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.
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.
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)?
Wolfram seems to think that the Lagarias-Odlyzko algorithm is the best to use: that's what's implemented in Mathematica.
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?
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)?
didn't Hardy and Littlewood prove that Li(x) occilates above and below Pi(x) an infinate number of times?
i got by solving the integral equation for Pi(x) a solution in the form: [tex]\frac{1}{2\pi}\int_{-\infty}^{\infty}t^{-iw}F(w)/G(w)dw [/tex] with [tex]F(w)=\int_{-\infty}^ {\infty}i(2+is)^{-iw-1}Ln\zeta(2+2s) [/tex] [tex]G(w)=\int_{-\infty}^ {2}e^{r}e^{ir}/exp(exp(ir))-1 [/tex] 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
sorry it was: [tex]F(w)=\int_{-\infty}^ {\infty}i(2+is)^{-iw-1}Ln\zeta(2+is) [/tex] [tex]G(w)=\int_{-\infty}^ {2}e^{r}e^{ir}/exp(exp(ir))-1 [/tex] [tex]\frac{1}{2\pi}\int_{-\infty}^ {\infty}t^{-iw}F(w)/G(w)=\pi(e^t) [/tex] with that you have an exact integral formula for solving it.......
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 [tex]\text{pi}(x)= \sum_{2}^{\floor{x}}s(x)[/tex] very simple and "solves" pi(x)....
Here is my favorite representation: [tex]\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[/tex] Not exactly efficient, but quite pretty
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: [tex]\zeta(1-s)=g(s)\zeta(s) [/tex] putting s=1/2+it you get: [tex]\zeta(1/2-it)=g(s)\zeta(1/2+it) [/tex] from that would,nt you get RH?...as you know that Zeroes are symmetrically placed
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)
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...
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.
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".
and what about studying the integral: [tex]\int_0^{\infty}\int_{c-i\infty}^{c+i\infty}\frac{x^{s-1-z}}{\zeta(s)s}dsdx[/tex] the singularities (the values when is infinite) would be the place where zeros are