Offizieller Riemann-Hypothese Faden

[petrw1]

For me it was mind-boggling to read about this and learn that an infinite sum of a function of integers is equal to (or appears to be) an infinite product of a function of prime numbers and they both equal (or converge to) a function of the value pi. surprisedsurprised:surprised

To me this states that prime numbers are anything but random but rather follow some "intriguing" pattern.

I created a simple Excel spreadsheet to demonstrate this and was further intrigued to find that the prime product series coverges MUCH quicker than the integer sum series.

My spreadsheet compares the following two series as defined by Euler.

The left side of the equation is the SUM of the fractions using the simple series:
1 + 1/2^n + 1/3^n + 1/4^n ...
The right side of the equation is the PRODUCT of the fractions using the sequence of primes starting with 2:
(2^n/((2^n)-1)) * (3^n/((3^n-)1)) * (5^n/((5^n)-1)) ...

For n=2 the result converges to pi^2/6 = 1.644934067...
The right equation is correct to 3 decimal places within 57 iterations while the left side requires 1070 iterations for the same precision.
The right side is correct to 4 decimal places within 679 iterations while the left side required 29354 iterations.

For n=4 the result converges to pi^4/90 = 1.082323234...
The right side is correct to 9 decimal places within 115 iterations while the left side required 1164 iterations.

[alpertron]

Originally Posted by petrw1

For me it was mind-boggling to read about this and learn that an infinite sum of a function of integers is equal to (or appears to be) an infinite product of a function of prime numbers and they both equal (or converge to) a function of the value pi.

The expressions are equal for the zeta argument s > 1, actually the proof is not complicated.

The functions of the value pi (more specifically a rational number times an even power of pi) correspond to even values of s.

[Raman]

Originally Posted by jasong

If I may ask a totally ignorant question:

In the very first post of this thread, someone said that if the Riemann(sp?) Hypothesis is proven, it would greatly speed up our search. My ignorant question is:

If the Riemann Hypothesis is wrong, what would be the consequences of using it? Note that I have no idea how it would be used, which is the reason I call this an ignorant question.

If the use of the Riemann Hypothesis causes the inclusion or exclusion of prime exponents in the Mersenne equation, perhaps the question of whether or not someone wants the Prime95 program to include the Hypothesis' findings in exponent selection could be an option.

If this is gibberish(I know some people like to use that word :) ) I apologize.

Hi,

Currently ECPP (Elliptic Curve Primality Proving) is the fastest method to prove whether a number is prime or not. AKS algorithm exist so as a deterministic polynomial time algorithm, that does not depend upon any unproved assumptions at all, but its average running time for a number is much greater than when compared up to the ECPP algorithm.

If the Riemann Hypothesis is true, (or the Extended Riemann Hypothesis), then there exist a very simple algorithm to prove whether a number is prime or not. The Miller's Test, it states that, if a number N is a-SPRP for all bases a from 1 to 2 (log₂ N)², then N is prime.

If the Riemann Hypothesis is false, then we have to go up within the traditional methods for proving primality of an integer, the ECPP test. However, that two facts about the Riemann Hypothesis are known to be true.

(1) There exist infinitely many zeros of the Zeta Function along the critical line Re(s) = 1/2.
(2) Atleast 40% of the zeros of the critical strip 0 < Re(s) < 1 lie up precisely on the critical line Re(s) = 1/2.

The Riemann Hypothesis claims that all of the non-trivial zeros of the Riemann Zeta Function across the critical strip 0 < Re(s) < 1 lie up precisely on the critical line Re(s) = 1/2.

[lavalamp]

Originally Posted by Raman

Currently ECPP (Elliptic Curve Primality Proving) is the fastest method to prove whether a number is prime or not.

If the Riemann Hypothesis is true, (or the Extended Riemann Hypothesis), then there exist a very simple algorithm to prove whether a number is prime or not. The Miller's Test, it states that, if a number N is a-SPRP for all bases a from 1 to 2 (log₂ N)², then N is prime.

Would performing 9.331 billion SPRP tests really be quicker than running ECPP on say ... a 20,562 digit number picked completely at random?

I mean sure, it's an embaressingly parallel operation and graphics cards are real sexy for stuff like that, but ECPP can be easily checked. The only way to be sure you have a good prime with the Miller test is to run the test again, perhaps somehow generating a residue in the process for comparison. So would running 18.663 billion SPRP tests be quicker than ECPP on a 20,562 digit number?

Is there a switch point below which one is more efficient than the other, and above which it is vice-versa? Are there multiple such switch points?

[R.D. Silverman]

Originally Posted by lavalamp

Would performing 9.331 billion SPRP tests really be quicker than running ECPP on say ... a 20,562 digit number picked completely at random?

?

The answer is unclear. Proof via ERH requires 2 log^2 N modular exponentiations. Each one takes O(log^(2 + epsilon)) N) time, using FFT's.
This makes it an O(log^(4 + epsilon) N) method. One can actually
replace fixed epsilon with o(1).

If certain conjectures about frequency of near-primes in short intervals
is correct, and if one uses FFT's for the arithmetic, ECPP is also an
O(log^(4 + o(1)) N) algorithm.

Crossover points will be machine and implementation dependent. There
will also be some statistical variation in the ECPP run time.

However none of this has anything to do with Mersenne primes.

[fivemack]

Is 'Faden' the right word here? It's 'thread' as in clothing and as in 'hanging by a thread' (as opposed to 'Gewinden' which is 'thread' as in screw) ; on the other hand I can't find German-language forums with the 'furniture' in German to check.

[kar_bon]

Originally Posted by fivemack

Is 'Faden' the right word here? It's 'thread' as in clothing and as in 'hanging by a thread' (as opposed to 'Gewinden' which is 'thread' as in screw) ; on the other hand I can't find German-language forums with the 'furniture' in German to check.

i think the best word in German for thread like these is 'Themenstrang' (see http://www.dict.cc/?s=thread)

[mart_r]

Actually, if the Germans were more eager to use their own language, then the right word would be 'Faden'.


On the RH: Did I get that right, if zeta(a±bi) with 0<a<1/2 is zero, then also zeta(1-a±bi) is zero? But isn't the value of, say, zeta(.6±bi) always more closely distributed around 1 than zeta(.4±bi)? With such a zero, what would the value of zeta(.5±bi) look like?

By the way, I looked at how the sums 1+1/2^s+...+1/n^s evolve, calculated interpolated values and found that the derivative(?) at the start (n=0) reach a minimum at every zero of zeta(s). I've never heard of that result before.

[mart_r]

Originally Posted by mart_r

By the way, I looked at how the sums 1+1/2^s+...+1/n^s evolve, calculated interpolated values and found that the derivative(?) at the start (n=0) reach a minimum at every zero of zeta(s). I've never heard of that result before.

Note: I was talking about the non-trivial zeroes here.