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The Riemann Hypothesis for Calculus Students: an interesting relative limit behavior

  1. Jul 15, 2009 #1
    Dear All,
    following up on my previous posting "The Riemann Hypothesis for High School Students" here are some further observations, which might instead be appreciated by students of typical II year undergraduate Calculus courses. You can get much background information by searching for Riemann Zeta function and Dirichlet Eta function on the websites of mathworld, wikipedia, planetmath.
    By studying my previous posting you can instead familiarise yourselves with the geometric meaning of "path described by the partial sums of the Dirichlet eta function", and which was based on the equivalent representation as vectors of the complex numbers that are addenda of the following infinite sum (called Dirichlet Eta Function):

    [tex] \eta(s) = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} =
    1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+-\ldots [/tex]

    which can be written as

    [tex] \eta(s) = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} =
    S_k(s) + R_k(s) [/tex]

    and where [tex] S_k(s) [/tex] is the [tex] k^{th} [/tex] partial sum and [tex] R_k(s) [/tex] the corresponding remainder term.
    As I have now managed to publish a manuscript on the arXiv (arxiv.org/abs/0907.2426), you might wish to inspect figures 5 and 6 at the end of said manuscript, and which explicitly draw the vectors representing examples of said partial sums and remainder terms ( [tex] S_{13}(s) [/tex] and [tex] R_{13}(s) [/tex] in fig. 5a, [tex] S_{292}(s) [/tex] and [tex] R_{292}(s) [/tex] in fig. 6b).
    A well known fact about the non-trivial zeros of the Riemann zeta function is that to a hypothetical zero located inside the critical strip, but not lying on the critical line, would always correspond a second zero located at a position symmetrical about the critical line. That is: non-trivial zeros located off the critical line always occurs in pairs
    [tex] \frac{1}{2}+ \alpha + it [/tex]

    [tex] \frac{1}{2} - \alpha + it [/tex]
    [tex] 0 < \alpha < \frac{1}{2} [/tex]
    (see the above mentioned web sites, or also the Introduction section of my manuscript).
    What I was not able to find in the published literature is a systematic study of the relative behavior of eta functions of critical line symmetrical arguments. In this respect it is very interesting to study the ratio of their respective modulus:

    [tex]P(\alpha,t) = \frac{\left|\eta(\frac{1}{2}+ \alpha + it)\right|}{\left|\eta(\frac{1}{2}- \alpha + it)\right|}=\left|\frac{\eta(1-(\frac{1}{2}- \alpha + it))}{\eta(\frac{1}{2}- \alpha + it)}\right| [/tex]

    where the right term can be obtained by observing that for any pair of critical line symmetrical arguments we have

    [tex]\overline{\frac{1}{2}+ \alpha + it}=\frac{1}{2}+ \alpha - it=1-(\frac{1}{2}- \alpha + it)[/tex]

    and recalling that

    [tex]\overline{\eta(s)} = \eta(\overline{s}) [/tex]

    and so resulting in

    [tex]\left|\eta(\frac{1}{2}+ \alpha + it)\right|=\left|\overline{\eta(\frac{1}{2}+ \alpha + it)}\right|=\left|\eta(1-(\frac{1}{2}- \alpha + it))\right|[/tex]

    A remarkable functional equation satisfied by the zeta function was originally proposed by Euler in 1749, and later proved by Riemann in his 1859 paper

    [tex]\zeta(1-s) = 2 (2\pi)^{-s} \cos\left(\frac{\pi s}{2}\right)\ \Gamma(s)\ \zeta(s)[/tex]

    which combined with the known relationship

    [tex]\zeta(s) = \frac{\eta(s)}{1-\frac{2}{2^s}}[/tex]

    finally allows to write

    [tex]P(\alpha,t) = \left|\frac{1-2^{\frac{1}{2}- \alpha + it}}{1-2^{\frac{1}{2}+ \alpha - it}}\right| \left|2\left(2\pi\right)^{\alpha-\frac{1}{2}-it}\right| \left|\cos\left(\frac{\pi}{2}\left(\frac{1}{2}- \alpha + it\right)\right)\right| \left|\Gamma\left(\frac{1}{2}- \alpha + it\right)\right|[/tex]

    which, inside the critical strip, is a continuos function always > 0, corresponding to a beautifully structured surface (see the 3D plot in fig. 2 of my manuscript).

    Let us now finally get to the point of that interesting limit behaviour mentioned in the title. The definition of partial sums given above allows us to write

    [tex]\frac{\left|\eta(\frac{1}{2}+ \alpha + it)\right|}{\left|\eta(\frac{1}{2}- \alpha + it)\right|}= \lim_{n\to\infty}
    \frac{\left|S_n(\frac{1}{2}+ \alpha + it)\right|}{\left|S_n(\frac{1}{2}- \alpha + it)\right|} = P(\alpha,t) [/tex]

    limit which exists, and that it is >0, irrespective of whether or not the pair of critical line symmetrical arguments might hypothetically correspond to a pair of zeros.
    Now, in case s=so is a zero of the eta function, the definition of partial sum and remainder term given above must imply |Sn (so)| = |Rn (so)| . In other words

    [tex]\eta(\frac{1}{2}+ \alpha + it) = 0 \; \; \; \Rightarrow \; \; \; \lim_{n\to\infty} \frac{\left|R_n(\frac{1}{2}+ \alpha + it)\right|}{\left|R_n(\frac{1}{2}- \alpha + it)\right|} = \lim_{n\to\infty} \frac{\left|S_n(\frac{1}{2}+ \alpha + it)\right|}{\left|S_n(\frac{1}{2}- \alpha + it)\right|} = P(\alpha,t) [/tex]

    Right ? ...... Wrong !

    By observing the pattern described by the partial sums, once they have already settled into that kind of bound crisscrossing orbit visible in fig. 1-5-6, it visually appear clear that the remainder terms [tex]|R_n(1/2 - \alpha + it)|[/tex] and [tex]|R_n(1/2 + \alpha + it)|[/tex], and which in Fig. 1 represent the distances of the points [tex]S_n(1/2 - \alpha + it) [/tex] and [tex]S_n(1/2+ \alpha + it)[/tex] to the points of convergence, [tex]\eta(1/2- \alpha + it)[/tex] and [tex]\eta(1/2+ \alpha + it)[/tex] respectively, is roughly of the same order of magnitude as the lengths,[tex]n^{\alpha-1/2}[/tex] and [tex]n^{-\alpha-1/2}[/tex], of their last segments. That this is indeed the case is the subject of Lemma 2 in my manuscript, whereby it is demonstrated also for the general case, by means of a geometric approach, that

    [tex] as \; \; n\rightarrow\infty \; \; \; \; \; \; |R_n(\frac{1}{2}- \alpha + it)| = O(n^{\alpha-1/2}) \; \; \; \; \; \; R_n(\frac{1}{2}+ \alpha + it)| = O(n^{-\alpha-1/2}) [/tex]


    [tex]\lim_{n\to\infty} \frac{\left|R_n(\frac{1}{2}+ \alpha + it)\right|}{\left|R_n(\frac{1}{2}- \alpha + it)\right|}= \lim_{n\to\infty} O(n^{-2\alpha}) = 0 \neq P(\alpha,t) [/tex]

    If I have not missed some basic logical flaw, hidden somewhere in the above described approach, this contradicting result appears to suggest that in fact there cannot exist any such pair of critical line symmetrical zeros lying off the critical line. And because inside the critical strip the zeros of [tex]\eta[/tex] coincide with the zero of [tex]\zeta[/tex], such a result would also confirm the validity of the Riemann Hypothesis.
    Of course, I am not at all sure that I have not missed some basic logical flaw, but I am sure that PhysicsForums is the right community for starting a very open discussion about it, and I would be very happy to receive comments and-or suggestions.

  2. jcsd
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