Riemann Hypothesis and Primes

[philosophking]

Hi,

When I hear about the Riemann hypothesis, it seems like the first thing I hear about it is its importance to the distribution of prime numbers. However, looking online this seems to be a very difficult thing to explain. I understand that the Riemann Hypothesis asserts that the zeroes of the zeta function all have real part equal to 1/2, but how is this related to prime numbers? I've looked online and the explanations seem very confusing.

Thank you.

 

[matt grime]

google the internet or search these very forums for the answer to this oft asked question

 

[keebs]

Basically Riemann gave a formula for the prime counting function that includes a sum over all zeros of the zeta function (well, not exactly, it's actually a sum of x to the power of all zeros of the zeta function), and if all of the zeros lie on the critical line than we can get a good estimate of the error between Li(x) and the prime counting function.

 

[eljose]

wouldn,t be easier to solve the interal equation for Pi(x) (i propesed a formula for solving this integral equation) to get the value of Pi(x)?

 

[Hurkyl]

Wolfram seems to think that the Lagarias-Odlyzko algorithm is the best to use: that's what's implemented in Mathematica.

 

[matt grime]

wouldn,t be easier to solve the interal equation for Pi(x) (i propesed a formula for solving this integral equation) to get the value of Pi(x)?

ui think the answer is almost certainly "no" still, cos you've never proven otherwise. but riemann stated something abuot the estimate of the difference between the *estiamte* via Li(x) compared to Pi(x). You understand Li(x) is an asymptotic estimate, right?

 

[eljose]

but as far as i know riemann hypothesis being true you stil have the error term that would go as O(x^{0.5}log(x)) so the error diverges, another question is supposed that the formula:

Pi(x)=Li(x)+O(x^{0.5}log(x)) is an asymptotic stimation isn,t it?...

so i don,t see why is better Riemann,s formula for Pi(x) rather than an exact method to calculate it,the approximation of Riemann would only be valid to compute Pi(x) for big x and don,t even that because we don,t know what the function is exactly...

Hurkyl..wher could i find the lagarias Odlyzko algorithm for calculating Pi(x)?

 

[matt grime]

no one has ever claimed that Li(x) tends to Pi(x), waht is the point of your question?

 

[Jonny_trigonometry]

didn't Hardy and Littlewood prove that Li(x) occilates above and below Pi(x) an infinate number of times?

 

[eljose]

i got by solving the integral equation for Pi(x) a solution in the form:

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}t^{-iw}F(w)/G(w)dw$$ with

$$F(w)=\int_{-\infty}^ {\infty}i(2+is)^{-iw-1}Ln\zeta(2+2s)$$

$$G(w)=\int_{-\infty}^ {2}e^{r}e^{ir}/exp(exp(ir))-1$$

so here you are a true function for PI(x),i have never understood mathematician and their "mania" to express always the formulae as f(x)=g(x)+O(h(x)) this does not prove anything, as there is no real way to calculate f(x) from the former expression,don,t even an approximation you only get that f(x)-g(x)<ch(x) this is useless..(at least in my opinion) unless for big x h(x) tends to a constant a and with that you get only that for big x f(x)-g(x)<ca...

Note:there is a mistake the upper limit in G(w) is 2 not oo

 

[eljose]

sorry it was:

$$F(w)=\int_{-\infty}^ {\infty}i(2+is)^{-iw-1}Ln\zeta(2+is)$$

$$G(w)=\int_{-\infty}^ {2}e^{r}e^{ir}/exp(exp(ir))-1$$


$$\frac{1}{2\pi}\int_{-\infty}^ {\infty}t^{-iw}F(w)/G(w)=\pi(e^t)$$

with that you have an exact integral formula for solving it...

 

[matt grime]

We do not have a mania (so far mathematicians are snobs and manics; i think we have shown you considerable patience and indulgence given your less than complimentary attitude). We do care about knowing the error in approximation via some method (something you could do with learning about). Again you appear to think that mathematicians are of the opinion that Li(x) is used to get Pi(x). It is a nice result, one of significant importance, and one that took many years (100) to prove from the original conjecture of Gauss (he made the conjecture when he was 15) based upon observations of tables of data. It is one of the high points of Analytic Number Theory. And as LeVeque says, the proof is too hard to include here.

Any time you actually want to "solve" those integrals go right ahead. I for one have a much simple method for "solving" Pi(x): let s(x)=1 if x is prime and 0 otherwise, then

$$\text{pi}(x)= \sum_{2}^{\floor{x}}s(x)$$

very simple and "solves" pi(x)...

 

[Zurtex]

Here is my favorite representation:

$$\pi(n) = -1 + \sum_{j=3}^n \left((j-2)! - j \Biggl{\lfloor} \frac{(j-2)!}{j} \Biggr{\rfloor}\right) \quad \text{for} \, \, n > 3$$

Not exactly efficient, but quite pretty

 

[eljose]

Matt and Zurtex: the problem you have with your approach is that either you don,t know what s(x) is (the analytic form) or it,s a sum hard to calculate (i don,t know why they published the formula zurtex gave but not mine :( ).

Another question is not RH proved (riemann hypothesis) as far as you know from the functional equation: $$\zeta(1-s)=g(s)\zeta(s)$$ putting s=1/2+it you get:

$$\zeta(1/2-it)=g(s)\zeta(1/2+it)$$ from that would,nt you get RH?...as you know that Zeroes are symmetrically placed

 

[matt grime]

and your integrals are easy to calculate are they? i haven't seen you do one yet. you do understand that we were being sarcastic? the formula zurtex gave, if it were ever published, would have been published because it was a new idea at the time. It may also have been one small result in one paper. yuor idea is not new, the method is not new, the result is not new, and it is very small. it may also have been published because its proof was of sigificant importance for a variety of reasons.

let me give you another example:
let f be any function, then we can work out f(x) via some transform...

that is all you have done, ok? that is not interesting. we know that many functions can be evaluated through inverse functions. you havwe not defined a new function or a new transform or a new way of evaluating anything! and you have royally pissed off anyone who might have wnated to help you make sense of your work.

and now you have a one line proof of the riemann hypothesis (yes i am tired of trying to be nice). and no, the RH does not follow from you observation. (g(s) being what? and why may that not be zero? by your logic ALL zeroes lie on the line Re(s) = 1/2 and we know there are ones that do not lie on the line)

 

[eljose]

you needn,t be sarcastic or unpolite,i have a proof for getting Pi(x) choose the method you want to solve the integral (yes you wll have a series but a finite series simpler than the series Zurtex gave) i give the formula for Pi(x) by means of a Fourier transform and you can use the algorithm of FFT (fast Fourier transform) to solve it so you can improve the time to solve the integrals, i gave an integral representation for Pi(x) and this is new.

In the formula R(1-s)=g(s)R(s) g(s) is a formula involving Gama function and pi^x and 2^{1-x} i,ve never said i had a proof of Riemann Hypothesis,i only gave a reasoning to prove it...

 

[matt grime]

you have given *a* different formula for pi(x). there are many. and an integral representation for it is not new as has been pointed out to you with a refereve to a paper by Ozlydsko. no one disputes what you have we are attempting to explain to you how it fits into the lagrer scheme of mathematics.

 

[shmoe]

Another question is not RH proved (riemann hypothesis) as far as you know from the functional equation: $$\zeta(1-s)=g(s)\zeta(s)$$ putting s=1/2+it you get:

$$\zeta(1/2-it)=g(s)\zeta(1/2+it)$$ from that would,nt you get RH?...as you know that Zeroes are symmetrically placed

No, all this shows (after examining the g term) is that if you have a zero at 1/2+it then you have a zero at 1/2-it (and vice versa), it does not guarantee that any non-trivial zero must be of this form. This is a trivial and well known fact. Actually more symmetry between the zeros is known-reflection principle plus the functional equation tells you non-trivial ones come in sets of 4 symmetric about s=1/2 (collapses to 2 zeros where the zero is on the critical line). This is why they only look for zeros in the upper half of the critical strip, the part below the real axis comes for "free".

 

[eljose]

and what about studying the integral:

$$\int_0^{\infty}\int_{c-i\infty}^{c+i\infty}\frac{x^{s-1-z}}{\zeta(s)s}dsdx$$

the singularities (the values when is infinite) would be the place where zeros are

 

[matt grime]

that doesn't make sense, does it?

 

[eljose]

why not you can check this integral is equal to 1/R(s) so when the integral is infinite so will be the value of 1/R(s)

another question: i have checked several webpages about Lagarias-odlyzko method for pi(x) and does not appear an integral by any side is only a small modification of Lehmer,s method to calculate Pi(x) and this implies calculatig a sum not an integral...

 

[shmoe]

another question: i have checked several webpages about Lagarias-odlyzko method for pi(x) and does not appear an integral by any side is only a small modification of Lehmer,s method to calculate Pi(x) and this implies calculatig a sum not an integral...

I really hope you aren't trying to imply again that an algorithm for computing pi by analytic methods doesn't currently exist. I've told you to look at "Computing pi(x):An Analytic Method", by Lagarias and Odlyzko. If you can't be bothered to actually do this and see how it differs from variants of the Meissel-Lehmer method (Odlyzko and Lagarias have written about the Meissel-Lehmer method as well) there's really no point at all in me wasting my time replying to you.


I just noticed this:

didn't Hardy and Littlewood prove that Li(x) occilates above and below Pi(x) an infinate number of times?

It was Littlewood who did this. There's an upper bound for where the first sign change is guaranteed to happen. I can't recall what the current best upper bound is, but it's still well out of computational reach (the first was by Skewes and was something like 10^34 digits, this has been greatly improved though).

 

[matt grime]

why not you can check this integral is equal to 1/R(s) so when the integral is infinite so will be the value of 1/R(s)

another question: i have checked several webpages about Lagarias-odlyzko method for pi(x) and does not appear an integral by any side is only a small modification of Lehmer,s method to calculate Pi(x) and this implies calculatig a sum not an integral...

i merely meant are you sure you're using valid expansions and relations for the zeta function with correct convergence issues.

 

[eljose]

it,s supposed Louis de Brange mathematician has solved it...but is it true?

another so called "proofs" are in the webpages:http://www.coolissues.com/mathematics/Riemann/riemann.htm

and: http://www.maths.ex.ac.uk/~mwatkins/zeta/RHproofs.htm

tiem will say us if they are correct or not...

 

[shmoe]

De Branges method is known to be flawed. Conrey and Li published a counter example years ago.

In your first link, he's attempting to claim that the real and imaginary parts of zeta(s) and zeta(1-s) cannot be equal (and hence both zero) if real part of s is not 1/2 by looking at their respective Dirichlet series (which don't quite give zeta anyways, this is the alternating guy from Hardy, there's a factor missing). He concludes that u and u' must not be equal because...? They 'look' different? This is actually false anyways, for s in the critical strip off the critical line it is false to claim the real parts of zeta(s) and zeta(1-s) cannot be equal (likewise for the imaginary parts).

One of my favorites is in your second link, by Pradas, which can prove RH on the very false assumption that Zeta is one to one. I've looked at a couple others on that list and have not found anything worthwhile. I'm going to hazard a guess that time will tell all on that page are false. It's likely that all have already been checked by mathematicians and found false. Don't take silence (in some of the cases) to mean that mathematicians are stunned into shame-it's more likely it's not worth the time to comment.

 

[eljose]

have you took a look to the geometric equivalent to riemann hypothesis?..it seems to be quite good..(at least is interesting and seems correcto)

 

[shmoe]

You don't mean Kadia Shi's 'geometric proof' of RH on that page do you? I looked at it a year (or two) ago and remember thinking it was pretty rubbish but I don't remember specifically what went wrong and I don't really want to waste my time (again). Bear in mind this is someone who also claims to have proven the generalized riemann hypothesis, fermat's last theorem, goldbach's conjecture, twin primes conjecture, the irrationality of gamma, and more, so you can guess whether or not I think he's a crackpot.

 

[nate808]

im not sure whether someone actually answered your question so if they did then this will be repetative, but the zeta function that was analyzed as 1/n^s, was transformed into p^s/(p^s-1), where p is a prime number--so the intial funtion look like this
1/2^s+1/3^s+1/4^s which = 2^s/2^s-1 3^s/3^s-1 4^s/4^s-1

 

[nate808]

http://primes.utm.edu/notes/rh.html
this site gives a better example

 

[eljose]

the only think i have been able to prove is that all the x values that satisfy $$\zeta(1/2+ix)=-\zeta(1/2-ix)$$ and the y values that satisfy $$\zeta(1/2-iy)=\zeta(1/2+iy)$$ are real..(with $$\zeta(1/2-ix)=c$$ and c is a complex number but not zero )but this will not be enough to prove RH..

i don,t know if this will be true the question is let,s suppose we have a complex number t+ib satisfying that:

$$\zeta(1/2-b+it)+\zeta(1/2+b-it)=0$$

as we know a zero of $$\zeta(1/2-b+it)$$ is also a zero of $$\zeta(1/2-b+it)$$ so we will have that the quotient 0/0 in the limit x->t+ib will be nonzero but if this happens will have that the quotient $$\frac{\zeta(1/2-b+it)}{\zeta(1/2+b-it)}=-1$$ but we have proved that all the roots that satisfy $$\zeta(1/2+ix)=-\zeta(1/2-ix)$$ are real so b=0..

(If my deduction would be true then i would have solved RH but i,m afraid is not true)...

 

[shmoe]

the only think i have been able to prove is that all the x values that satisfy $$\zeta(1/2+ix)=-\zeta(1/2-ix)$$ and the y values that satisfy $$\zeta(1/2-iy)=\zeta(1/2+iy)$$ are real..

See https://www.physicsforums.com/showthread.php?t=84445 for why your assertion involving y is false. Your statement with an x is false as well and a counterexample is similar, only you want $$\chi(s)=-1$$ for some s off the critical line.

 

[mathwonk]

Riemann's philosophy that a meromorphic function is a global object, associated with its maximal domain, and determined in any subregion, "explains" why the analytic continuation of the zeta function and the Riemann hypothesis help understand primes.

I.e. Euler's product formula shows the sequence of primes determines the zeta function, and such functions are understood best by their zeroes and poles. Since there is only one pole, at z=1, the location of zeroes must be intimately connected with the distribution of primes!

More precisely, in paper VII of his collected works, Riemann says Gauss's logarithmic integral Li(x) actually approximates the number <pi>(x) of primes less than x, plus 1/2 the number of prime squares, plus 1/3 the number of prime cubes, etc..., hence over - estimates <pi>(x). He inverts this relation, obtaining a series of terms Li(x^[1/n]) as a better approximation to <pi>(x), whose proof apparently requires settling the famous "hypothesis".

Empirical data do in fact show that Riemann's approximate formula is better than Gauss's, for finite sets of primes up to reasonable bounds.

So the famous "prime number theorem" was apparently just to show that Gauss's formula was asymptotically valid, and the Riemann hypothesis is required to verify the validity of Riemann's formula, which nonetheless is visibly better than Gauss's for numbers within range of computer evidence.

It is very instructive to read Riemann's original paper, now available in English translation from Kendrick Press, or the book by Edwards that tries to explain it. I myself find Riemann himself clearer and certainly more brief than others coming after him. The material above is taken from the first and last pages of his paper.

(By the way, in Riemann's original version, there was a 90 degree rotation of the z variable, so he speaks of the zeroes of zeta all being on the real axis.)

 

[eljose]

But Shmoe then if $$\zeta(1/2+it)+\zeta(1/2-it)=0$$ and the both zeta function are non-zero the only possible solution is $$\zeta(1/2+it)=-\zeta(1/2-it)$$ so taking the quotient (we can take it as none of them is zero) we would have that $$\frac{\zeta(1/2+it)}{\zeta(1/2-it)}=-1$$ so the modulus of the quotient would be one isn,t that true...( i don,t know the key in my computer to make the modulus of a complex variable,sorry about that)

 

[shmoe]

But Shmoe then if $$\zeta(1/2+it)+\zeta(1/2-it)=0$$ and the both zeta function are non-zero the only possible solution is $$\zeta(1/2+it)=-\zeta(1/2-it)$$ so taking the quotient (we can take it as none of them is zero) we would have that $$\frac{\zeta(1/2+it)}{\zeta(1/2-it)}=-1$$ so the modulus of the quotient would be one isn,t that true...( i don,t know the key in my computer to make the modulus of a complex variable,sorry about that)

I don't understand what you're trying to say. From your last post it looked like you were hoping that if $$\zeta(1/2+it)+\zeta(1/2-it)=0$$ then t had to be real. This is equivalent to $$\zeta(s)+\zeta(1-s)=0$$ implying that s is on the critical line, which I've explained to be false.

 

[eljose]

but you are forgeting that:

$$\zeta(1/2-is)+\zeta(1/2+is)=\zeta(1/2-is)[1+\chi(1/2-is)]=0$$

as i have said that zeta is non-zero then we have s satisfy the roots of the equation

$$\chi(1/2-is)=-1$$ so the modulus of the function $$\chi(1/2-is)$$ would be equal to 1.