What is the inequality for prime numbers in the Prime Number Theorem proof?

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    Inequality Prime
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Discussion Overview

The discussion revolves around understanding an inequality related to prime numbers as presented in the proof sketch of the Prime Number Theorem. Participants explore the notation and implications of the inequality, as well as its context within number theory.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about the inequality involving primes and logarithms, specifically questioning the meaning of "O" in the context of the inequality.
  • Another participant suggests that "O" likely refers to zero, while also speculating on various forms of big O notation that could apply.
  • Some participants note potential typos in the original inequality presented, particularly regarding the summation and the logarithmic terms.
  • A participant points out that the inequality can be understood by taking logarithms of both sides, indicating a relationship between the variables involved.
  • There is a discussion about the conditions under which the inequality holds, particularly the requirement that p must be greater than a certain threshold.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the inequality or the notation used. Multiple viewpoints and uncertainties remain regarding the details of the inequality and its implications.

Contextual Notes

Limitations include potential typos in the original inequality, ambiguity in the notation used, and varying interpretations of the mathematical expressions involved.

PsychonautQQ
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I believe this is probably a high level undergraduate question, but i could easily be underestimating it and it's actually quite a bit higher than that.

I'm reading the Prime number theorem wikipedia page and I'm in part 4 under Proof sketch where sometime down they give in inequality:

x is a natural number, p is for prime's obviously: ##\sum_{x^{1-\epsilon }\geq p\geq x}^{} \log p \geq \sum_{p\leq x}^{} logx##

Where epsilon is any value greater than O. (O is some special value that they use in computer science a lot apparently, I might need to understand this value better to understand this inequality, I'm not sure.

Can somebody help me understand this inequality? C'mon I know there are some really smart people here!
 
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PsychonautQQ said:
O is some special value that they use in computer science a lot apparently
Hi PsychonautQQ:

I am not into number theory, but I am almost certain that the "O" is zero, i.e., "0".

Regards,
Buzz
 
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Buzz Bloom said:
Hi PsychonautQQ:

I am not into number theory, but I am almost certain that the "O" is zero, i.e., "0".

Regards,
Buzz
My guess is ##O(something)##, e.g. ##O(1)\, , \,O(\log x) \, , \, O(\log \log x)## or even ##o(1)##. In any case I assume there is more than one typo in it, e.g. the ##\log x## in a summation over ##p## looks suspicious.
 
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Here is a link to the article, the inequality discussed is the 6th formula in this section.
"for any ε > 0" is a comparison with the number 0. The big O notation (with the letter O) is mentioned because it is used later in the same line.
 
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mfb said:

I found the equation in post #1 in the the linked article, although the post #1 version seems to have omitted a detail. It seems clear that the intent is
ε > zero​
and
O(x1-ε)​
is using the "big O" notation.

Regards,
Buzz
 
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I thought it is the 6th formula there, but the right hand side is not identical.
 
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Wow, I am so smart for confusing 0 and O haha.

I still don't understand the inequality though, can somebody explain why it makes sense intuitively or link a proof to me?
 
With the correct inequality from the Wikipedia article: We know ##x^{1-\epsilon} < p##. Take the logarithm on both sides and you get the inequality used in the article.
 
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mfb said:
With the correct inequality from the Wikipedia article: We know ##x^{1-\epsilon} < p##. Take the logarithm on both sides and you get the inequality used in the article.
And we know that x^(1-epsilon) < p because of the domain which is being summed over? p is required to be bigger?
 
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We know it because the sum only runs over p satisfying this condition, right.
 

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