primework123 wrote:Please see https://en.wikipedia.org/wiki/Collatz_conjecture for explanation.
We let [tex]n_{1 }[/tex] = [tex]3n_{0 }[/tex]+ 1 where [tex]n_{0}[/tex] is any odd integer > 1;
[tex]l_{1 }[/tex] = [tex]Log( n_{1 }[/tex] / (greatest odd factor > 1 of [tex]n_{1 })) / Log(2)[/tex];
[tex]n_{2}[/tex] = [tex]3((3n_{0 } + 1[/tex]) / [tex]2^ {l_{1 }} ) + 1[/tex] = [tex]3n_{1 }[/tex] /[tex]2^ {l_{1 }} + 1[/tex];
[tex]l_{2}[/tex] = [tex]Log( n_{2 }[/tex] / (greatest odd factor > 1 of [tex]n_{2 })) / Log(2)[/tex];
...
[tex]n_{m}[/tex] = [tex]3n_{m-1 }[/tex] / [tex]2^ {l_{m-1 }} + 1[/tex];
[tex]l_{m}[/tex] = [tex]Log( n_{m }[/tex] / (greatest odd factor > 1 of [tex]n_{m})) / Log(2)[/tex];
Note: [tex]n_{m}[/tex] = [tex]2^{l_{m }}[/tex] * (greatest odd factor > 1 of [tex]n_{m }[/tex]);
Probability( [tex]n_{1}[/tex] is a multiple of 2) = 1;
Probability( [tex]n_{1}[/tex] is a multiple of [tex]2^{2}[/tex] | [tex]n_{1}[/tex] is a multiple of 2) = 1/2;
Probability( [tex]n_{1}[/tex] is a multiple of [tex]2^{3}[/tex] | [tex]n_{1}[/tex] is a multiple of [tex]2^{2}[/tex]) = 1/2;
...
Probability( [tex]n_{1}[/tex] is a multiple of [tex]2^{2}[/tex])
= Probability( [tex]n_{1}[/tex] is a multiple of [tex]2^{2}[/tex] | [tex]n_{1}[/tex] is a multiple of 2)
* Probability( [tex]n_{1}[/tex] is a multiple of 2) = (1/2) * 1 = 1/2;
Probability( n1 is a multiple of [tex]2^{3}[/tex]) =
Probability( n1 is a multiple of [tex]2^{3}[/tex] | n1 is a multiple of [tex]2^{2}[/tex])
* Probability( n1 is a multiple of [tex]2^{2}[/tex]) = (1/2) * (1/2) = 1/4; ...
So, Probability([tex]n_{1 }[/tex] is at least a multiple of [tex]2^{2}[/tex]) = [tex]\sum_{i=0}^{l_{1 } - 1 }(1/2)^{i+1}[/tex]
...
Probability([tex]n_{m }[/tex] is at least a multiple of [tex]2^{2}[/tex]) = [tex]\sum_{i=0}^{l_{m-1 } - 1 }(1/2)^{i+1}[/tex]
Hence,
Probability(Collatz sequence does not converge to 1)
[tex]=\prod_{m=1}^{m=\infty} \sum_{i=0}^{l_{m-1 } - 1 }(1/2)^{i+1} \rightarrow 0.[/tex]
So, the Collatz conjecture is true!
Furthermore,
Let [tex]N_{0 }[/tex] be the event that [tex]3n_{0 }[/tex]+ 1 is a multiple of 2 for any odd integer, [tex]n_{0 }[/tex] > 1;
Let [tex]N_{1 }[/tex] be the event that [tex]n_{1 }[/tex] = [tex]3n_{0 }[/tex]+ 1 is at least a multiple of [tex]2^{2}[/tex];
Let [tex]N_{2 }[/tex] be the event that [tex]n_{2 }[/tex] = [tex]3n_{1 }/2^{l_{1 }}[/tex]+ 1 is at least a multiple of [tex]2^{2}[/tex];
...
Let [tex]N_{m }[/tex] be the event that [tex]n_{m }[/tex] = [tex]3n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1 is at least a multiple of [tex]2^{2}[/tex];
Note: [tex]l_{j }[/tex] = [tex]Log( n_{j }[/tex] / (greatest odd factor > 1 of [tex]n_{j })) / Log(2)[/tex];
Note: Probability([tex]N_{0 }[/tex]) = 1.
Prob([tex]N_{1 }[/tex]) = Prob([tex]N_{1}[/tex] | [tex]N_{0 }[/tex]) * Prob([tex]N_{0 }[/tex]) = Prob([tex]N_{1}[/tex] | [tex]N_{0 }[/tex])
[tex]= \sum_{i=0}^{l_{1 } - 1 }(1/2)^{i+1}[/tex];
Prob([tex]N_{2 }[/tex]) = Prob([tex]N_{2 }[/tex] | [tex]N_{1 }[/tex]) * Prob([tex]N_{1 }[/tex])
= Prob([tex]N_{2 }[/tex] | [tex]N_{1 }[/tex]) * Prob([tex]N_{1 }[/tex] | [tex]N_{0 }[/tex])
= [tex]\sum_{i=0}^{l_{2 } - 1 }(1/2)^{i+1}[/tex] * [tex]\sum_{i=0}^{l_{1 } - 1 }(1/2)^{i+1}[/tex];
...
...
Prob([tex]N_{m }[/tex])
[tex]=\prod_{j=1}^{m} \sum_{i=0}^{l_{j } - 1 }(1/2)^{i+1}[/tex];
Again, we have the following result:
Probability(Collatz sequence does not converge to 1)
= Prob([tex]N_{m }[/tex] as [tex]m \rightarrow \infty[/tex])
[tex]=\prod_{j=1}^{\infty} \sum_{i=0}^{l_{j } - 1 }(1/2)^{i+1} \rightarrow 0.[/tex]
Recall: Basics of Conditional Probability
Prob([tex]N_{2 }[/tex]|[tex]N_{1 }[/tex])Prob([tex]N_{1 }[/tex]) = Prob([tex]N_{2 }[/tex] ∩ [tex]N_{1 }[/tex]) = Prob([tex]N_{2 }[/tex]).
In context of our proof of the Collatz conjecture, [tex]N_{2 }[/tex] ∩ [tex]N_{1 }[/tex] means that the greatest odd factor of events, N2 and N1 is greater one, and the events, [tex]N_{2 }[/tex] and [tex]N_{1 }[/tex] are at least multiples of 4.
And since the prior existence of event N1 in the Collatz sequence indicates the existence of the succeeding event N2 ... all according to our prior assumption that there exists a Collatz sequence that does not converge one.
And thus, [tex]N_{2 }[/tex] ∩ [tex]N_{1 }[/tex]= [tex]N_{2 }[/tex].
Please share your comments too. Thanks!
David Cole
aka primework123
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!
Guest wrote:In his argument,https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjF7rzEnpfPAhVMHD4KHQOnDHgQFggeMAA&url=https%3A%2F%2Fwww.researchgate.net%2Ffile.PostFileLoader.html%3Fid%3D5712d7a1615e274be90d8662%26assetKey%3DAS%253A351655620038656%25401460852641147&usg=AFQjCNEqoCrtTMmOxT90KZdo0GaipII_Cg&sig2=66w9oubBjvBGTA5zb14Utw, Dr. G. Lassmann claims to refute David Cole's proof of Collatz Conjecture.
But, he does not understand that David's proof of the Collatz Conjecture is conditional:
Important Notes:
I. 3[tex]n_{m }[/tex] + 1 will always be a multiple of 2 for all positive odd integers, [tex]n_{m }[/tex] or that the
Probability[3[tex]n_{m }[/tex] + 1 is a multiple of 2] = 1.
II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some [tex]n_{m }[/tex] or that the Collatz sequence never converges to one.
III. [tex]N_{m }[/tex] be the event that [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1 = is at least a multiple of [tex]2^{2}[/tex]
= 3 * f([tex]n_{0 }[/tex]) + 1 where f([tex]n_{0 }[/tex]) is a function of [tex]n_{0 }[/tex] according to the Collatz Process and according to the definition of [tex]n_{0 }[/tex] in David Cole's proof of the Collatz Conjecture.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 3 * 2 = 6, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.
So, for the above example, the Probability[3[tex]n_{0 }[/tex] + 1 is at least a multiple of 4] is 1/2.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 5 * 2 = 10, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 8 | 3[tex]n_{0 }[/tex] + 1 is multiple of 4] = 1/2.
So, for this latest example, the Probability[3[tex]n_{0 }[/tex] + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 11 * 2 = 22, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 8 | 3[tex]n_{0 }[/tex] + 1 is multiple of 4] = 1/2.
And furthermore, Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 16 | 3[tex]n_{0 }[/tex] + 1 is multiple of 8] = 1/2.
So, for this latest example, the Probability[3[tex]n_{0 }[/tex] + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.
So, [tex]2^{l_{m }}[/tex] depends on the size of the largest odd factor of [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1, and they, [tex]2^{l_{m }}[/tex] and [tex]l_{m }[/tex] , will vary generally too.
So, the Probability [ 4 divides [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1] varies:
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 1] = 1/2;
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 2] = 3/4;
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 3] = 7/8;
...
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = k] = [tex]\sum_{i=1}^{k }(1/2)^i[/tex];
Therefore, 1/2 [tex]\le[/tex] Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] < [tex]\infty[/tex]] < 1.
Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest wrote:Guest wrote:In his argument,https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjF7rzEnpfPAhVMHD4KHQOnDHgQFggeMAA&url=https%3A%2F%2Fwww.researchgate.net%2Ffile.PostFileLoader.html%3Fid%3D5712d7a1615e274be90d8662%26assetKey%3DAS%253A351655620038656%25401460852641147&usg=AFQjCNEqoCrtTMmOxT90KZdo0GaipII_Cg&sig2=66w9oubBjvBGTA5zb14Utw, Dr. G. Lassmann claims to refute David Cole's proof of Collatz Conjecture.
But, he does not understand that David's proof of the Collatz Conjecture is conditional:
Important Notes:
I. 3[tex]n_{m }[/tex] + 1 will always be a multiple of 2 for all positive odd integers, [tex]n_{m }[/tex] or that the
Probability[3[tex]n_{m }[/tex] + 1 is a multiple of 2] = 1.
II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some [tex]n_{0 }[/tex] or that the Collatz sequence never converges to one.
III. [tex]N_{m }[/tex] be the event that [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1 = is at least a multiple of [tex]2^{2}[/tex]
= 3 * f([tex]n_{0 }[/tex]) + 1 where f([tex]n_{0 }[/tex]) is a function of [tex]n_{0 }[/tex] according to the Collatz Process and according to the definition of [tex]n_{0 }[/tex] in David Cole's proof of the Collatz Conjecture.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 3 * 2 = 6, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.
So, for the above example, the Probability[3[tex]n_{0 }[/tex] + 1 is at least a multiple of 4] is 1/2.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 5 * 2 = 10, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 8 | 3[tex]n_{0 }[/tex] + 1 is multiple of 4] = 1/2.
So, for this latest example, the Probability[3[tex]n_{0 }[/tex] + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 11 * 2 = 22, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 8 | 3[tex]n_{0 }[/tex] + 1 is multiple of 4] = 1/2.
And furthermore, Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 16 | 3[tex]n_{0 }[/tex] + 1 is multiple of 8] = 1/2.
So, for this latest example, the Probability[3[tex]n_{0 }[/tex] + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.
So, [tex]2^{l_{m }}[/tex] depends on the size of the largest odd factor of [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1, and they, [tex]2^{l_{m }}[/tex] and [tex]l_{m }[/tex] , will vary generally too.
So, the Probability [ 4 divides [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1] varies:
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 1] = 1/2;
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 2] = 3/4;
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 3] = 7/8;
...
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = k] = [tex]\sum_{i=1}^{k }(1/2)^i[/tex];
Therefore, 1/2 [tex]\le[/tex] Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] < [tex]\infty[/tex]] < 1.
Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest wrote:Guest wrote:Guest wrote:In his argument,https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjF7rzEnpfPAhVMHD4KHQOnDHgQFggeMAA&url=https%3A%2F%2Fwww.researchgate.net%2Ffile.PostFileLoader.html%3Fid%3D5712d7a1615e274be90d8662%26assetKey%3DAS%253A351655620038656%25401460852641147&usg=AFQjCNEqoCrtTMmOxT90KZdo0GaipII_Cg&sig2=66w9oubBjvBGTA5zb14Utw, Dr. G. Lassmann claims to refute David Cole's proof of Collatz Conjecture.
But, he does not understand that David's proof of the Collatz Conjecture is conditional:
Important Notes:
I. 3[tex]n_{m }[/tex] + 1 will always be a multiple of 2 for all positive odd integers, [tex]n_{m }[/tex] or that the
Probability[3[tex]n_{m }[/tex] + 1 is a multiple of 2] = 1.
II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some [tex]n_{0 }[/tex] or that the Collatz sequence never converges to one.
III. [tex]N_{m }[/tex] be the event that [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1 = 3 * f([tex]n_{0 }[/tex]) + 1 is at least a multiple of [tex]2^{2}[/tex] where f([tex]n_{0 }[/tex]) is a function of [tex]n_{0 }[/tex] according to the Collatz Process and according to the definition of [tex]n_{0 }[/tex] in David Cole's proof of the Collatz Conjecture.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 3 * 2 = 6, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is a multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.
So, for the above example, the Probability[3[tex]n_{0 }[/tex] + 1 is at least a multiple of 4] is 1/2.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 5 * 2 = 10, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is a multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 8 | 3[tex]n_{0 }[/tex] + 1 is a multiple of 4] = 1/2.
So, for this latest example, the Probability[3[tex]n_{0 }[/tex] + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.
Consider for example, if 3[tex]n_{0 }[/tex] + 1 = 11 * 2 = 22, then the Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 4 | 3[tex]n_{0 }[/tex] + 1 is a multiple of 2] = 1/2. Why? Because 3[tex]n_{0 }[/tex] + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 8 | 3[tex]n_{0 }[/tex] + 1 is a multiple of 4] = 1/2.
And furthermore, Probability [3[tex]n_{0 }[/tex] + 1 is a multiple of 16 | 3[tex]n_{0 }[/tex] + 1 is multiple of 8] = 1/2.
So, for this latest example, the Probability[3[tex]n_{0 }[/tex] + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.
So, [tex]2^{l_{m }}[/tex] depends on the size of the largest odd factor of [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1, and they, [tex]2^{l_{m }}[/tex] and [tex]l_{m }[/tex] , will vary generally too.
So, the Probability [ 4 divides [tex]n_{m }[/tex] = 3[tex]n_{m-1 }/2^{l_{m-1 }}[/tex]+ 1] varies:
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 1] = 1/2;
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 2] = 3/4;
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = 3] = 7/8;
...
Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] = k] = [tex]\sum_{i=1}^{k }(1/2)^i[/tex];
Therefore, 1/2 [tex]\le[/tex] Probability [ 4 divides [tex]n_{m }[/tex] | [tex]l_{m }[/tex] < [tex]\infty[/tex]] < 1.
Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest wrote:Probability(d = [tex]2^{l_{m}}[/tex]) = Probability([tex]l_{m }[/tex] = m) = [tex]2^{-m}[/tex] over the set, M, of all possible random variables, m, or any positive integer where d is any divisor in the Collatz processing.
Guest wrote:Interesting! You have used the probabilistic method to prove the Collatz Conjecture... How sound is that?
Guest wrote:Guest wrote:Interesting! You have used the probabilistic method to prove the Collatz Conjecture... How sound is that?
FYI: Please read the relevant article on the mathematical concept of, 'Almost Surely', at
https://en.m.wikipedia.org/wiki/Almost_surely.
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