Kainui
  • Kainui
Trying to prove an inequality in the prime number theorem. I wanna understand the whole proof of the prime number theorem! https://en.wikipedia.org/wiki/Prime_number_theorem#Proof_sketch
Mathematics
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SOLVED
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chestercat
  • chestercat
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Kainui
  • Kainui
for any \(\epsilon > 0\), $$\sum_{x^{1-\epsilon} \le p \le x} \ln p \ge \sum_{x^{1-\epsilon} \le p \le x}(1-\epsilon) \ln x$$ Where p are primes, so really the summation is just on all the primes between \(x^{1-\epsilon}\) and \(x\).
Kainui
  • Kainui
I found that I could write the inequality in terms of the primorial and prime counting function if anyone's curious: \[\ln (x \#) - \ln (x^{1-\epsilon} \#) \ge (1- \epsilon) *\ln(x) *[\pi(x)-\pi(x^{1-\epsilon})]\] Still doesn't seem "obviously true" to me or anything hmm.
Kainui
  • Kainui
@dan815 let's do this proof, I wanna prove that: \[\lim_{x \to \infty} \frac{\pi(x)}{x/\ln x} = 1\]

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dan815
  • dan815
what is pi(x)?
Kainui
  • Kainui
\[\pi(x) = \sum_{p \le x} 1\] Just means it counts the total number of primes less than or equal to x. Here's an example: \[\pi(6) = 3\] since 2, 3, 5 are the 3 primes less than or equal to 6.
dan815
  • dan815
ok gotcha
Kainui
  • Kainui
So like for large numbers, \[\pi(x) \approx \frac{x}{\ln x}\] Which is kinda cool.
dan815
  • dan815
interesting
dan815
  • dan815
e is also seen in derrangement formula,
Kainui
  • Kainui
Yeah so that's what I wanna prove, it's supposed to be like the most important thing to number theory and like related to the Riemann hypothesis.
Kainui
  • Kainui
but I don't care if we get side tracked like I'll eventually get there, what's the derrangement formula?
dan815
  • dan815
http://prntscr.com/9a44eq
dan815
  • dan815
|dw:1449200879517:dw|
Kainui
  • Kainui
What's !n ? lol
Kainui
  • Kainui
W/e ur 2 addicted to chess you're about to leave anyways lol
dan815
  • dan815
no lol im still thinking
kohai
  • kohai
Kai you sound like a woman
dan815
  • dan815
lets think about how ln and es naturally pop up
dan815
  • dan815
d/dx(e^x)=e^x
Kainui
  • Kainui
dan always leaves me, he breaks all the girls at os's hearts, mine most of all @kohai
dan815
  • dan815
oh brother
kohai
  • kohai
Omg kai you poor thing :(
Kainui
  • Kainui
\[e^{\ln x} = x\] \[\frac{d}{dx} (e^{\ln x}) = 1\] \[\frac{d}{dx} \ln x = \frac{1}{x}\] Hmmm
dan815
  • dan815
thats cool hehe
dan815
  • dan815
okay lets think about it like branching
dan815
  • dan815
i can kind of see why e and ln have to be showing up for primes
dan815
  • dan815
think about divsibility having to do with exponential spirals
Kainui
  • Kainui
Alright sure, I guess we could do that. I think it might be easier to think about the 'why' after learning the steps of the prime number theorem first though, it's not by any means some simple little thing lol.
dan815
  • dan815
as your increase in manitude of the number, you will keep seeing more and more numbers dividing more and more numbers
dan815
  • dan815
okay lets try to work on some interesting solved number theory problems
dan815
  • dan815
like some number theory question, where people have found cool theorems on
Kainui
  • Kainui
Like this one right here: https://en.wikipedia.org/wiki/Prime_number_theorem#Proof_sketch
dan815
  • dan815
we will try to figure our the theorem we need to solve it ourself, and we can see the ones they found
dan815
  • dan815
i cant understand what he is saying
dan815
  • dan815
|dw:1449202026809:dw|
dan815
  • dan815
what does this look like if u write out a few terms.... what is this thing summer over
dan815
  • dan815
nvm the next couple lines indirectly explains how that notation works then so
dan815
  • dan815
where p and k are intergers, and satisfy the equation p^k<=x
Kainui
  • Kainui
Yeah there are sorta three repesentations that mean the exact same thing there, this is called the second Chebyshev function: \[\psi(x) = \sum_{p^k \le x} \log p = \sum_{n \le x} \Lambda(n) = \sum_{p \le x} \log p \left\lfloor \frac{\log x}{\log p} \right\rfloor \] \(\Lambda(n)\) is called the von Mangoldt function.
ganeshie8
  • ganeshie8
cancel log p and get log x
ganeshie8
  • ganeshie8
\(\sum\limits_{p\le x} \log x = \log x\cdot \sum\limits_{p\le x}1 = \log x\cdot \pi(x)\)
dan815
  • dan815
hey lets go back to the reimann zeta function
Kainui
  • Kainui
This is all related to the Riemann zeta function so we're getting there pretty soon
ganeshie8
  • ganeshie8
\(\zeta(s) =\sum\limits_{n=1}^{\infty}n^{-s}\) \(\zeta'(s) =-\sum\limits_{n=1}^{\infty}n^{-s}\log n\) \(\sum\limits_{n=1}^{\infty}\Lambda(n)n^{-s} = \begin{array}{}&~~~\log( 2)(2^{-s}+2^{-2s}+2^{-3s}+\cdots )\\&+\log(3)(3^{-s}+3^{-2s}+3^{-3s}+\cdots)\\&+\log(5)(5^{-s}+5^{-2s}+5^{-3s}+\cdots)\\ &~~~\vdots \end{array} = \sum\limits_{p} \dfrac{\log p}{p^s-1}\)
Kainui
  • Kainui
Alright I see, I keep messing myself because I am wrongly assuming that \(\Lambda(n)\) is multiplicative!
Kainui
  • Kainui
So @ganeshie8 are you familiar with this equation? \[\psi(x)=x - \sum_\rho \frac{x^\rho}{\rho} - \log(2 \pi)\]
dan815
  • dan815
|dw:1449206042485:dw|
dan815
  • dan815
what will u do about this now?
Kainui
  • Kainui
Ok I see that this statement must be true: \[\zeta'(s) =- \sum_{n=1}^\infty \frac{(u \star \Lambda)(n)}{n^s}\] since \[(u \star \Lambda)(n) = \sum_{d|n} \Lambda(d) = \ln n\]
ikram002p
  • ikram002p
=D easy proof!

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