x}{\log(x)} \).
  • A participant claims to have proven \( Li(x) \sim \frac{x}{\log(x)} \) but expresses uncertainty about proving \( P(x) \sim Li(x) \).
  • One participant suggests that if \( \pi(x) \sim \frac{x}{\log(x)} \) and \( Li(x) \sim \frac{x}{\log(x)} \) are established, then by transitivity, \( \pi(x) \sim Li(x) \) follows.
  • Another participant emphasizes the distinction between \( P(x) \) and \( \pi(x) \), asking for suggestions on how to prove \( P(x) \sim Li(x) \).
  • A subsequent reply suggests that \( P(x) \) can be viewed as \( \pi(x) \) plus some insignificant terms, proposing that it suffices to show \( P(x) = \pi(x) + O(\sqrt{x}) \).
  • Another participant expresses confusion about how to proceed with this proof.

Areas of Agreement / Disagreement

Participants appear to agree on the definitions and relationships between \( P(x) \), \( Li(x) \), and \( \pi(x) \), but there is no consensus on how to prove the equivalence \( P(x) \sim Li(x) \). Multiple competing views and approaches remain in the discussion.

Contextual Notes

There are unresolved mathematical steps regarding the proof of \( P(x) \sim Li(x) \) and the significance of the terms involved in the relationship between \( P(x) \) and \( \pi(x) \).

peteryellow
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We have that

$P(x) = \sum_{k=1}^{\infty} \frac 1k \pi(x^{1/k})$
and
$Li(x) = \int_2^n \frac {dt}{\log t}$

And the prime number theorem is:

$$\pi(n) \sim \frac{n}{\log n }$$
I want to show that $$P(x) \sim Li(x)$$ is equivalent to prime number theorem.

Can some body please help me with this.
 
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\sim is transitive, so it suffices to show that
[tex]\operatorname{Li}(x)\sim x/\log(x)[/tex]
 
I have shown this but still I need to show that P(x) \sim Li(x).
 
Then you're done:
1. [tex]\pi(x)\sim x/\log(x)[/tex] (Prime Number Theorem)
2. [tex]\operatorname{Li}(x)\sim x/\log(x)[/tex] (you said you proved it already)
3. [tex]\pi(x)\sim\operatorname{Li}(x)[/tex] (by transitivity of ~)
 
I wish I was done but I am not how can I prove that P(x) \sim Li(x).

P is different from /pi. Give any suggestion how can I prove this.
 
peteryellow said:
P is different from /pi. Give any suggestion how can I prove this.

Ah, sorry, I forgot your notation. But P(x) is just pi(x) plus some insignificant terms. It suffices to show that P(x) = pi(x) + O(sqrt(x)).
 
yes, but how.
 

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