Zeta of "1"

Discussion in 'Physics & Math' started by BrianHarwarespecialist, Jun 11, 2015.

  1. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    Is there any method(s) available using analytical continuation to solve for the zeta of 1? Or even any speculations?
     
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  3. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    Also what is the value for Zeta of 6
    zeta of 2= pi^2/6
    zeta of 4= pi^4/90
     
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  5. origin Trump is the best argument against a democracy. Valued Senior Member

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    This type of questioning seems oddly familiar.
     
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  7. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    If you can answer my question please do, it will be greatly appreciated.

    The countdown begins...
     
  8. mathman Valued Senior Member

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  9. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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  10. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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  11. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    "Since it is a pole of first order, its principal value exists and is equal to γ."

    So is this article essentialy saying the zeta of 1 is equal to "γ" I just want to make sure I am comprehending the message correctly?
     
  12. Kristoffer Giant Hyrax Valued Senior Member

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    Was lurking here for ages before I signed up and I agree.
     
  13. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    What is the main difference between the actual value and the principal value?
     
  14. rpenner Fully Wired Staff Member

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    If \(\gamma_n = \lim_{m\to\infty} \left( \sum_{k=1}^{m} \frac{(\ln k)^n}{k} - \frac{(\ln m)^{n+1}}{n+1} \right) \)
    then
    \(\zeta(1 + x) = \frac{1}{x} + \sum_{n=0}^{\infty} \frac{ (-1)^n \gamma_n x^n}{n!}\)
    and
    \(\zeta(1 - x) = -\frac{1}{x} + \sum_{n=0}^{\infty} \frac{ \gamma_n x^n}{n!}\)
    so
    \(\frac{1}{2} \left( \zeta(1 + x) + \zeta(1 - x) \right) = \sum_{n=0}^{\infty} \frac{ \gamma_{2n} x^{2n}}{(2n)!} = \gamma_0 + \sum_{n=1}^{\infty} \frac{ \gamma_{2n} x^{2n}}{(2n)!} \)
    thus
    \(\lim_{x\to 0} \frac{1}{2} \left( \zeta(1 + x) + \zeta(1 - x) \right) = \gamma_0 = \lim_{m\to\infty} \left( \sum_{k=1}^{m} \frac{1}{k} - \ln m \right) = \gamma\)

    So the Wikipedia article is saying that the value of zeta taken from two points equally distant from 1 on the real line has an average which gets arbitrarily close to the Euler-Mascheroni Constant, \(\gamma\), as the distance from 1 shrinks. Another way to say this is \(\zeta(1 + x) \approx \frac{1}{x} + \gamma ; \quad | x | < < 1\).

    This has nothing to do with "solving" zeta at 1.

    \( \begin{array}{c|c} n & \zeta(n) = \frac{ (-1)^{\frac{n}{2}-1} B_{n} 2^{n-1} \pi^n}{n!} \\ \hline \\ 0 & - \frac{1}{2} \\ 2 & \frac{\pi^2}{6} \\ 4 & \frac{\pi^4}{90} \\ 6 & \frac{\pi^6}{945} \\ 8 & \frac{\pi^8}{9450} \\ 10 & \frac{\pi^{10}}{93555} \\ 12 & \frac{691 \, \pi^{12}}{638512875} \\ 14 & \frac{2 \, \pi^{14}}{18243225} \\ 16 & \frac{3617 \, \pi^{16}}{325641566250} \end{array}\)
    (The formula holds only for non-negative, even values of n)
    \(\sqrt{\sqrt{\sqrt{\sqrt{\frac{325641566250}{3617}}}}} \approx \pi\) because \(\lim_{n\to\infty} \zeta(n) = 1\)

    (But there is nothing remarkable about that since \(\sqrt{\sqrt{\sqrt{\sqrt{\frac{17196154181}{191}}}}} \) is both simpler and much closer.)
     
    Last edited: Jun 12, 2015
  15. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    813
    "So the Wikipedia article is saying that the value of zeta taken from two points equally distant from 1 on the real line has an average which gets arbitrarily close to the Euler-Mascheroni Constant, \(\gamma\), as the distance from 1 shrinks. Another way to say this is \(\zeta(1 + x) \approx \frac{1}{x} + \gamma ; \quad | x | < < 1\)."
    Rpenner-

    Yes that makes perfect sense since 1 is a pole the range of those two points will decrease closer and closer to 1 or "γ" but they will never converge. This makes me both happy and sad but it gives me some fun math to investigate.

    But thanks for the response Rpenner this constant "γ" is quickly attracting more attention from me and I feel happy.
     
  16. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    813
    This is arbituary but still please still comment
    pi/sqrt{2} - pi^2/6= γ

    this is accurate to three decimal places...
     
  17. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    I was hoping Rpenner would have blessed this comment with his extensive Knowledge in math but I guess this result is too trivial, I guess am going to have to do "all the heavy lifting myself" then return with an actual mathematical argument that has a real significance but I was really trying to brainstorm.
     
  18. Fednis48 Registered Senior Member

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    Look, I'm glad that you're so interested in higher-level math, and I encourage you to keep exploring it, but there's something fundamentally wrong with your method of thinking that you need to correct. What you're doing is sometimes referred to as "numerology" in scientific circles: you mix and match arbitrary constants with arbitrary operators, and you mistakenly assume significance when you find something that's close to an equality. The expression from post 13 is a great example. You take the difference of two ratios involving different powers of pi, a square root, and a random 6 thrown in for good measure, and the result matches γ to 3 decimal places. That means nothing. With that many degrees of freedom to play with, one could probably come up with an expression that matches any physical or mathematical constant to 3 places. If you could find an exact equality, that would probably be more interesting, and rpenner would probably be able to give you some insight into why the equality worked. But if the equality only holds to 3 (or even 30) decimal places, it's just a sign that you mashed together enough numbers to find two unrelated values that happen to be close to each other.
     
  19. Dr_Toad It's green! Valued Senior Member

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    I think he's quite aware of what he's doing: I don't know who it is, but my troll-detector has been going off on this one for a few days now.

    Please Register or Log in to view the hidden image!

    We know him by some other name...
     
  20. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    813
    This is just a small part of a bigger picture. I have my reasons for posting it, this result is hinting on something important to my work but I have to expand it to broader parameters. Its going to take a lot more mathematical investigation for me to isolate the significant position of this result.

    I suspect there is a relationship with these ratio's just a hunch so far I was getting a confirmation on my assumptions.
     
  21. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    Fednis48 the 6 is not random pi^2/6 is the solution for the zeta of 2 I actually had a hunch I would get this result before I did the calculation for other reasons I am still investigating and have not discussed.
     
  22. rpenner Fully Wired Staff Member

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    I think so because there are always close relationships between numbers.

    Take 135/43 and 5/26.

    \(\left( \frac{ \frac{135}{43} }{ \sqrt{2} } - \frac{ \left( \frac{135}{43} \right)^2 }{6} \right)^3 = \frac{ 1033945529625 \sqrt{2} - 1452494683125}{50570904392} \approx \frac{ 5 - \frac{1}{29339} }{26} \approx \frac{5}{26}\)

    Now \(\frac{135}{43} + \frac{1}{486} \approx \pi \) and \(\sqrt[3]{\frac{5}{26}} \approx \gamma\) for no other reason than an infinite set of numbers are close to any number. Finding close, non-exact simple relations between numbers is more of an art form or recreation than mathematics. Assigning significance to "close" relations is pseudo-mathematics.

    http://mathworld.wolfram.com/Euler-MascheroniConstantApproximations.html (not included: \(\frac{\ln 9}{\ln 45}\) )
    http://mathworld.wolfram.com/PiApproximations.html
    http://mathworld.wolfram.com/AlmostInteger.html
    http://xkcd.com/1047/
     
  23. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    813
    Rpenner I will continue this discussion once my work has matured passed trivial results at the moments I cannot make the time commitment hence my sloppy introductions...
     

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