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Is this a correct statement? If the twin prime conjecture is correct, then we may conclude that the [prime] integers are not constructed randomly. Another int

by sytelus 12y ago

Is this a correct statement? If the twin prime conjecture is correct, then we may conclude that the [prime] integers are not constructed randomly. Another interesting tid-bit.. Looks to me that mathematics research is getting overwhelmingly complex... For the next two semesters, I organized a seminar with five graduate students (including Yitang) on Prof. Hironaka’s monumental papers on 2 the theory of resolutions of singularities. I believed that we doubled the world population of those who had studied the papers after we finished two semesters. Prof. Grothendick once described those papers as among the most complicated thesises in the human history.

darkmighty 12y ago
> Is this a correct statement? Of course it is. By 'randomly' he refers to a maximum entropy ("uniform") distribution conforming Gauss' Prime Number Theorem. Since by that theorem the gap rises, the probability of twin primes would go to 0 and, crucially, the expected number of twin primes would be finite (or any finite gap really). That's an interesting reason for Yitang's proof being so significant without actually reaching the small gap -- it shows primes violate eloquently the random distribution in some ways.
- Retric 12y ago
  If the probably of a number being prime is ~1 / ln (N) The probably of x and x + 2 being prime is ~ (1 / ln (N)) ^2. For large N, (ln (N)) ^2 < N so there are infinitely many twin primes. Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x). So, the real issue is if the number of twin primes significantly larger than (1/ln(x)) ^ 2.
  - darkmighty 12y ago
    Ah yes you're right I just naively assumed the gap going to infinity implied the number of small gaps has to be finite, thanks. I believe your argument argument sightly miscounts the twin primes, since you're giving a new "chance" after you had already drawn a prime/not-prime. A lower bound assuming the probabilistic model would be E[Twin Primes] > Sum ( 1/( ln(4k) ln(4k+2) ), which is still divergent so your point stands. > so there are infinitely many twin primes http://en.wikipedia.org/wiki/Twin_prime http://en.wikipedia.org/wiki/Twin_prime Note the twin prime conjecture is not yet proven (your argument does not follow directly from PNT, but from the approximate probabilistic model that (almost surely?) displays PNT-like count). > Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x). Here's a cute counterexample: x/(2*sign(sin(x))) < x, but the sum < infinity (you're missing an absolute value |f(x)| in your statement).
    - Retric 12y ago
      Yea, I was playing fast a loose. Though, for positive infinity you need 0 < f(x) < (x) not just |f(x)| < x. EX: f(x) = -0.5 * x
- ilyagr 12y ago
  To expound on Retric's comment (I expounded much more in a cousin thread): I also first assumed that what you said must be true. When I tried to check with a calculation, it turned out that the opposite is the case.
  - darkmighty 12y ago
    Yea, my bad. I implicitly made the rookie mistake of assuming that f(x)->0 implies sum(f(x))->0, as shown by Retric :)

by sytelus 12y ago

darkmighty 12y ago

> Is this a correct statement? Of course it is. By 'randomly' he refers to a maximum entropy ("uniform") distribution conforming Gauss' Prime Number Theorem. Since by that theorem the gap rises, the probability of twin primes would go to 0 and, crucially, the expected number of twin primes would be finite (or any finite gap really). That's an interesting reason for Yitang's proof being so significant without actually reaching the small gap -- it shows primes violate eloquently the random distribution in some ways.

Retric 12y ago
If the probably of a number being prime is ~1 / ln (N) The probably of x and x + 2 being prime is ~ (1 / ln (N)) ^2. For large N, (ln (N)) ^2 < N so there are infinitely many twin primes. Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x). So, the real issue is if the number of twin primes significantly larger than (1/ln(x)) ^ 2.
- darkmighty 12y ago
  Ah yes you're right I just naively assumed the gap going to infinity implied the number of small gaps has to be finite, thanks. I believe your argument argument sightly miscounts the twin primes, since you're giving a new "chance" after you had already drawn a prime/not-prime. A lower bound assuming the probabilistic model would be E[Twin Primes] > Sum ( 1/( ln(4k) ln(4k+2) ), which is still divergent so your point stands. > so there are infinitely many twin primes http://en.wikipedia.org/wiki/Twin_prime http://en.wikipedia.org/wiki/Twin_prime Note the twin prime conjecture is not yet proven (your argument does not follow directly from PNT, but from the approximate probabilistic model that (almost surely?) displays PNT-like count). > Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x). Here's a cute counterexample: x/(2*sign(sin(x))) < x, but the sum < infinity (you're missing an absolute value |f(x)| in your statement).
  - Retric 12y ago
    Yea, I was playing fast a loose. Though, for positive infinity you need 0 < f(x) < (x) not just |f(x)| < x. EX: f(x) = -0.5 * x
ilyagr 12y ago
To expound on Retric's comment (I expounded much more in a cousin thread): I also first assumed that what you said must be true. When I tried to check with a calculation, it turned out that the opposite is the case.
- darkmighty 12y ago
  Yea, my bad. I implicitly made the rookie mistake of assuming that f(x)->0 implies sum(f(x))->0, as shown by Retric :)