reciprocals of twin primes
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- Опубликовано: 11 фев 2024
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Fun fact is that trying to approximate Brun’s constant discovered a bug in Intel CPUs in 1994.
B_3 is exactly 0.7 Phew, that was exhausting!😜
The twin prime conjecture is my second favourite unproved mathematical conjecture after Riemann's zeta conjecture.
My favorite conjecture is the collatz conjecture
@@kpopalitfonzelitaclide2147same
My favorite confecture is the Napoleon
🍰
😏
Brun’s Thm is my favorite result in number theory!! i haven’t dug into it in about 25 years since undergrad so I’m jazzed for this vid…thanks!!!
Very interesting. Thank you, Michael.
For natural numbers i, j and n>j there exist a product function from i=2 to i=n-1 [sin (pi*n/i)] which is different than 0 and it is a prime number.
For such function there is also true that:
P= sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
where P+1 is the order of prime number (O.O.P.N.)
which also can be written as
O.O.P.N. =1 + sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
We also must add number 2 as a 1st prime as an axiom to complete this formula.
mind blowing bit there
Can we assume p+2 being prime is independent of p being prime? I ask because when writing the probabilities Michael assumed so at 7:26
With a heuristic of looking at different moduli, p+2 should be less likely to be prime given p is prime, meaning the sum should converge faster. But I don't know if that's a good heuristic to make.
@@galoomba5559 Thanks, this is one of those things you need to think carefully about.
Problem assumes we are dealing with twin primes, e.g. 11,13 , 17, 19
For natural numbers i, j and n>j there exist a product function from i=2 to i=n-1 [sin (pi*n/i)] which is different than 0 and it is a prime number.
For such function there is also true that:
P= sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
where P+1 is the order of prime number (O.O.P.N.)
which also can be written as
O.O.P.N. =1 + sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
We also must add number 2 as a 1st prime as an axiom to complete this formula.
@@MyOneFiftiethOfADollar For natural numbers i, j and n>j there exist a product function from i=2 to i=n-1 [sin (pi*n/i)] which is different than 0 and it is a prime number.
For such function there is also true that:
P= sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
where P+1 is the order of prime number (O.O.P.N.)
which also can be written as
O.O.P.N. =1 + sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
We also must add number 2 as a 1st prime as an axiom to complete this formula.
Don't know if this is useful for this proof, but the sum of twin primes is congruent to 0 modulo 12.
Example: 29+31 = 60==0 (mod 12). True because twin prime are of the form 6k +- 1
Could you please link the previous video?
Convergence is not a matter of infinite terms. It's about a limiting sum while the number of terms can still remain infinite!!!
7:30 Do you need to use Baye's Theorem when combining the probabilities? (I know the probabilities of two numbers being prime is independent, but is the probably of twin primes is dependent?)
This should be interesting!
Since these B_n are only nonzero for even n, I'd wonder whether there's some connection to Bernoulli numbers hiding in there. This isn't an outrageous thought to have either, since we already know the zeta function values are related to Bernoulli numbers, and the log of the zeta function is related to sums of reciprocal of primes...
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I'm not entirely sure why having an infinite amount of terms makes it obvious that this series diverges.
15:09
7:55 the problem is for example the only primes that differ by 1 is obviously 2,3 which you can still make an argument that the probability of n and n+1 being prime is still≈1/ln(n)^2
Well obviously we know what B_1 is then - it’s 5/6 😂
You mean n and n+2
@@NathanSimonGottemer It's also fairly easy to get a closed for for B_n for all odd numbers n.
@@normanstevens4924 yep, it’s 0 for all the other odds :P
Observe: There's a constant that can be obtained by studying the edge cases of paradoxes. That constant has a binary representation.
Fortune's Conjecture:
There's a logical ordering of all possible conjectures such that the provability (1) or non-provability (0) of that conjecture corresponds to the binary representation of that constant.
Are probabilities of numbers being prime independent? If not, multiplying probabilities doesn't work.
I thought this at first, too, but I think the assumption that the primes are uniformly distributed precisely means that we're just testing primeness for each number in the set regardless of their place in the actual order of numbers. Obviously if n>2 is prime, then n+1 isn't , but considered as a set with uniform probability, the normal ordering doesn't matter.
I may be mistaken - if so apolologies ...
As it seems difficult to get exact results then creating bounds using real numbers and functions of real numbers allows finite and infinite bounds as estimates. And until situation improves that is best math can do at this level for the moment ∞∃∞∀∞
no, of course the probabilities are not independent
and of course this proof "doesn't work", he was saying that the entire time
it's evidence why it *should* be true, for the proof you need to be much more careful
@@Mmmm1ch43l really? I would think that the probability that 2 numbers are prime would be independent.
@@trogdor20X6 not when conditioned on their difference
for example: if you know that p is prime, what's the probability that p+7 is also prime? zero, cause there are no primes which are exactly 7 apart
nobody knows what happens with a difference of 2, which is why this proof is only a heuristic
Interesting!
what about reciprocals of consecutive primes with a gap of 4
As of today we.know that there are infinitely many prime pairs with a difference of 246 or less. There has been further work to reduce the gap to 6 but for the moment it relies on other conjectures that haven't been proven. For 2 and 4, we simply don't know
Can anyone explain to me how, at around 11:00, he goes from the sum over all the p's which are the first numbers in a twin prime pair of 2/p, to the sum over all n between 2 and t of P_2(n) / n ?? I'm not seeing how it follows, even asymptotically?
Set p_n the nth prime number, then, by the prime number theorem (in fact, it is an equivalent statement) p_n ~ n*log(n)
Anyways this is not a trivial result, i suppose Penn skips all the technicalities of that step and goes on on bona fide...
@@r.maelstrom4810 Thank you, I appreciate the answer.
I know the reciprocals of primes diverge, I even understand the proof, but it'll never make sense to me. It's the harmonic minus like all the terms, why tf wouldn't it converge
For natural numbers i, j and n>j there exist a product function from i=2 to i=n-1 [sin (pi*n/i)] which is different than 0 and it is a prime number.
For such function there is also true that:
P= sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
where P+1 is the order of prime number (O.O.P.N.)
which also can be written as
O.O.P.N. =1 + sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
We also must add number 2 as a 1st prime as an axiom to complete this formula.
Wouldn't it be much faster to prove convergence by Cauchy condensation test?
Doesn't seem like this really works?
I don't see how to use that test here.
Yes indeed, from Cauchy's condensation test and the p-series criteria it would be a straight forward coclusion.
@@CesarDiaz-zy5yy Could you please show the steps how to do that?
@@bjornfeuerbacher5514 I was thinking on the proof for the convergence of the series of the last step, the one that begins on minute 12:30. The convergence of the reciprocals of the twin primes certainly is not consequence solely of Cauchy's condensation test. The part of obtaining an estimate for the quantity of twin primes less or equal than n is the important step.
You are too modest, Michael! You could present it as stronger proof: in the 1/n/log(n)^2 sum, one factor log(n) is already certain from the prime number theorem, and you actually do not need log(n)^2 to get convergence, for any positive eps, the sum 1/n/log(n)^(1+eps) is already convergent. To be non-convergent the primes for large n would have to come almost exlusively in twins! And most likely that is ruled out as well by proven results (the work that has already been done on the twin-prime conjecture, I mean...)
Congratulations you are a physicist now
11:05
did you just prove here that the sum of reciprocals of primes converges???
I love this. Dancing on the boundary. We don't know if it's true, maybe it is both true and false, maybe we just can't prove it yet, but we can clearly do some stuff using it. I wonder if the twin prime conjecture will fall to a diagonal argument (i.e. it's both)
Someone out there needs to write a song about the amazing properties of Twin Primes, to the tune of Blurred Lines. Michael will perform in on this channel!
For natural numbers i, j and n>j there exist a product function from i=2 to i=n-1 [sin (pi*n/i)] which is different than 0 and it is a prime number.
For such function there is also true that:
P= sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
where P+1 is the order of prime number (O.O.P.N.)
which also can be written as
O.O.P.N. =1 + sum j=3 to j=n from [product function from i=2 to i=n-1 psin (pi*n/i)]/product function from i=2 to i=n-1 [sin (pi*n/i)]
We also must add number 2 as a 1st prime as an axiom to complete this formula.
@@bozydarziemniak1853 Did you mean to post this somewhere else? It doesn't seem like you meant it as a response to my comment.
@@jerrysstories711 I just showed that I have found the prime number generator.
I don't understand that bit around 5:30. The probability that "some number from the set {1,2,...,x} is a prime" is 100%. It's right there. 2 is a prime. So p(x) = 1 if x >= 2, and zero otherwise. What am I missing? ELI5 pls ^_^
he means:
"if you randomly choose some number from the set {1,2,...,x} what's the probability that it's prime?"
not:
"what's the probability that at least one number in the set {1,2,...,x} is prime?"
@@Mmmm1ch43l ohhhh thanks that makes sense.