Is the following heuristic probabilistic argument correct?

Probability theory and statistics

Is the following heuristic probabilistic argument correct?

Postby Guest » Tue Jan 28, 2020 2:08 pm

"The following heuristic probabilistic argument support the 3x + 1 (Collatz) Conjecture ... Pick an odd integer [tex]n_{0 }[/tex] at random and iterate the

function T until another odd integer [tex]n_{1}[/tex] occurs. Then 1/2 of the time, [tex]n_{1} = \frac{3(n_{0}+1)}{2}[/tex], 1/4 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{4}[/tex], 1/8 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{8}[/tex] , and so on. If one supposes that the function T is sufficiently 'mixing'' that successive odd integers in the trajectory of n behave as though they were drawn at random (mod [tex]2^{k}[/tex]) from the set of odd integers (mod [tex]2^{k}[/tex]) for all k, then the expected growth in size between two consecutive odd integers in such a trajectory is the multiplicative factor,

[tex]( \frac{3}{2} )^{\frac{1}{2}} * ( \frac{3}{4} )^{\frac{1}{4}} * ( \frac{3}{8} ) ^{\frac{1}{8}} *[/tex] ... = [tex]\frac{3}{4} < 1[/tex].

Consequently, ... "

Source Links: http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/node3.html;

'Collatz (3x+ 1) Con jecture',

https://en.wikipedia.org/wiki/Collatz_conjecture.

Does the above argument makes sense? Is it correct?

If that argument is correct, then it suggests a counterexample to Collatz Conjecture may exist. Do you agree?
Attachments
Algorithm for the Collatz Conjecture.jpg
Algorithm for the Collatz Conjecture.jpg (7.11 KiB) Viewed 609 times
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Jan 28, 2020 2:50 pm

Hmm. We suspect the argument is flawed. We must first quantify time with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.
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Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Jan 28, 2020 6:32 pm

Relevant Reference Link:

'Proof of Collatz Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1485.
Attachments
collatz_conjecture.png
Collatz Conjecture is true!
collatz_conjecture.png (23.48 KiB) Viewed 598 times
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Sun Feb 09, 2020 6:17 pm

David Cole wrote:Hmm. We suspect the argument is flawed. We must first quantify time with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ?

The answer depends on t and k.


"I am the author of the above post." -- David Cole, https://www.researchgate.net/profile/David_Cole29.

P.S. Please try different values of t and k to see if the original heuristic probabilistic argument is correct. I am convinced the argument is wrong!
Attachments
collatz_conjecture.png
The Collatz Conjecture is true!
collatz_conjecture.png (23.48 KiB) Viewed 501 times
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Wed Feb 19, 2020 1:55 pm

Guest wrote:Hmm. We suspect the argument is flawed. We must first quantify time (or the number of trials) with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Wed Feb 19, 2020 5:47 pm

Guest wrote:
Guest wrote:Hmm. We suspect the argument is flawed. We must first quantify time (or the number of trials) with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.


Remarks: The number of times (t) implies the number of appropriate trials (t) in any Collatz sequence as described by the stated argument... Time alone is a bit incoherent and inappropriate here. -- Dave.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Sep 15, 2020 9:27 am

Guest wrote:Hmm. We suspect the argument is flawed. We must first quantify time (or the number of trials) with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.

Remarks: The number of times (t) implies the number of appropriate trials (t) in any Collatz sequence as described by the stated argument... Time alone is a bit incoherent and inappropriate here. -- Dave.


We let t be the number of trials it takes for the Collatz sequence to converge.

We can determine k by the largest positive even integer, [tex]e_{max }[/tex], in the Collatz sequence.

[tex]k = \frac{log(e_{max })}{log(2)}[/tex].

Therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *...*( \frac{3}{k} ) ^{\frac{t}{k}} \rightarrow 0[/tex]. :D
Attachments
A Collatz Sequence.jpg
The Collatz conjecture is true!
A Collatz Sequence.jpg (9.6 KiB) Viewed 175 times
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Sep 15, 2020 10:04 am

Oops! We corrected the k value in the previous post.

[tex]k = \left \lfloor{\frac{log(e_{max })}{log(2)}}\right \rfloor[/tex].
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Sep 15, 2020 11:44 am

Added Update: We let t be the number of trials it takes for the Collatz sequence of odd integers to converge.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Sep 15, 2020 12:14 pm

Added Update: That convergent value, 0, represents the probability that the Collatz sequence of odd integers does not converge to one.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Tue Sep 15, 2020 3:54 pm

Please vote on the truth of the Collatz conjecture at the link below.

Link: https://theory-of-energy.org/2020/09/15/please-vote-on-the-truth-of-the-collatz-conjecture-thank-you/.

Thank you! :)
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Thu Sep 17, 2020 3:01 am

Relevant Reference Link:

Our Response to 4.2 A probabilistic heuristic:

https://en.wikipedia.org/wiki/Talk:Collatz_conjecture#Our_Response_to_4.2_A_probabilistic_heuristic:.

Dave.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Fri Oct 16, 2020 8:53 am

Guest wrote:"The following heuristic probabilistic argument support the 3x + 1 (Collatz) Conjecture ... Pick an odd integer [tex]n_{0 }[/tex] at random and iterate the

function T until another odd integer [tex]n_{1}[/tex] occurs. Then 1/2 of the time, [tex]n_{1} = \frac{3(n_{0}+1)}{2}[/tex], 1/4 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{4}[/tex], 1/8 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{8}[/tex] , and so on. If one supposes that the function T is sufficiently 'mixing'' that successive odd integers in the trajectory of n behave as though they were drawn at random (mod [tex]2^{k}[/tex]) from the set of odd integers (mod [tex]2^{k}[/tex]) for all k, then the expected growth in size between two consecutive odd integers in such a trajectory is the multiplicative factor,

[tex]( \frac{3}{2} )^{\frac{1}{2}} * ( \frac{3}{4} )^{\frac{1}{4}} * ( \frac{3}{8} ) ^{\frac{1}{8}} *[/tex] ... = [tex]\frac{3}{4} < 1[/tex].

Consequently, ... "

Source Links: ...;

'Collatz (3x+ 1) Conjecture',

https://en.wikipedia.org/wiki/Collatz_conjecture.

Does the above argument makes sense? Is it correct?

If that argument is correct, then it suggests a counterexample to Collatz Conjecture may exist. Do you agree?


Oops! Thus far, the heuristic probabilistic argument is apparently correct! More analysis may be needed.

Dave,
https://theory-of-energy.org/2020/09/17/a-brief-analysis-of-the-collatz-conjecture/.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Fri Oct 16, 2020 2:31 pm

Dave wrote: ...

Oops! Thus far, the heuristic probabilistic argument is apparently correct! More analysis may be needed.

Dave,
https://theory-of-energy.org/2020/09/17/a-brief-analysis-of-the-collatz-conjecture/.


More analysis is required to determined if the heuristic probabilistic argument is true for all positive odd integers greater than one.

Dave.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Sat Oct 17, 2020 8:35 am

Final Remark: The heuristic probabilistic argument is true for infinitely many positive odd integers greater than one, but it is also false for infinitely many positive odd integers greater than one.

Dave.
Guest
 

Re: Is the following heuristic probabilistic argument correc

Postby Guest » Sat Oct 17, 2020 10:38 am

Update:

Final Remark: The heuristic probabilistic argument is true!

Dave.
Guest
 

Re: Is the following heuristic probabilistic argument...

Postby Guest » Sat Oct 17, 2020 6:35 pm

Dave wrote:Update:

Final Remark: The heuristic probabilistic argument is true!

...


Dave also wrote:
Dave's Conjecture: [tex]r = r(n_{0 }, t) =[/tex] O[tex](t)[/tex] or [tex]r = c_{t } * t[/tex] for some real number, [tex]c_{t }[/tex], such that either [tex]c_{t } >1[/tex] or [tex]0< c_{t } < 1[/tex]. (Update)

Example: If we have [tex]t = 1[/tex] for some [tex]n_{0 }[/tex], then as [tex]k \rightarrow \infty[/tex],

[tex]r = \frac{1}{ \prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}}} \rightarrow \frac{4}{3 } = c_{1 }[/tex].

Therefore, [tex]r = r(n_{0 }, 1) \le \frac{4}{3}[/tex].

[tex]r = r(n_{0 }, 1) \approx \frac{4}{3}[/tex] for infinitely many positive odd integers, [tex]n_{0} > 1[/tex].

Remark: [tex]\prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}} \rightarrow \frac{3}{4 }[/tex] as [tex]k \rightarrow \infty[/tex].


Relevant Reference Link:

"The Collatz equation that supports the truth of the Collatz conjecture",

https://www.math10.com/forum/viewtopic.php?f=63&t=9166.
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