Question

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

Answer 1

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 \(x\).

Answer 2

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.

Answer 3

@dan815 let's do this proof, I wanna prove that:

\[\lim_{x \to \infty} \frac{\pi(x)}{x/\ln x} = 1\]

Answer 4

what is pi(x)?

Answer 5

\[\pi(x) = \sum_{p \le x} 1\]

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.

Answer 6

ok gotcha

Answer 7

So like for large numbers,

\[\pi(x) \approx \frac{x}{\ln x}\]

Which is kinda cool.

Answer 8

interesting

Answer 9

e is also seen in derrangement formula,

Answer 10

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.

Answer 11

but I don't care if we get side tracked like I'll eventually get there, what's the derrangement formula?

Answer 12

http://prntscr.com/9a44eq