May 15th, 2017, 10:26 AM  #181 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,142 Thanks: 2382 Math Focus: Mainly analysis and algebra  
May 16th, 2017, 12:39 AM  #182 
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 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 
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 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 
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 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 $$x_n=\frac{3^nx_0+3^{n1}+3^{n2}2^{a_1}+3^{n3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n2}a_i}+2^{\sum\limits_{n=1}^{n1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Theorem If $x_n=x_0$, then $x_n=1$. Proof Let $x_n=x_0$, hence $$x_0=\frac{3^nx_0+3^{n1}+3^{n2}2^{a_1}+3^{n3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n2}a_i}+2^{\sum\limits_{n=1}^{n1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Let $\sum\limits_{n=1}^na_i$ be $k$ $$2^kx_03^nx_0=3^{n1}+3^{n2}2^{a_1}+3^{n3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n2}a_i}+2^{\sum\limits_{n=1}^{n1}a_i}$$ $2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$ $$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n\{3\!\sqrt[n]{x_0}\}^n=3^{n1}+3^{n2}2^{a_1}+3^{n3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n2}a_i}+2^{\sum\limits_{n=1}^{n1}a_i}\tag{1}$$ When factoring, $$a^nb^n=(ab)(a^{n1}+a^{n2}b+a^{n3}b^2+\cdots +ab^{n2}+b^{n1})$$ This is same as $$a^nb^n=(ab)(b^{n1}+b^{n2}a+b^{n3}a^2+\cdots +ba^{n2}+a^{n1})$$ 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}\}^{n1}+\{3\!\sqrt[n]{x_0}\}^{n2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n2}\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n1}) \\ \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n2}\{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}\}^{n3}\{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}\}^{n2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}3\!\sqrt[n]{x_0})\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n1}(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^{n1}&=\{3\!\sqrt[n]{x_0}\}^{n1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}3\!\sqrt[n]{x_0})\\ 1&=(\sqrt[n]{x_0})^{n1}\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^{n2}2^{a_1}&=\{3\!\sqrt[n]{x_0}\}^{n2}\{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}^{n2}\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  
Member Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 
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:
 
May 22nd, 2017, 11:24 AM  #186 
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 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^(n1) = something but that something doesnt have to be 3^(n1) * t that t part can also have be a multiple of 3 so you what you are doing is in fact 3^(n1) = 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  
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory  Quote:
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.. $$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n\{3\!\sqrt[n]{x_0}\}^n=3^{n1}+3^{n2}2^{a_1}+3^{n3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n2}a_i}+2^{\sum\limits_{n=1}^{n1}a_i}$$ $$a^nb^n=(ab)(b^{n1}+b^{n2}a+b^{n3}a^2+\cdots +ba^{n2}+a^{n1})$$ if $ab=1$ then $$a^nb^n=b^{n1}+b^{n2}a+b^{n3}a^2+\cdots +ba^{n2}+a^{n1}$$ 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 $$3^{n1}=\{3\!\sqrt[n]{x_0}\}^{n1}$$ 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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