
Re: SpencerBrown's purported proof of Riemann Hypothesis
Posted:
Aug 6, 2006 6:05 AM


In article <je2dnSC6K7IEWVbZnZ2dnUVZ_qWdnZ2d@comcast.com>, Tim Peters <tim.one@comcast.net> wrote:
>He claims that these first few pages prove: > > pi(n)  li(n) < sqrt(li(n)) [3] > for all n > 1
>Anyway, /if/ [3] holds RH follows, because: > >a. sqrt(li(n)) < sqrt(n) for n > 1 (which he notes) > >b. so pi(n)  li(n) < sqrt(n) for n > 1 would follow from [3] > >c. RH is equivalent to the statement (which he alludes to, but doesn't > give) that there's some constant C for which: > > pi(n)  li(n) < C sqrt(n) log(n) > > holds for all sufficiently large n; or, IOW, > > pi(n) = li(n) + O(sqrt(n) log(n)) > >So [3], if true, is in fact a substantially tighter bound than required to >prove RH.
At first I thought it might be too tight to be true  but now I guess such a bound has not yet been ruled out.
In Bombieri's article on the Riemann Hypothesis:
http://www.claymath.org/millennium/Riemann_Hypothesis/Official_Problem_Description.pdf
he writes:
"The validity of the Riemann hypothesis is equivalent to saying that the deviation of the number of primes from the mean Li(x) is
pi(x) = Li(x) + O(sqrt(x) log x);
the error term cannot be improved by much, since it is known to oscillate in both directions to order at least Li(sqrt(x)) log log log x (Littlewood)."
But, unless I'm making a mistake, Li(sqrt(x)) log log log x is eventually much smaller than SpencerBrown's claimed bound sqrt(Li(x)).
Of course this has nothing to do with whether his "proof" makes sense.

