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Can someone explain zeros and zeta function for Riemann Hypothesis? (Yr13)

  1. Sep 22, 2009 #1
    Hi,

    I'm Yr 13 and just wanted to do some further reading/exploring.

    So i understand that the zeta function is something to do with summing up like this:

    1/ (1^s) + 1/(2^s) etc etc

    Now, I just want to know what are non-trivial zeros and trivial zeros? I just want to be able to understand this concept. And that graph where all non-trivial zeros lie on the x=1/2 line. How is that graph plotted? Btw, im only Yr 13, so layman terms please >__>

    Thanks!

    PS: Also, IF EVER one zero was found OFF of the critical line...would the e-commerce community collapse? Coz I heard that many companies use prime numbers to construct the encryption of their security. Or is that just the RSA Number Challenge? Hmm not so sure.
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2
    See https://www.physicsforums.com/showthread.php?t=322847&highlight=Riemann+Hypothesis Also the imaginary part (which is n* the square root of minus 1 or "i") + 1/2 = s in this case. Even though the square root of -1 is imaginary, it can be multiplied in algebraic equations and the like. The answer will be either real or have an imaginary component B*i. In the Riemann Hypothesis the sum of the real parts and the imaginary parts for non-trival zeros will both equal zero.
     
  4. Sep 22, 2009 #3
    I inadvertently posted the wrong link which was a followup on the thread I wanted to cite. It is https://www.physicsforums.com/showthread.php?p=2235996#post2235996 The two components which sum to zero are the real component which corresponds to the x axis and the imaginary component which corresponds to the "y" axis. If you draw a unit circle on such a plane, and draw a radius at the angles 120 degrees or 240 degrees the x and y component (which correspond to the sine and cosine ) of the meeting point with the circle correspond respectively to the real and imaginary components of a complex number which if cube would be equal to 1, other points on the unit circle when raised to the power of three would give other points on the unit circle. I'm not sure if they would be again spaced at +/- 120 degrees, but the cube roots of -1 are also -1 or imaginary points spaced 120 degrees on the unit circle from -1.
     
    Last edited: Sep 22, 2009
  5. Sep 23, 2009 #4
    Thanks ! :D
     
  6. Oct 2, 2009 #5
    REPLY. I am not a grammarian but have been in conn special subject of Topology at oxon in the past. I am permitted ethically a shot at this. Have you read DR PENROSE BOOK ON ROAD TO REALITY? He mentions this subject. I am not satisfied myself that it would answer your quest but there is a section on (1 minus third root of z) empowered minus 1. Terrible for yourstage but take note. Suppose the minus one root had a particular route associated with it so that all such paths went through it to yield a power view for higher logic ie an iterative implication. Could it not be that you yourself would become a philosopher of the moment and show that a Riemann Zeta function is on offshoot of this line of reasoning? I am afraid I do not have status myself to say much about this view of grammarians but you should continue to look around at new books on the subject. Iassume you have read Dr Riemanns Zeroes by Dr Sabach;a famous writer known to Dily Telegraph. They get wind sometimes.
     
  7. Oct 7, 2009 #6
    Um...sorry lol, I dont really understand your post. The only book Ive read so far is Marcus Du Sautoy's Music of the Primes. Ive heard about Road to reality and how notoriously difficult it is to read (bearing in mind its length) Lol. But I will definitely give more books a shot.

    Thanks
     
  8. Oct 7, 2009 #7
    Ok so I've read all of it. All I understand now is how non-trivial zeros converge at your starting point when you follow the rules. So, where is the critical 'strip' in these graphs? And how do these link to the primes? Apparently when you choose a correct value for T, it will bring you back to where you started, but what has any of this got to do with primes?

    Apparently the first zero happens at t=14.134725 .......um...not a prime number... yeh sorry for being stupid, but I cant find it anywhere on the internet. On wikipedia they just go on to talk about it in another language (ie, im stupid).

    Thanks
     
  9. Oct 7, 2009 #8

    CRGreathouse

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    The number of primes up to x can be approximated, as Gauss guessed as a child, by x / log x. A better estimate turns out to be the logarithmic integral Li(x), which is the integral of 1/log(t) up to x.

    Riemann's explicit formula is a way to transform this estimate into an exact value by considering not only the logarithmic integral of x, but the logarithmic integrals of x raised to the power of the zeros of the zeta function. Try this page for an overview of that:
    http://www.math.ucsb.edu/~stopple/explicit.html
     
  10. Jun 12, 2011 #9
    Yes, you are correct, except that gauss guessed it during prebirth while he was still in his mothers womb.
     
  11. Jun 17, 2011 #10
    Hi 2710.
    A very good book you can read in order to understand what Riemann's Hypothesis is about is John Derbyshire's "Prime Obsession".

    It is a brilliant book, and I don't know any other book that can explain Riemann's Hypothesis in a more comprehensive way.
     
  12. Jun 18, 2011 #11

    phyzguy

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    I second this recommendation. Derbyshire's book is a great read.
     
  13. Jul 29, 2012 #12

    CKM

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    I think I know where the person who asked this question is coming from.

    The zeros (such as 14.13 ...) and so on, are not primes.

    CRGreathouse is very helpful, above.

    But I would ask then:
    a) Is the position of these non-trivial zeros (say, 14.13 ...) the power to which the primes must be raised (i.e., their log) relative to the Li(x)? Is that what the position is?

    b) And what makes the -2, -4 ... happen? Lots of books say they're well understood but what do they come from, basically?

    I, too, cannot find these answers anywhere answered just as this person asked them.
     
  14. Aug 18, 2012 #13
    Hi.

    When Riemann introduced his zeta function, he started to explore its properties and found many new interesting zeta function features. For instance, he could tell zeta is infinite at point 1, and he could tell that zeta vanishes at negative even integers. So those first and obvious facts are called "obvious" or "trivial".

    Then Riemann and other mathematicians continued to explore properties of zeta function. They found that zeta is connected to number of primes. They were able to show that number of primes [itex]\pi(x)[/itex] is

    [itex]\pi(x)=Li(x) + \mathcal{O}[/itex]

    We know everything there is to know about function [itex]Li(x)[/itex]. But what do we know about [itex]\mathcal{O}[/itex]? Well, we know that

    [itex]\mathcal{O}=\sum_{\rho} \frac{x^\rho}{\rho}[/itex]

    Here sum ranges all over zeta function zeros [itex]\rho[/itex] that are not "obvious" or "trivial". We do not know a lot about these not-so-obvious zeros [itex]\rho[/itex]. And yet, we need those not-obvious zeros in order to calculate number of primes less than x as exactly as possible. These not-obvious and yet misterious zeta zeros are called non-trivial.

    All non-trivial zeta function zeros are inside a critical strip. Complex plane is like a big sheet, and critical strip is like an infinitely long ribbon in it. Critical strip stretches infinitely up and down, but is not very wide. Critical strip is confined between 0 and 1 horizontally, and stretches ad infinitum vertically.

    Riemann hypothesis is that all non-trivial zeros are on critical line. Critical line is exactly in the middle of critical strip.

    We know how to calculate location of non-trivial zeros on critical line. The formula to calculate position of non-trivial zeros on critical line is long known. It's name is Rieman-Siegel formula. Siegel was a mathematician who happened to have read Riemann's scribblings long after Riemann died. Reading Riemann's notes, Siegel found an interesting equation. He played with that newly discovered equation, and found Riemann-Siegel formula. So we can calculate location of zeros we need to know in order to calculate number of primes [itex]\pi(x)[/itex].

    So, how many zero locations we have to calculate? How many zeros are there in critical strip?

    Well... Infinitely many. And we don't really know much about them.

    So, in order to calculate number of primes between 1 and 2, or number of primes between 1 and 10000000000000000000000000000000000000000, our formula tells us to do a sum over all infinitely many non-trivial zeros we know nothing about?

    Yes.

    And this is how primes and zeta function zeros are connected.

    So what's the use of Riemann zeta function if we do not know much about it anyway?

    Well, we know nothing much about primes at all. However, we do know something about Riemann zeta function: we know how to calculate zeta function if only we have powerful computers. So, we can calculate Riemann zeta function. And Riemann showed that zeta function is connected to primes. This means that we can also calculate primes now. That's why is Riemann zeta function such a blast. It enables us to calculate number of primes we couldn't really calculate before Riemann.

    So, what happens if there are zeros off critical line?

    Several unpleasant things happen then. First: calculations become harder now. Second: there are huge gaps between large primes then, empty spaces without primes stretching almost infinitely. If Riemann hypothesis holds, then primes appear more regularly, and "there is music in primes".

    Will the universe end if Riemann hypothesis is false? No. Will economy collapse? No. Will anything happen at all? Yes. If RH stands, many many theorems in analytic number theory will come to life, and number theory will flourish. If RH is false, then... theorems will die and number theory will become dead. For a long, long time to come...

    I hope this explained it a bit.

    Cheers.
     
    Last edited: Aug 18, 2012
  15. Aug 18, 2012 #14
    Hi.

    CKM asked: Where do trivial zeros come from?


    When Riemann discovered and explored zeta function, he found zeta function satisfies functional equation

    [itex]\pi^{-s/2} \Gamma\left(\frac{s}{2} \right) \zeta(s)=\pi^{-(1-s)/2} \Gamma\left(\frac{1-s}{2} \right) \zeta(1-s)[/itex]

    He also discovered [itex]\zeta(s) [/itex] had no zeros to the right of 1, for [itex]\Re s >1[/itex]. It was long known that Gamma function is infinite at negative integers. So left-hand side is finite and has no zeros at positive integers. However, if [itex]s[/itex] is positive integer, [itex]s >0[/itex], then [itex]1-s[/itex] is a negative integer, [itex]1-s <0[/itex]. So at positive integers Gamma function to the right is infinite, but only at positive odd integers [itex]s[/itex]: odd because of denominator 2, and because 1 minus odd number is even number. Therefore zeta function on the right must be zero at positive odd integers in order to counter infinities of Gamma on the right. Now, if [itex]1-s [/itex] is positive odd integer, then [itex] s[/itex] is negative even integer. Therefore [itex]\zeta(s)[/itex] is zero at negative even integers. Only this way are left and right sides both finite and zero-free for positive integral [itex]s[/itex].

    One more thing: [itex]\zeta(s)[/itex] is infinite at [itex]s=1[/itex]. This explains why there is no zero at origin.

    I hope this explained it a bit.

    Cheers.
     
  16. Aug 18, 2012 #15
    Hi 2710,

    I follow the already given advice: read John Derbyshire's "Prime Obsession"

    In my opinion, it's the best book to start in the field of the Riemann zeta

    it's really a must!
     
  17. Aug 19, 2012 #16

    CKM

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    Thank you, Kraflyn.

    When I read your response to this thread (which I didn't start), I was delighted you were so thorough.

    1. I mean, what is the Gamma function for here? I read somewhere it was something to do with a differential ... what would that mean, here? You say they're infinite? I think I'm confused.

    2. You say: "He also discovered ζ(s) had no zeros to the right of 1, for Rs>1. It was long known that Gamma function is infinite at negative integers. So left-hand side is finite and has no zeros at positive integers. However, if s is positive integer, s>0, then 1−s is a negative integer, 1−s<0. So at positive integers Gamma function to the right is infinite, but only at positive odd integers s: odd because of denominator 2, and because 1 minus odd number is even number. Therefore zeta function on the right must be zero at positive odd integers in order to counter infinities of Gamma on the right. Now, if 1−s is positive odd integer, then s is negative even integer. Therefore ζ(s) is zero at negative even integers. Only this way are left and right sides both finite and zero-free for positive integral s."

    So ... Left and Right sides have positives? I thought it was only left side. Do you mean in 3D or 4D representation? (I read that it requires a 4D process to really capture the details of what's going on.) ------
    And you say there are no zeroes at positive evens on right, but infinite at positive ODDS on right. ------
    And there are negative numbers on the right as well as the left? ------
    You mention that denominator 2 makes it odd, when 1 minus odd is even. But divided by 2, that's even again ... so did you mean IF odd and divided by 2 it's negative?

    Thanks. :) And sorry. I'm trying to get some more detailed idea than I've gotten from the partial explanations (so admitted, by the authors) of several books for the general audience.
     
    Last edited: Aug 19, 2012
  18. Aug 19, 2012 #17
    Hello.

    Than You, CKM.

    When Hadamard and de la Vallee-Poussin independently proved the prime number theorem, they actually proved the following identity:

    [itex]\psi_{0}(x)=x-\ln(2 \pi)-\frac{1}{2}\ln \left(1-\frac{1}{x^2} \right)-\sum_\rho \frac{x^\rho}{\rho}[/itex]

    Here [itex]\psi(x)[/itex] is Chebyshev psi function [itex]\psi(x)=\sum_{p \leq x} k \ln p[/itex]. And Chebyshev psi is related to prime counting function [itex]\pi (x)[/itex] as [itex]\pi(x)=\frac{\psi(x)}{\ln x}+\int_2^x \frac{\psi(t) dt}{t \ln^2 t}[/itex].

    Term [itex]-\frac{1}{2}\ln \left(1-\frac{1}{x^2} \right)[/itex] in first equation is really [itex]-\frac{1}{2}\ln \left(1-\frac{1}{x^2} \right)=-\sum_\tau \frac{x^\tau}{\tau}[/itex]; a sum over trivial zeta zeros [itex]\tau[/itex]. Sum over non-trivial zeros [itex]\rho[/itex] is the only hard part here...

    So, trivial zeros do get involved in the mix. It's just that they are trivial: we handle them with ease!

    OK, let's discuss Gamma now. [itex]\Gamma(z)[/itex] has poles at negative integers [itex]-n, \; n \in \bf{N}_0[/itex] and at zero. This means

    [itex]z_{pole} = -n[/itex]

    So where are poles of [itex]\Gamma(s/2)[/itex]? Substituting [itex]z=s/2[/itex], one finds

    [itex]s_{pole} = -2n[/itex]

    Now consider [itex]\Gamma((1-w)/2)[/itex]. Where are its poles located at? Obviously [itex]z=(1-w)/2[/itex], and so

    [itex]w_{pole} = 1+2n[/itex]

    So, function [itex]\Gamma(s/2)[/itex] has singularities at negative even integers, and function [itex]\Gamma((1-s)/2)[/itex] has singularities at positive odd integers. And so on, You can handle all the rest this way Yourself. Good luck. Please do not hesitate to ask anything You find of interest.

    Cheers.
     
  19. Aug 19, 2012 #18

    CKM

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    You're very kind!!

    I just edited my question while you were typing this reply. Ha ha! Oops.
    I'll try to work through your reply very thorough above, a bit (with my limited experience), but do check out my edited question, if you don't mind. It will tell you where I got confused more ... in regular language. I do pick up some things from what you've written and from the equations sometimes, too, since you're interspersing explanatory prose, too.

    I always wondered, by the way, what is the T ------ is that T for Trivial? I assume not, but am not sure.

    Thanks! I'm not an idiot; I am expanding on my readings about it.

    And OH SO YOU USE TRIVIALS AS WELL ... is that towards counting the Gamma function? I.e., what part of the process do they come in? They are not byproducts, but USED for something?
     
    Last edited: Aug 19, 2012
  20. Aug 19, 2012 #19
    Hello.

    Ha ha, go get it, tiger!

    You can handle it Yourself, CKM, i believe in You. This is the easy part of proving something about zeta. Try it!

    Let me tell You something about first steps into zeta instead.

    Everything begins with:

    Fundamental theorem of arithmetic: For every [itex]n \in \bf{N}[/itex] there exists a unique sequence of natural numbers [itex]h(p)\in \bf{N}_0[/itex] such that
    [itex]n=p_1^{h(p_1)} p_2^{h(p_2)} ... =\prod_p p^{h(p)}[/itex]
    Factorization is unique for every [itex]n[/itex] and product runs over all primes [itex]p[/itex].

    One can easily prove this.

    Let us continue by defining
    [itex]\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}[/itex]
    One can also rewrite this as another definition of zeta,
    [itex]\zeta(s)=\prod_p \frac{1}{1-{1}/{p^s}}[/itex]
    Last two equations can be converted one into another just by applying fundamental theorem of arithmetic. So this is all very basic.

    OK. Consider definition [itex]\zeta(s)=\prod_p \frac{1}{1-{1}/{p^s}}[/itex]. None of factors vanishes nor is unbounded if [itex]\Re s >1[/itex]. This means zeta has no singularities nor zeros in region [itex]\Re s >1[/itex]. So, You see, we can get basic informations on zeta rather easily.

    In order to proceed, we need complex analysis: we would next deploy Gamma function! However, this will do for now. The point is: first steps into zeta are easy and fun!

    Cheers.
     
    Last edited: Aug 19, 2012
  21. Aug 19, 2012 #20
    Hi.

    If I got You correctly, You ask what is T: do You mean what is [itex]\Gamma[/itex]? [itex]\Gamma[/itex] is capital greek letter Gamma. Function [itex]\Gamma(s)[/itex] is the factorial Gamma function:

    [itex]\Gamma(n)=(n-1)!=(n-1)(n-2)(n-3)\dots3\times 2 \times 1[/itex]

    It is defined for all complexes [itex]s[/itex] except negative integers and zero, where it becomes unbounded.

    I seriously hope I got You right, because otherwise... I'm doomed c:

    Cheers.
     
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