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Stieltjes

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  • Dutch mathematician T.J. Stieltjess claim to have solved Riemanns hypothesis has

    received attention in the scholarly literature , especially after the 1985 proof of the falsity of the

    Mertens conjecture, a stronger version of Stieltjes 1885 claim to have proven, in todays

    notation, that ()/is bounded. [Borwein, 2009, 69] Number theorists today are skeptical of

    Stieltjes claim that he had a proof of an equivalent of Riemanns hypothesis, and with good

    reason. And yet, revisiting his attempt at proof, something that has never been done, can yield

    insights on the history of understanding the problem. His letters offer a little studied resource on

    this first attempt to solve Riemanns conjecture, an attempt that, while lacking rigor by

    contemporary standards, still has not been entirely rejected. Studying Stieltjes letters can still

    shed light,not only on the assumptions of his 19th century mathematical contemporaries but also

    on those assumptions we have inherited from them.

    Reconstruction of Stieltjes proof of RH:

    Definition 1.1 The Liouville function is defined by

    () = () ,

    where is the number of, not necessarily distinct, prime factors of .

    Theorem 1.2 The Riemann hypothesis is equivalent to the statement that

    for every fixed > 0,

    () + () + + ()

    +;

    =

    This translates to the following statements: The Riemann hypothesis is equivalent to the

    statement that an integer has equal probability of having an odd number of or an even

    number of distinct prime factors (in the precise sense given above). The sequence

    {()}= = {, , , , }

  • behaves more or less like a random sequence of s and -s in that the difference between

    the number of s and -s is not much larger than the square root of the number of terms.

    (Borwein, 2008, 6)

    We will use the equivalent statement The Riemann hypothesis is equivalent to

    () + () + + ()

    +

    (Borwein, 2008, 46)1

    Cauchys Cours dAnalyse (1827) ; Note II, Theorem XV: Let , , , be any real

    quantities. If these quantities are not equal to each other, then the numerical value of the sum

    + + +

    is less than the product

    2 + 2 + 2 +

    so that we have

    val.num. ( + + + ) < 2 + 2 + 2 +

    [Cauchy, 2009, 302]

    The last part is equivalent to | + + + | < 2 + 2 + 2 +

    Let = (1) , k being the number of prime factors, either distinctive, for the Mbius

    function () or not, for the Liouville function ().

    |1 + 2 + + | 12 + 2

    2 + + 2

    1 if we interpret the statement in terms of the frequencies, this is just the statement that the sequences of the averages of the partial sums have the limit 1/2" (italics in the original). [Varadajaran 126 Cesaro. ] Jahnke 171 says its true!. Euler Ayoub has lambda

  • From the above we have that substitution gives us

    |1 + 2 + + | (11)2 + (12)2 + + (1)2

    |1 + 2 + + | < (11)2 + (12)2 + + (1 )2

    And if we let = or we are done since absolute convergence implies convergence and

    =1

    1

    =1

    =1

    (0)

    =1

    1

    2

    QED2

    More rigorously (?),we recall the AM-QM inequality (Nahin, 334), generalized in Jensens inequality for any real 1, 2, ,

    1 + 2 + +

    1

    2 + 22 + + 2

    2 Note that Cauchy gives us that

    |1 + 2 + + |

    = (2 ,

    2,

    2 )

    = ((1)2, (1)2 , (1)2 )

    =1

    as Stieltjes claimed.

  • with equality iff 1 = 2 = , = . Now, multiplying by we get

    1 + 2 + + 1

    2 + 22 + + 2

    Following Cauchys Course we have

    1 + 2 + + 12 + 2

    2 + + 2

    So now we can prove that the following is a constant.

    12 + 2

    2 + + 2 =

    Let = (1) . From the above we have that substitution gives us

    1 + 2 + + (1)2 + (1)2 + (1)2 + We then have that

    1 + 2 + + < (11)2 + (12)2 + + (1)2

    1 + 2 + + < (0)

    1 + 2 + + < 1

    2

    And if we let = or 1we are done.

    QED.

    Version 2: Recall the quadratic mean, or root mean square.

    1

    1

    2

    =1

    =1

    Now, let

    =1 be the partial sums of the Mbius function and multiply both sides by n.

  • 1 ()

    1

    2()

    =1

    =1

    () 1

    2()

    =1

    =1

    () (2)1

    2()

    =1

    =1

    () 2()

    =1

    =1

    ()

    =1

    2()

    =1

    ()

    =1

    |()|

    =1

    ()

    =1

    1

    =1

    Via zeta regularization (or Grandis series /Diracs comb , Eulers characteristic, (0) = 1

    2 )

    ()

    =1

    1

    2

    QED

    Note: For (0) = 1

    2 see also (Kaneko 2003) and Eulers Beau rapport

    http://eulerarchive.maa.org/pages/E352.html

    Juan Marin