Guest wrote:FYI: 'Mathematician Proves Huge Result on ‘Dangerous’ Problem
By
KEVIN HARTNETT
December 11, 2019
Mathematicians regard the Collatz conjecture as a quagmire and warn each other to stay away. But now Terence Tao has made more progress than anyone in decades.'
https://www.quantamagazine.org/mathematician-terence-tao-and-the-collatz-conjecture-20191211/.
Hah! Any 'almost proof' of Collatz Conjecture by T. Tao (a so-called math expert on Collatz Conjecture, etc.) et al. is not a true proof! And any 'almost proofs' of the Riemann Hypothesis or the Goldbach Conjecture are not true proofs!
The Collatz Conjecture, Riemann Hypothesis, and the Goldbach Conjecture are true!!!
-- David Cole,
https://wwww.researchgate.net/profile/David_Cole29/amp.
"Don't pay attention to "authorities" (so-called experts)!
Think for yourself!" -- Richard Feynman.
Guest wrote:An Important Note Update:
The value, [tex]l_{m} = 1[/tex], which implies division by two in the Collatz process will occur as often as all other values combined, i.e., [tex]l_{m} > 1[/tex], which imply division either by four or by eight or by sixteen or by thirty-two, … in the Collatz process according to the our probability calculations and according to the Law of Large Numbers. Thus, we expect the division by two will occur 50% of all possible divisions in the Collatz processing. That fact explains any growth in the Collatz sequence, and it also explains why the Collatz sequence will always converge to one.
Another Relevant Reference Link:
'Is the Collatz Conjecture solvable?'
https://www.math10.com/forum/viewtopic.php?f=63&t=1708.
Guest wrote:An Update: ( Oops! index [tex]n_{m }[/tex] is wrong! We revert back to m instead.)
Probability(Collatz sequence does not converge to 1 or [tex]n_{m < \infty} \ne 1[/tex] for any initial positive odd integer, [tex]n_{0}[/tex] )
[tex]=\prod_{j = 1}^{m } < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.[/tex]
where [tex]n_{1} = 3 * n_{0} + 1[/tex] ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).
So, the Collatz conjecture is true!
Remarks: In our Collatz sequence from
[tex]n_{0}[/tex] to [tex]n_{m < \infty}[/tex], we have,
[tex]1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)[/tex]
for [tex]0 \le k \le m - 1[/tex] where [tex]1 \le j \le m < \infty[/tex].
Dave.
Guest wrote:Guest wrote:An Update: ( Oops! index [tex]n_{m }[/tex] is wrong! We revert back to m instead.)
Probability(Collatz sequence does not converge to 1 or [tex]n_{m < \infty} \ne 1[/tex] for any initial positive odd integer, [tex]n_{0}[/tex] )
[tex]=\prod_{j = 1}^{m < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.[/tex]
where [tex]n_{1} = 3 * n_{0} + 1[/tex] ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).
So, the Collatz conjecture is true!
Remarks: In our Collatz sequence from
[tex]n_{0}[/tex] to [tex]n_{m < \infty}[/tex], we have,
[tex]1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)[/tex]
for [tex]0 \le k \le m - 1[/tex] where [tex]1 \le j \le m < \infty[/tex].
Dave.
Dave wrote:Remarks: Probability [ [tex]l_{j } = 1[/tex] ] = .5 for [tex]1 \le j \le m < \infty[/tex] in accordance with the Law of Large Numbers.
We had assumed the Probability [ [tex]l_{m } = 0[/tex] ] = 0. However, it becomes clear that our assumption is wrong!
Relevant Reference Link:
'Law of Large Numbers',
https://en.wikipedia.org/wiki/Law_of_large_numbers.
Guest wrote:Remark: For our previous remarks, we assume and we refer to the ordered set of all positive even integers, E = {2, 4, 6, 8, 10, 12, ..., [tex]\infty[/tex]}.
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