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Topic: The 3n+1 Conjecture is Unprovable?! (Read 551 times) |
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Barukh
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The 3n+1 Conjecture is Unprovable?!
« on: Nov 3rd, 2004, 4:33am » |
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ThudnBlunder
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Re: The 3n+1 Conjecture is Unprovable?!
« Reply #1 on: Nov 3rd, 2004, 5:25am » |
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Yet he admits the possibility of the conjecture being false, either provably or not.
Actually, this paper has been rubbished on sci.math, particularly by mathematicians Robin Chapman and Torkel Franzen. Expert opinion seems to indicate that his approach is fundamentally flawed. For example:
Try reading the Feinstein paper. "Theorem 2" states
that "For any m, n in N, if each T^{(k)}(n) is distinct
(for k = 0, 1, ..., n-1) then it is impossible to determine
the value of T^{(m)}(n) without performing at least m computations.
Leaving aside the private language (what does Feinstein mean by
"m computations"? He doesn't tell us) note the universal
quantifiers on the m and n ("For any"). The family of counterexamples
m arbitrary and n = 2^m immediately come to mind. Each of these
satsifies the hypothesis, and it takes no "computations" to prove
that T^m(n) = 1 for these m and n.
(Chapman)
AND
> Feinstein proved in theorem 2 that it takes m steps to prove that
> T^m(n)=1 for some m.
By all means let us stipulate that he did. The idea is that "since
the value of F(m_0) is determined in showing that there exists an m
such that F(m)=1", where m_0 is the smallest m such that F(m)=1,
proving that there exists an m such that F(m)=1 must involve at least
as many steps as proving that F(m_0)=1. This idea is mistaken, or
perhaps just confused. Proving the existence of an m such that F(m)=1
does not in general involve any "determination" of F(m_0) in the sense
of actually computing F(m_0).
(Franzen)
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| « Last Edit: Nov 3rd, 2004, 8:48am by ThudnBlunder » |
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