• The Reimann "Zeta" function: How can it ever converge?

    From Kenny McCormack@21:1/5 to All on Tue Mar 25 19:15:55 2025
    So I was reading in Wikipedia about the Zeta function, which is defined as:

    Z(s) = 1/(1**s) + 1/(2**s) + 1/(3**s) + ...

    Both the domain and range are specified as the complex numbers.

    And it says that if s is a negative integers (-2, -4, -6, etc), then Z(s)
    is zero. But that can't be right. But first, a little manipulation:

    Suppose s is -2:

    1/(n**s), where s = -2

    is:

    1/(1/(n**2))

    is:

    n**2

    so, the sum is like:

    1+4+9+16+25+...

    Which just grows without bounds. And is certainly never zero.

    So, is Wikipedia wrong? Or just a typo?

    --
    The randomly chosen signature file that would have appeared here is more than 4 lines long. As such, it violates one or more Usenet RFCs. In order to remain in compliance with said RFCs, the actual sig can be found at the following URL:
    http://user.xmission.com/~gazelle/Sigs/Reaganomics

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From efji@21:1/5 to All on Tue Mar 25 21:57:53 2025
    Le 25/03/2025 à 20:15, Kenny McCormack a écrit :
    So I was reading in Wikipedia about the Zeta function, which is defined as:

    Z(s) = 1/(1**s) + 1/(2**s) + 1/(3**s) + ...

    Both the domain and range are specified as the complex numbers.

    And it says that if s is a negative integers (-2, -4, -6, etc), then Z(s)
    is zero. But that can't be right. But first, a little manipulation:

    Suppose s is -2:

    1/(n**s), where s = -2

    is:

    1/(1/(n**2))

    is:

    n**2

    so, the sum is like:

    1+4+9+16+25+...

    Which just grows without bounds. And is certainly never zero.

    So, is Wikipedia wrong? Or just a typo?


    I think it is far beyond the scope of this group, but let's try :

    The Riemann zeta function (please, Riemann and not Reimann...) is
    defined as the following series for any complex s such that Re(s) > 1:
    Z(s) = 1/(1**s) + 1/(2**s) + 1/(3**s) + ..

    The condition Re(s)>1 ensures the convergence of the series. Elsewhere,
    there is an smooth continuation and thus it can be rewritten in the
    whole complex plane as

    Z(s) = 2^s \pi^{s-1} \sin(\pi s/2) \Gamma(1-s) Z(1-s)

    Where \Gamma denotes the gamma-function, i.e. the regular function
    defined on the half complex plane Re(z)>0 such that \Gamma(n) = (n-1)!
    forall n integer ("n!" denotes "factorial n").

    On this expression you can check that Z is then well defined for every
    complex number s.

    You can also check that if n is an negative even integer, then

    Z(n) = 2^n \pi^{n-1} \sin(\pi n/2) \Gamma(1-s) Z(1-n) = 0
    since
    \sin(\pi n/2) = 0

    Be careful, you cannot use the same idea for a positive even integer n,
    since Γ has also a smooth continuation of the genuine gamma-function for Re(z)<0, and this continuation has a pole (i.e. it goes to infinity) for negative integers. Thus you get a product 0*\infty that has to be
    properly treated (and in the end Z(n) is not 0 for n a positive odd
    integer).

    That's only the "trivial part" of the study of the Riemann zeta function :)


    --
    F.J.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Kenny McCormack@21:1/5 to [email protected] on Wed Mar 26 01:37:13 2025
    In article <vrv5ci$3rj4$[email protected]>, efji <[email protected]> wrote:
    ...
    I think it is far beyond the scope of this group, but let's try :

    Interesting. I didn't think this group would shy away from this. Seemed
    right up your alley, if nothing else.

    The Riemann zeta function (please, Riemann and not Reimann...) is
    defined as the following series for any complex s such that Re(s) > 1:
    Z(s) = 1/(1**s) + 1/(2**s) + 1/(3**s) + ..

    Got it. Thanks. The point is that the series expansion only works for
    Re(s) > 1. One has to use one of the more complicated expressions for
    other input values. Note: This is there in Wikipedia; I must have glossed
    over it.

    Again, thanks for your long and very helpful explanation.

    P.S. (to the other responder on this thread). ** means exponentiation in
    many (but not all) computer languages (*). Maybe I should have written pow(...) instead, but the point is that there is no clear-cut, 7 bit ASCII representation for superscripting notation on Usenet. I did the best I
    could.

    (*) Some other languages use ^, but in most/all C-like languages, that
    means bitwise XOR.

    --
    He must be a Muslim. He's got three wives and he doesn't drink.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)