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Disproof of Riemann Hypothesis

  1. Jun 28, 2011 #1
    I think I have managed to disprove the Riemann Hypothesis. Hope I have not made a calculation error.

    Attached Files:

  2. jcsd
  3. Jun 28, 2011 #2
    So, if the Riemann hypothesis isn't true, can you give me a zero not on the critical line?

    Anyway, you say in your proof:

    I don't quite see how you obtained the equation that comes after it. Your numerator and denominator are the same thing, thus that would imply [itex]\Gamma(\frac{1}{2}+it)=1[/itex] which isn't true for all t.

    And I don't quite see why


    Fine, your calculator says it is, but you need to prove that this is the case. (and by the way, I think it converges).
  4. Jun 28, 2011 #3
    The biggest problem I see with this proof is that you are treating infinities as numbers instead of examining the limits of the expressions. I also agree with Micromass that some of the expressions you introduce are non sequitur.
  5. Jun 29, 2011 #4
    Simply in the equations, you have not proved the hypothesis wrong, and you have made a couple of errors.
    When you write( i have put three dots instead because i don't wish to copy down the whole thing)


    You are in essence not correctly solving for the gamma function. Your derivation of one for the gamma function is quite impossible indeed. If you look at the next equation involving the cosine, your gamma function does not make sense at all. Your equations are contradicting themselves. The previous solution for the gamma function(the equation before you 'solve' the gamma function) contradicts your solution of one.
    Furthermore, the summation is definitely not divergent, you can test that out yourselves.
    Finally, give me an actual example. At this point, pi has nothing to do with it.
    The riemann hypothesis maybe unproven as of yet, but nor has it been disproved. The paper simply does not disprove the hypothesis.
    Last edited: Jun 29, 2011
  6. Jun 29, 2011 #5

    Where is the contradiction?

    A calculator showed it correct. And, a function can converge to infinity. This series does become infinity.

    Riemann Hypothesis has to do with the Gamma function, Z-function, Zeta function, Theta function... And these have to do with pi.
  7. Jun 29, 2011 #6
    Even taking it as a limit, you get infinite.
  8. Jun 29, 2011 #7
    That is about the Theta function. Instead of the theta function itself, I used its definition. That part is a direct implication of the second equation there.
  9. Jun 29, 2011 #8
    Ok. I understood. I'll try to do that. Thanks.
  10. Jun 29, 2011 #9
    Well, I don't see how it is a direct implication. Can you elaborate? Do you agree that your implication implies that


    Also, I have a qual about [itex]i\infty[/itex] that you seem to use. That notation doesn't really make sense in the complex numbers. The only way you can talk about [itex]\infty[/itex] in the complex numbers is by looking at the Riemann-sphere. But [itex]e^x[/itex] is not a meromorphic function, so you can't just say that [itex]e^{i\infty}[/itex]=0.

    What you could perhaps say is that


    but I'm afraid that this limit will not exist...
  11. Jun 29, 2011 #10
    Don't bother, the series converges.
  12. Jun 29, 2011 #11
    I don't understand how this is a justification for what you are doing. By letting infinity act as a number you end up with nonsensical conclusions such as the exponential function equaling zero. The exponential function is never equal to zero. The exponential function only approaches zero as its argument decreases without bound. This is only properly stated as a limit.

    Let us suppose that infinity and negative infinity are numbers and that the exponential function does equal zero at negative infinity. The only reason you say this in your proof is because the expression:


    is equal to infinity when t is equal to infinity. You then go on to conclude that the denominator in your equations must necessarily equal zero because zero times a number is zero, right? Not exactly. If you let t equal infinity in the other expressions of your denominator you will ultimately end up with the expression:


    This expression is undefined! Don't believe me? You implicitly assume the following identity in your proof:

    [itex]\infty = \frac{c}{0}[/itex]

    I'm going to ignore the fact that this expression makes absolutely no sense and is undefined for the moment. You are implicitly allowing division by zero to be defined by the normal rules of division. This means I can multiply both sides of the equation by zero and cancel out the zeros on the right side of the equation. This means that zero times infinity can equal anything we desire since c is any number. I could also multiply both sides of the equation by zero and let the right side just equal zero since the normal rules of multiplication apply. This is a contradiction to the previous result. Ultimately the expression must be nonsense if we let the normal rules of multiplication and division apply.

    I hope this helps you understand the problems of treating infinity as a number (and the problems of division by zero).
  13. Jun 29, 2011 #12
    Things like


    are actually ok is complex analysis if we are working in the Riemann sphere (and if c is not zero). What it actually is saying is that the function


    is meromorphic and takes on the value [itex]\infty[/itex] in 0.

    However, I don't think the OP really meant all of this...
  14. Jun 29, 2011 #13
    I have heard of the Riemann sphere, but have no background in complex analysis. Would zero times infinity be defined on the Riemann sphere, if so what would it be? I would think it would still give answers which are undefined.
  15. Jun 29, 2011 #14
    Things like [itex]0\cdot \infty[/itex], [itex]\infty+\infty[/itex], [itex]\infty\cdot \infty[/itex] and [itex]0/0[/itex] will remain undefined. See http://en.wikipedia.org/wiki/Riemann_sphere
  16. Jun 29, 2011 #15
    Ah, thank you, Micromass!
  17. Jun 30, 2011 #16
    Isn't [tex]\infty\times \infty=\infty[/tex] and [itex]\infty+\infty=\infty[/itex]
  18. Jun 30, 2011 #17


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    Well, what are you referring to by "[itex]\infty[/itex]"? In the extended reals, only the latter is true.
    Last edited: Jun 30, 2011
  19. Jun 30, 2011 #18
    Ok, I don't know what's wrong with my calculator but with excell sheets, I got a graph which strongly suggests convergence. I have attached that file here. The sum seems to be around [tex]t /times /sqrt{3}[/tex]

    Attached Files:

    Last edited: Jun 30, 2011
  20. Jun 30, 2011 #19
    I also made a discovery ! This graph here shows that clearly [tex]\sum_{k=1}^{\infty }\frac{1}{k} = 0 [/tex]



    You simply cant deduse facts from simple looking graphs, even if the graph looks like one thing, the truth could be something completely different. What you has to do, is prove it! Not just say that it is true, because of a graph that might be true.
  21. Jun 30, 2011 #20
    Not in the complex numbers.
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