"Yod" function

Discussion in 'The Cesspool' started by Prox"Y", May 24, 2015.

  1. Prox"Y" Registered Member

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    {Re(zeta(i x + 0.8)), Im(i x+0.8))}

    Yod function vs Xi function...could the Riemann hypothesis be false help me further investigate the Yod function, will post more...
     
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  3. origin Heading towards oblivion Valued Senior Member

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    That would be helpful to figure out what you are talking about.
     
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  5. Prox"Y" Registered Member

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    I dont think the site gives me permission to upload what I need to yet am just assuming this I cant seem to locate the functionable option.
     
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  7. rpenner Fully Wired Valued Senior Member

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    If we are talking about the Riemann zeta function, then since \(\overline{\zeta(s)} = \zeta( \bar{s} )\) it follows that \( \Re \zeta( ix + \frac{4}{5} ) = \frac{1}{2} \left( \zeta( ix + \frac{4}{5} ) + \zeta( -i \bar{x} + \frac{4}{5}) \right)\).
    And then as you have written it, \(\textrm{Yod}(x) = \Re \zeta( ix + \frac{4}{5} ) + i \Im ( ix + \frac{4}{5} ) = \frac{1}{2} \left( \zeta( ix + \frac{4}{5} ) + \zeta( -i \bar{x} + \frac{4}{5}) + i x + i \bar{x} \right)\).
    You haven't said any thing about the domain of x. But if x is restricted to real numbers, this simplifies to:
    \(\textrm{Yod}(x) = \frac{1}{2} \left( \zeta( ix + \frac{4}{5} ) + \zeta( -i x + \frac{4}{5}) \right) + i x\)
    with the property: \(\textrm{Yod}(-x) = \overline{\textrm{Yod}(x)}\).

    \(\textrm{Yod}(0) = \zeta(\frac{4}{5}) \approx \gamma - \frac{3059}{610}\) where \(\gamma\) is the Euler-Mascheroni constant.

    The Riemann hypothesis isn't necessary to derive these properties. Yod doesn't seem to have any obvious application.
     
    Last edited: May 24, 2015
  8. Prox"Y" Registered Member

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    \(\textrm{Yod}(0) = \zeta(\frac{4}{5}) \approx \gamma - \frac{3059}{610}\) where \(\gamma\) is the Euler-Mascheroni constant.

    The Riemann hypothesis isn't necessary to derive these properties. Yod doesn't seem to have any obvious application.[/QUOTE]

    Using the Yod function I was able to locate at least the first two nontrivial zeros of the Zeta function "14", and "21" And these nontrivail zeros was not on the critical line of "1/2" they were on the line of "4/5" or "1/1.25". So at first glance the graph " [{Re[Zeta[1/1.25+I x]], Im[Zeta[1/1.25+I x]]}" seems to suggest that the Riemann hypothesis is false, since nontrivial zeros was found on a line other than the "critical line" contrary to the RH.

    Here is the graph...
    http://www.wolframalpha.com/input/?i=[{Re[Zeta[1/1.25+I x]], Im[Zeta[1/1.25+I x]]}
     
    Last edited: May 25, 2015
  9. rpenner Fully Wired Valued Senior Member

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    Ok -- you are posting on a topic well in advance of your understanding. You also have the arrogance of posting a trivial discovery which you found using online tools written by teams of mathematicians without considering that if it was so trivial, that it may not be the answer long-searched for. Such arrogance has shown up many times before, such as the idiot who took out an advertisement in Scientific American to proclaim that he solved a problem of antiquity didn't understand the first thing about geometric constructions in Euclidean geometry.

    The (analytic continuation of the) Riemann zeta function maps complex numbers to complex numbers, except for the pole at 1. So you have to understand complex numbers. 0 is a complex number -- it's just the most trivial example. 0 is simultaneously a purely real number and a purely imaginary number because those subsets of the complex numbers hold zero as their unique point of intersection.
    A zero of the Riemann zeta function is a complex number that the zeta function maps to zero.
    Thus if a + ib is a zero of the Riemann zeta function, then \(\zeta(a + ib) = 0\), thus naturally, \(\Re \zeta(a + ib) = \Im \zeta(a +ib) = 0\). This is not the case for 4/5 + 14 i or 4/5 + 21 i. These are merely points where the value of the Riemann zeta function is purely imaginary or purely real. Or rather, mostly purely imaginary or purely real.
    You only came close to actual answers -- when you have a tool which when properly used can extract approximate answers with arbitrary precision.
    Indeed, it's not "14" or "21" but numbers closer to "14.0885767746660" and "21.0974794379181".

    So these are not zeros of the Riemann zeta function, but much less interesting points along the path of \(z(x) = \zeta( \frac{4}{5} + i x ) ; \quad x \in \mathbb{R}\) where the value of z(x) has 0 for the imaginary or real part of the complex number.
    Im[Zeta[4/5±0 I]] (exact)
    Re[Zeta[4/5± 0.5609 I]]
    Im[Zeta[4/5±14.0886 I]]
    Im[Zeta[4/5±21.0975 I]]
    Re[Zeta[4/5±110.549 I ]]
    Re[Zeta[4/5±110.676 I ]]

    Actual examples of zeros are:
    Zeta[-2] (trivial, exact)
    Zeta[1/2 ± 14.1347251417346937904572519835624702707842571156992431756855674601499634298092567649490103931715610127792 I] (non-trivial, approximate)

    Finally, your bizarre insistence of writing 4/5 as 1/1.25 is detrimental to both Wolfram Alpha's ability to distinguish exact rational expressions from floating point approximation and your reputation.

    //Postscript: Because you didn't explain yourself well in post #1 we had to wait for your third post for the trivial definition of \(\textrm{Yod}(x) = \zeta \left(\frac{4}{5} + i x\right)\) where the \(\frac{4}{5}\) has no theoretical motivation.
     
    Last edited: May 25, 2015
  10. Prox"Y" Registered Member

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    So according to your logic this "14.134" is the approximation of a 0 and this is approximately the exact quantity "14.088" of a zero? If this is not your claim let us move on...To further explain my position I was not making any claims thus I stated in the op
    ..."Help me further investigate the Yod function"
    The quantities I posted, I never claimed to be exact as I am aware I was using a free sofware "14" and "21" was just an estimation I could see vaguely from looking at the graph [{Re[Zeta[1/1.25+I x]], Im[Zeta[1/1.25+I x]]} at "first glance".
    I do not know the exact quantities Yod yields yet, I have not worked it out that is why I posted it here to get help in yielding the exact quantities and comparing it to already known quantities of the zeta function nontrivial zeros. The graph "[{Re[Zeta[1/1.25+I x]], Im[Zeta[1/1.25+I x]]}" seems to suggest even at first glance something closer to "14.1" for the first nontrivial zero and so on.
    "You also have the arrogance of posting a trivial discovery which you found using online tools written by teams of mathematicians without considering that if it was so trivial, that it may not be the answer long-searched for."
    r penner quote above
    The reason I posted this is because it seemed like a trivail discovery to me, I am aware I was using free sofware, but none the less at first glance the graph appears to yield close enough approximations to "nontrivial zeros" of the zeta function that was not located on the crictical line.
    So my last question I will end with, is what is the exact quantity for the first nontrivail zero of the zeta function; or you can just define it in the appropriate terms thank you very much for your investment of time, rpenner.
     
    Last edited: May 25, 2015
  11. Prox"Y" Registered Member

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    This is not all there is these were just my trivial examples a starting point, but a thank you for your time investment I was looking forward to having this discussion in as much detail as I can possible get.
     
  12. rpenner Fully Wired Valued Senior Member

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    No, I'm saying near a point like 1/2 + 14.1347 i, the function zeta gives zero. While near 4/5 + 14.0886 I , the function zeta gives a number close to 0.213 which is purely real, with 0 imaginary part.

    This is not your private journal -- you have to write for an audience and spell out your motivations and assumptions. And please don't disparage "free software" -- using that same free software, I got the above high-precision results because I understood it well enough to polish the answer to 13 decimal places.

    You should
    • Stop calling it Yod -- it's far too trivial to deserve a unique name. It has no known motivation or application.
    • Stop writing 4/5 as 1/1.25 which also lacks any motivation. 4/5 and 0.8 take the same number of keystrokes, but Wolfram Alpha recognizes 4/5 as an exact rational quantity and can use higher-precision math with it than either 1/1.25 or 0.8.
    Better to call your work a "study of Riemann's zeta function on the line \(\frac{4}{5} + i x\)" than to introduce unique names and weird ways of specifying rational numbers.

    4/5 + 14.1 i is not a zero of the Riemann zeta function.

    Huh?

    That's largely because you didn't then understand what a "zero of the zeta function" was.

    You don't have a "first" non-trivial zero because the complex plane doesn't come with a natural ordering. However, the above example, 1/2 + 14.1347 i, is the lowest magnitude example of a non-trivial zero in the critical strip.

    Please look at this graph, where four lines are plotted: http://www.wolframalpha.com/input/?i=plot { Re[Zeta[x + iy]] = 0, Im[Zeta[x + iy]] = 0, x = 1/2, x = 4/5 } for x from 0 to 1, y from 0 to 22

    Please Register or Log in to view the hidden image!

    The four lines on the complex plane are:
    1. The locus of points where \(\Re \zeta(z) = 0\)
    2. The locus of points where \(\Im \zeta(z) = 0\)
    3. The locus of points where \(\Re z = \frac{1}{2}\)
    4. The locus of points where \(\Re z = \frac{4}{5}\)

    The zeros of the Riemann zeta function are only the points where the first two curves intersect. Many of those happen to be on the third line. The fourth line is much less interesting, and while it may cross either of the first two lines, it doesn't seem to cross them both at the same time. That's one of the reasons why Yod(x) is not interesting.

    Pro-tip: if you reduce the number of equations plotted to two, Wolfram Alpha will draw the intersections with red dots and if you hover the mouse over the dots, you will get approximate coordinates.
     
    Last edited: May 25, 2015
  13. Prox"Y" Registered Member

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    Thank you for the graph this is what I came here for.
     
  14. Prox"Y" Registered Member

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    Further to note the graph you provided confirms my intial suspicions this is the outcome I first predicted myself by hand but I did not reproduce the clear results as you did witht the graph.
     
  15. Kittamaru Ashes to ashes, dust to dust. Adieu, Sciforums. Valued Senior Member

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    Mod Note:

    Hello again Jason.Marshall... yes, we know it is you, we aren't as stupid as you seem to think. As you are well aware, the use of sock-puppets to evade a ban is, itself, a bannable offense; one that, as you are also aware, results in a permanent ban...

    Given your past behavior on this site, I see no reason to allow you to return at this time, especially given the rules of the forum regarding ban-evasion. If you are looking to return, you would be far better off contacting the administration (such as JamesR) and requesting your account be reinstated; creating another account simply amounts to little more than digging your hole a little bit deeper.
     
  16. Kristoffer Giant Hyrax Valued Senior Member

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    The guy that told me I'd burn in hell and in 20-30 years I'd be easy to take down? Paraphrased a bit, but that's essentially what he wrote.

    In 20-30 years I probably don't give a shit what 36 year old me wrote on a forum.
     
  17. rpenner Fully Wired Valued Senior Member

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