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Math Help - Riemann-zeta function

  1. #1
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    Question Riemann-zeta function

    for the sum of 1/n^x
    (1) For which vlues of x is the fiemann-zeta function diverges?
    (2) For x greater than 1 find the fiemann-zeta function/
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  2. #2
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    Quote Originally Posted by savetra View Post
    for the sum of 1/n^x
    (1) For which vlues of x is the fiemann-zeta function diverges?
    (2) For x greater than 1 find the fiemann-zeta function/
    First you mean Riemann.

    The Infinite series determined by 1/n^x is a continous,postive,decreasing function.
    We can use the Integral Test for series.

    It is equivalent to,
    INTEGRAL_1 ^ (infty) n^{-x}dx
    This integral diverges for x<=1
    Thus, the Riemann zeta function converges for x>1.
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  3. #3
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    May I suggest a great book on the topic of the Riemann Zeta-function?. It's called "Prime Obsession" by John Derbyshire.
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    Quote Originally Posted by galactus View Post
    May I suggest a great book on the topic of the Riemann Zeta-function?. It's called "Prime Obsession" by John Derbyshire.
    Is it really good?
    I was thinking about buying it.

    (But it does not contain any math )
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    Yeah, it has math in it. He talks about the Prime Number theorem, Prime distribution, Riemann's functional equation, etc. Toward the back it has more on the Mobius function, Big O, etc. It's a good read. He does not go into depth on how to derive the non-trivial zeros of the Zeta function. That's beyond the scope of the book. It involves analytic continuation and the like. Go ahead and read it, PH, it's OK.

    What may be more to your liking on the subject is "Riemann's Zeta Function" by H.M. Edwards. Now, THAT CONTAINS MATH!!.
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    Quote Originally Posted by ThePerfectHacker View Post
    Is it really good?
    I was thinking about buying it.

    (But it does not contain any math )
    I own it. I don't think you'd appreciate it because it tries hard to put everything in layman's terms. The analytic continuation, which I'm most interested in, is presented but not derived.
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    Quote Originally Posted by galactus View Post
    What may be more to your liking on the subject is "Riemann's Zeta Function" by H.M. Edwards. Now, THAT CONTAINS MATH!!.
    Too difficult for me.
    Not to offend anyone here very few poeple on this forum can actually understand it. It used analytic number theory, very difficult stuff.

    I would not buy such a book. From experience I know that such books are bad if you want them as an introduction to a new subject. Textbooks are always the best (tough expensive).
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  8. #8
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    I believe you're correct, PH. I have Hardy's book. For fun though, on page 157

    it says, "Using the Riemann-Siegel formula it is quite easy to locate the first

    few non-trivial zeros(those that lie on the line 1/2+it)by computation".

    He then goes on to show how to find t=14.134.

    Easy for the author, apparently.

    On page 114, he shows how to find roots on the line by Euler MacLaurin

    summation.

    You could look through the book, Jameson, and see what you think.

    No matter how you cut it, the Riemann Zeta is a complex subject if one wants to delve into it.
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    Quote Originally Posted by galactus View Post
    No matter how you cut it, the Riemann Zeta is a complex subject
    Arg! You're killin' me!

    -Dan
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    Re: Riemann-zeta function

    (1) The series quoted is the Euler Zeta Series. It is convergent for all integers > 1 It is equivalent to the Riemann Zeta series for these positive integers only. For 0 & all negative integers the Euler Zeta Series diverges, but the Riemann Zeta Series converges.
    (2)RZ(x)= Lim as n=> infinity{sum n=1 to n=n of 1/n^x +1/{(x-1)n^(x-1)} + sum of s=1 to s=infinity of {B(s)/s} .(x+s-2)C(s-1).n^1-k-x}
    This works for both positive & negative x values for RZ(x), where B(s) are Bernoulli nos. & (x+s-2)C(s-1) is standard Combinations notation= (x+s-2)!/(s-1)!(x-1)! The even Bernoulli Nos are the values: 1/6,-1/30,1/42,-1/30, 5/66, etc. The new previously undefined ODD B(s)numbers can be newly defined.(See if you can do this.)
    RZ(2)=pi^2/6, RZ(4)= pi^4/90, RZ(6)= pi^6/945 RZ(8)=pi^8/9540...RZ(3)={pi^3/28}.[4k-1]P(3)= 1.202056903 approx. where
    [4k-1]P(3)= infinite product of (p^3+1)/(p^3 -1) taken over all (4k-1)primes ie over p= 3,7,11,19,23,31.....Alan Walter.
    Here [4k-1]P(3) = 1.085509029 approx. Note: B(3) = - (3/56).[4k-1]P(3).The general result for integral n>1 for both Euler's Z(n) series &
    for Riemann's RZ(n) series is (2pi)^n.[ B(n) ]/ {2.(n)!} Works for n ODD or n EVEN values, where [ B(n) ] is the positive value of B(n)
    For n=3, RZ(3)= (2pi)^3.(3/56).[4k-1]P(3)/{2.(3)!} Hence RZ(3) = {pi^3/28}.[4k-1]P(3)= Z(3)= 1.202056903 approx = Apery's constant.
    Last edited by nala; September 7th 2012 at 10:19 PM. Reason: Typo in RZ(3) result quoted & in RZ(8)
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    Re: Riemann-zeta function

    Just about the Riemann Zeta function and the corresponding Riemann Hypothesis: Briefly stated, the Riemann Hypothesis states that all non-trivial zeros to the riemann zeta function are on the line whose real component is 1/2. Not one zero should lie outside this line if the hypothesis is true. Now, I have been doing research into various papers, books, etc, to see if there are any valid papers or even just techniques that could be utilized in solving this connjecture. However, in my endeavours, I found a paper on vixra.org which seemed to be so simple and logical in its approach to solving the riemann hypothesis that when I saw that the author arrived at the real number component 1/2 for an arbitrary zero to the zeta function in the critical strip, I thought the author might just be right. Perhaps we can take a look at the paper and challenge it a bit. It can be found at vixra.org, and the paper is called "On the Resolution to the Riemann Hypothesis". The link to the paper is the following: viXra.org e-Print archive, viXra:1210.0026, On the Resolution of the Riemann Hypothesis . If you have trouble opening the PDF, just right-click and open in a new window.
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