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Math Help - Zeta function convergence

  1. #1
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    Zeta function convergence

    I have tried to write a program to compute the Zeta function for complex arguments. In particular, I wanted to look at non trivial zeros.

    My HP calc and my program agree on some spot checked terms.

    For example, 1 / (2 ^ s) where s is equal to (.5 + 14.1347), my program and my HP calculator agree on the answer.

    But after adding the first 1000 terms, I don't see any obvious convergence.

    How fast does the Zeta function converge? Can you think of any obvious way for me to have made mistakes when adding the terms, or mistakes in my understanding?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Given  s = \sigma +it ,

    1.  |\zeta(s)| < A_1 \log(t)\;\;\;(\sigma \geq 1, \; t \geq 2) ,
    2.  |\zeta ' (s)| < A_2 \log^2(t)\;\;\;(\sigma \geq 1, \; t \geq 2) ,
    3.  |\zeta(s)| < \frac{A_3}{\sigma(1-\sigma)}t^{1-\delta}\;\;\;(1 > \sigma \geq \delta, \; t \geq 1) ,
    where  0 < \delta < 1 .
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  3. #3
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    Oh man, this is stuff I have not seen in a LONG time.

    I will have to study what you wrote and try to understand it. Thanks.

    Seems the first two bounds should not come into play with critical strip zeros because sigma should be .5, right?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by MichaelJHuman View Post
    I have tried to write a program to compute the Zeta function for complex arguments. In particular, I wanted to look at non trivial zeros.

    My HP calc and my program agree on some spot checked terms.

    For example, 1 / (2 ^ s) where s is equal to (.5 + 14.1347), my program and my HP calculator agree on the answer.

    But after adding the first 1000 terms, I don't see any obvious convergence.

    How fast does the Zeta function converge? Can you think of any obvious way for me to have made mistakes when adding the terms, or mistakes in my understanding?
    Your efforts are admirable... but I suppose you have neglect a little minor particular...

    ... the particular you have neglect is that the expression...

    \zeta(s)= \sum_{n=1}^{\infty} \frac{1}{n^{s}} (1)

    ... holds only for Re(s)>1. If you instruct your HP calculator to compute (1) for s_{0}=.5 + i \cdot 14.1347 may be you don't find as result \zeta(s_{0})= 0 + i\cdot 0 ...

    Kind regards

    \chi \sigma
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  5. #5
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    Gotcha! I was worried that might be the case. So I need to use a version which extends the answer to the proper domain.

    I will do more research, thanks
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  6. #6
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    Looks like this is the function I need?



    The part I don't know is how to estimate or calculate the gamma function for 1 - s.
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  7. #7
    MHF Contributor chisigma's Avatar
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    The forumula you have written is nothing else that this...

    \zeta (1-s) = 2 \cdot (2\pi)^{-s}\cdot \cos(\frac{\pi s}{2})\cdot \Gamma(s)\cdot \zeta(s) (1)

    ... that was 'discovered' by Leonhard Euler in XVIII° century and is commonly known as 'Reflection relation'. Effectively once you know \zeta(s) and \Gamma(s) for a given s, you automatically know \zeta(1-s). The trouble is that a formula like this...

    \zeta(s)= \sum_{n=1}^{\infty} \frac{1}{n^{s}} (2)

    ... holds only for Re(s)>1, so that (1) and (2) are useful to you only to arrive to compute \zeta(s) for Re(s)<0. If I undestand correctly you are searching the zeroes of \zeta(s) in the 'critical strip', where is 0<Re(s)<1 ...

    It is evident that the problem is not so easy to approach... anyway... never say never again! ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 30th 2009 at 06:00 AM. Reason: one spelling error
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    The extension of  \zeta(s) to  \Re(s) > 0 is  \zeta(s) = \frac{s}{s-1}-s\int_{1}^\infty \frac{x-\lfloor x \rfloor}{x^{s+1}} dx .
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  9. #9
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    Ok.

    So do I need to write a program to approximate the integral given in the most recent post?

    That seems manageable from what I can see?
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    You will need a program to approximate  \zeta(s) still.
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