My Math Forum Proof to Collatz conjecture.

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May 15th, 2017, 10:26 AM   #181
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Quote:
 Originally Posted by Mariga I realised I might need to prove that for $4x_1-3x_0=1$, why the only values that could satisfy the equation is 1 for both $x_1$ and $x_0$.
$x_1=7,x_0=9$

 May 16th, 2017, 12:39 AM #182 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory I indeed I made an error on that statement. Probably I was a little stuck and frustrated by the need to prove everything which I later realised was necessary. I couldn't have meant if $x_0=x_1$ because that would have been easy to do. The latter proof is solid.
 May 19th, 2017, 03:51 AM #183 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Mathematics might not be my destiny. This is just one of those things that I have done as part and process of my self discovery. But I hope that one day I shall be remembered as the man who proved Collatz conjecture. I have an arxiv account but no endorsers. Any endorser can help in publishing this and include their name as a collaborator. Send me a private message on my profile. Also I can post the full proof from which more editing can be done.
 May 22nd, 2017, 02:32 AM #184 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory 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 $$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Theorem If $x_n=x_0$, then $x_n=1$. Proof 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 $k$ $$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$ $2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$ $$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$ When factoring, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$ This is same as $$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$ We therefore notice equation (1) holds iff the polynomial is regular. We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$. \begin{align*} \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}) \\ \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\ &\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0}))\tag{2} \end{align*} Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2). Hence \begin{align*} 3^{n-1}&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 1&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\ x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{3} \end{align*} The next pair of equal terms, \begin{align*} 3^{n-2}2^{a_1}&=\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 2^{a_1}&=\sqrt[n]{x_0}^{n-2}\sqrt[n]{x_0}\sqrt[n]{x_0}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\ 2^{a_1}&=x_02^{\frac{k}{n}}(2^{\frac{k}{n}}-3) \end{align*} Substituting (3) \begin{align*} 2^{a_1}&=\frac{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 $2^{a_1}$ in equation (3), $$x_0=\frac{1}{(2^{a_1}-3)}$$ The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.
May 22nd, 2017, 05:10 AM   #185
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sorry but this is the first time i looked at your worked and i only read the last post which it seem to have all the proof you need to read so i didnt read all the rest of your posts

i think you have a mistake

Quote:
 Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2). Hence
why?

 May 22nd, 2017, 11:24 AM #186 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory What specifically isn't clear? I have shown that the polynomial in equation 1 is regular and hence expressed equation 1 in a different way.. as shown in equation 2 such that each terms are equal. take this as an example.. lets say $$z=3x^4+3x^3+3$$ and $$z=ax^4+ax^3+a$$ isn't it obvious that a=3?
 May 22nd, 2017, 09:52 PM #187 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 what dont you understand? there could be a case where the "next pair" can add to a diffrent multiple of 3 you wrote 3^(n-1) = something but that something doesnt have to be 3^(n-1) * t that t part can also have be a multiple of 3 so you what you are doing is in fact 3^(n-1) = 3^(n+r) * d same goes for all the rest Last edited by isaac; May 22nd, 2017 at 10:42 PM.
 May 22nd, 2017, 10:05 PM #188 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 if you want somthing that will help you check try to test this for negative cycles positive cycles: (1,4,2) negative cycles: (-1,-2) (-5,-14,-7,-20,-10) (−17,−50,−25,−74,−37,−110,−55,−16 4 ,−82,−41,−122,−61,−182,−91,−272,−1 36,−68,−34) Last edited by isaac; May 22nd, 2017 at 10:43 PM.
May 22nd, 2017, 10:42 PM   #189
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 Originally Posted by isaac what dont you understand? there could be a case where the "next pair" can add to a diffrent multiple of 3 you wrote 3^(n-1) = something but that something doesnt have to be 3^n(n-1)*t that t part can also have be a multiple of 3 so you what you are doing is in fact 3^(n-1) = 3^(n+r)*d same goes for all the rest
If the t part is a multiple of 3 then it would be a multiple of 3 for the rest of the terms since it is a constant for each term and hence the first term, which would have the highest power, would still be equal to the first term in the series.

Also still, it cannot be a multiple of 3. the equation would not hold. see how the factoring rules applies.

If this is still not clear for you then we can try this other method..

$$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$

$$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$
if $a-b=1$ then
$$a^n-b^n=b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1}$$

So it is clear that since the polynomial in our example is regular, then
$$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=1$$
and
$$3^{n-1}=\{3\!\sqrt[n]{x_0}\}^{n-1}$$

Also, testing this for negative cycles gives cycles for those integers you have shown.

Last edited by Mariga; May 22nd, 2017 at 10:53 PM.

 May 22nd, 2017, 11:05 PM #190 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 please check again until you will get what i wrote i dont think you understand what i wrote .... i told you where is the problem in your proof ... check that transition again

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