Proof to Collatz conjecture.

Proof to Collatz conjecture.

Postby Mariga » Fri Mar 31, 2017 3:12 am

Consider any positive integer n from which the sequence is formed. n has a probability of 0.5 of either being odd or even. If even, we divide it by two. If odd, we multiply it by three, add one and then divide the result by two since the resulting number must be even. This is same as multiplying n by 1.5 and adding 0.5. The resulting integer, say m, will hence either be n/2 or 1.5n+0.5 and also has a probability of 0.5 of being either even or odd. The 0.5 that is added has considerable effect on outcome only if n is 1 (which can explain the repeating cycle when the sequence reaches 1). In this case we will neglect it.
Since n has equal chances of being odd or even, it therefore has equal chances of being divided by 2 or being multiplied by 1.5, and so is m and the rest of the outcomes. The factor by which the outcomes are divided by is greater than the factor they are multiplied with and hence, the sequence will converge.





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Re: Proof to Collatz conjecture.

Postby Guest » Wed Apr 05, 2017 6:33 pm

You have to prove that 3x+1 is always even :D

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Re: Proof to Collatz conjecture.

Postby Guest » Sat Apr 08, 2017 4:14 am

That part of the "proof" is actually correct. If x is odd it can be written as 2y+1 (for some integer y) so 3x+1 becomes 3(2y+1)+1 = 6y+4 = 2(3y+2), which is obviously even.

The final line of the proof is wrong. Firstly proving things about how things work on average is not the same as proving that something can never happen. For example if I continually flip a coin the percentage of flips that land on heads will converge to 50%, that doesn't mean it is impossible for me to flip all tails (and get an observed percentage of 0%). Secondly even if at each stage you multiply by 1.5 or 0.5 with equal probability, the expected value doesn't go down to 0, it actually stays the same as what you started with (although in lots of cases you will end up slightly below the original number you started with, in a few cases will end up with a number much bigger than what you started with, so they balance out).

Anything to do with the collatz conjecture, riemann hypothesis, anything by David Cole (aka primework123), and any posts claiming to have discovered an amazing new result (or disproof of an existing result) should be ignored. No legitimate amazing new discovery will ever appear on this forum or similar forums. The result should be ignored, unless it appears in a reputable peer reviewed journal (there are plenty of non-peer reviewed journals which will print anything and should be ignored) or has the legitimate backing of a reputable researcher employed in a reputable well known university (and never take the poster's word that this is the case).

Check out
https://www.quora.com/Why-do-math-crank ... rish-exist
and David Joyce's answer (which explains that some people are just misguided, whilst others are blind to their mistakes)

It is best to ignore posts like these, rather than engage the author's in a pointless discussion about why they're wrong, neither side will convince the other. If I'm totally honest, the site admin (aka Math Tutor) should be removing these posts, as I've seen people on this site genuinely believing what's being claimed.

The only reason I'm responding to this, is because this is probably going to be my last post on this site (so if the original poster, or David Cole wants to argue with me, feel free, I won't be here).

I don't really get any benefit out of posting, so it is starting to feel like I'm working for the site without getting paid, so it is time for me to go.

Hope this helped,

R. Baber.

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Re: Proof to Collatz conjecture.

Postby Guest » Sat Apr 08, 2017 10:23 am

The great philosopher, Aristotle (https://en.wikipedia.org/wiki/Aristotle), had warned us about the danger of always appealing to 'authorities or experts' for answers to important questions. I ask the reader to think for himself or herself concerning my work posted here and elsewhere, and to make good decisions too. I expect and accept critical review of my work. But it is unfair to make an unfounded personal attack against me ... I have relevant research experience, and I have university degrees in mathematics and in electrical engineering too.

Sincerely,

David Cole (author of several works posted on math10.com)

Reference link:
https://www.researchgate.net/profile/David_Cole29.

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Re: Proof to Collatz conjecture.

Postby MM » Wed Apr 12, 2017 11:21 am

To R. Baber - I would like to say that your departure of the site will be a great loss for it, and for the Mathematics in it. You are one of the last (if not the last) people in this site who are competent and interested in answering questions. Giving interesting solutions and points of view. I hope you reconsider your decision!

Best Regards

S. Apostolov

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Re: Proof to Collatz conjecture.

Postby Guest » Sun Apr 23, 2017 10:20 pm

Baber there is something sincerely wrong with you I guess. You and I both know that the real proof can be built from this work. You must be hiding your insecurities. Either way, this is on so many sites, so even if it gets deleted here, you won't escape plagiarism.

Would you like me to answer your questions or they just wouldn't make any sense?

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Re: Proof to Collatz conjecture.

Postby Mariga » Thu May 11, 2017 1:31 pm

Let $x_n$ be an odd positive integer. From the sequence’s formula, $x_{n+1}=\frac{3x_n+1}{2^k}$, $x_{n+2}=\frac{3x_{n+1}+1}{2^m}$ and so forth, $m,k\in \mathbb{Z^+}$\\
For there to exist a cycle in the sequence, there must exist an odd integer $x_0$ such that $x_n=x_0$.\\
Let $x_1$, the next odd integer after $x_0$,be given by
$$x_1=\frac{3x_0+1}{2^{a_1}}$$
$x_2$ will be given by
$$x_2=\frac{3x_1+1}{2^{a_2}}$$
$x_2$ in terms of $x_0$ will be
\begin{align*}
x_2&=\frac{3(\frac{3x_0+1}{2^{a_1}})+1}{2^{a_2}}\\
&=\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}}
\end{align*}
$x_3$ will be given by
\begin{align*}
x_3&=\frac{3x_2+1}{2^{a_3}}\\
&=\frac{3(\frac{3x_1+1}{2^{a_2}})+1}{2^{a_3}}\\
&=\frac{3(\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}})+1}{2^{a_3}}\\
&=\frac{27x_0+9+3.2^{a_1}+2^{a_1+a_2}}{2^{a_1+a_2+a_3}}
\end{align*}
From the three examples,we can generate a formula for $x_n$ in terms of $x_0$,which will be
$$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $x_n=x_0$, hence
$$x_0=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $\sum\limits_{n=1}^na_i$ be $k$
$$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$$x_0(2^k-3^n)=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$2^k$ can be expressed as $(2^{\frac{k}{n}})^n$
$$x_0[(2^{\frac{k}{n}})^n-3^n]=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$
When factoring,
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +a.b^{n-2}+b^{n-1})$$
This is same as
$$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +b.a^{n-2}+a^{n-1})$$
We therefore notice the polynomial in equation (1) is regular.
Hence, $x_0[(2^{\frac{k}{n}})^n-3^n]$ can be expressed as,
$$x_0[(2^{\frac{k}{n}})^n-3^n]=x_0(2^{\frac{k}{n}}-3)(3^{n-1}+3^{n-2}2^{\frac{k}{n}}+3^{n-3}(2^{\frac{k}{n}})^2+\cdots +3(2^{\frac{k}{n}})^{n-2}+(2^{\frac{k}{n}})^{n-1})$$
$$=3^{n-1}(2^{\frac{k}{n}}-3)x_0+3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)x_0+3^{n-3}(2^{\frac{k}{n}})^2(2^{\frac{k}{n}}-3)x_0+\cdots +3(2^{\frac{k}{n}})^{n-2}(2^{\frac{k}{n}}-3)x_0+(2^{\frac{k}{n}})^{n-1})(2^{\frac{k}{n}}-3)x_0$$
\\
Each term in this polynomial corresponds to the terms in the polynomial in equation (1).
\begin{align*}
3^{n-1}\quad &:\quad 3^{n-1}(2^{\frac{k}{n}}-3)x_0\\
3^{n-2}2^{a_1}\quad &:\quad 3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)x_0\\
3^{n-3}2^{a_1+a_2}\quad &:\quad 3^{n-3}(2^{\frac{k}{n}})^2(2^{\frac{k}{n}}-3)x_0\\
\vdots\qquad &:\qquad \vdots \\
3.2^{\sum\limits_{n=1}^{n-2}a_i}\quad &:\quad 3(2^{\frac{k}{n}})^{n-2}(2^{\frac{k}{n}}-3)x_0\\
2^{\sum\limits_{n=1}^{n-1}a_i}\quad &:\quad (2^{\frac{k}{n}})^{n-1})(2^{\frac{k}{n}}-3)x_0
\end{align*}
Hence, each of these corresponding terms will be equal. Hence,
\begin{align*}
3^{n-1}&=3^{n-1}(2^{\frac{k}{n}}-3)x_0\\
1&=(2^{\frac{k}{n}}-3)x_0\\
x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{2}
\end{align*}
Substituting this in the next pair of terms,
\begin{align*}
3^{n-2}2^{a_1}&=3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\frac{1}{(2^{\frac{k}{n}}-3)}\\
2^{a_1}(2^{\frac{k}{n}}-3)&=2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\
2^{a_1}&=2^{\frac{k}{n}}
\end{align*}
Substituting this in equation (2),
$$x_0=\frac{1}{(2^{a_1}-3)}$$
The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.

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