The Lumber Room

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Equivalent forms of the Riemann hypothesis

with 3 comments

  1. Let H_n be the nth harmonic number, i.e. H_n = 1 + \frac12 + \frac13 + \dots + \frac1n. Then, the Riemann hypothesis is true if and only if

    \displaystyle \sum_{d | n}{d} \le H_n + \exp(H_n)\log(H_n)

    The left-hand side, which is the sum of the divisors of n, is also denoted \sigma(n).
    See Jeffrey Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis [PDF, arXiv.

  2. Define the Redheffer matrix A = A(n) to be the n \times n 0-1 matrix with entries A_{ij} = 1 \text{ if } j=1 \text{ or } i \text{ divides } j (and 0 otherwise). Then the Riemann hypothesis is true if and only if \det(A) = O(n^{1/2+\epsilon}) for all \epsilon.

For more, see Equivalences to the Riemann hypothesis (by J. Brian Conrey and David W. Farmer), and Consequences of the Riemann hypothesis (MathOverflow)

For fun: claimed [dis/]proofs.

Quote found via rjlipton:

The Riemann Hypothesis is the most basic connection between addition and multiplication that there is, so I think of it in the simplest terms as something really basic that we don’t understand about the link between addition and multiplication.
Brian Conrey

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Written by S

Tue, 2010-07-06 at 22:03:54 +05:30

Posted in mathematics

3 Responses

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  1. #1 is really beautiful. Wow! Something like e^x*log(x) is inherently interesting, [valleygirl] like you know, the product of taking something ‘higher’ with taking it ‘lower’. [/valleygirl] I wonder if there’s some name to a function g(x) = f'(x)*Integral(f(x),0,x).

    Can #2 be derived from #1 by any simple way?

    KVM

    Sun, 2011-01-16 at 06:10:57 +05:30

    • Wait why did this post from 7th July 2010 appear in Reader now?

      KVM

      Sun, 2011-01-16 at 06:11:50 +05:30

      • Because it was ‘private’ and I just set it ‘public’… and the former probably because there’s no context or completeness or anything.

        I actually don’t know how to prove either of these, or whether it’s possible to derive one from the other… just a pair of curiosities to me, for now at least!

        S

        Sun, 2011-01-16 at 07:14:38 +05:30


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