I have the proof to Collatz conjecture.

I have the proof to Collatz conjecture.

Postby Mariga » Wed Mar 22, 2017 3:14 am


Hi people, yes I have the proof. In fact it is quite simple and the technique used can be used to prove other sequences.. I don't know who to trust with this information or where to submit it.
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Re: I have the proof to Collatz conjecture.

Postby Guest » Thu Mar 23, 2017 12:30 pm

Please share your proof...
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Re: I have the proof to Collatz conjecture.

Postby Mariga » Fri Mar 31, 2017 3:08 am

Mariga wrote:Hi people, yes I have the proof. In fact it is quite simple and the technique used can be used to prove other sequences.. I don't know who to trust with this information or where to submit it.


Here it is..


Proof to Collatz conjecture.

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.

Mariga
 
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Re: I have the proof to Collatz conjecture.

Postby Guest » Thu May 11, 2017 1:22 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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Re: I have the proof to Collatz conjecture.

Postby Guest » Thu Sep 28, 2017 4:45 pm

I'm not sure about equating the powers of three.

Your term by term equation implies all the a's are equal, when I think the point is that should be arbitrary.

To put it another way, I think equating terms is not always right.
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Re: I have the proof to Collatz conjecture.

Postby Guest » Mon Jan 22, 2018 8:53 pm

Equating the coefficients the way you do it is not correct, because they depend on the same variable, [tex]x_{0 }[/tex]
You do that way only by looking the powers of 3, but for the entire solution you must do all the sum...
You also have numbers in your expressions that can be irrational, there is no proof of k can be divided by n.

Also assuming x_{0 }=x_{n } is not correct, there can be odd numbers generating a crescent sequence (it was calculated that an ipotetica number with this property if exist, must be extremely large, but this is not a "it doesn't exist"

I do not understand you initial assumption about the N odd and the 1.5n+0.5 with 0.5 being irrelevant as you say. It is a part of the next result, big or small.
Take n=23 and we go to 35, odd so we must repeat the 3n+1/2 operation, that increase the value. With n=21 we obtain 32, a perfect power of 2, so we rapidly go to 1 by the sequence rules.
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Re: I have the proof to Collatz conjecture.

Postby Guest » Sat Sep 29, 2018 1:26 pm

The "proof" isn't correct. People often say that the Collatz conjecture intuitively ought to be true, because half the time you multiple by 1.5 and half the time you divide by 2, so it seems like on average it should be going down. But that isn't a proof. It's just an intuitive argument. And that kind of argument sometimes gives the wrong result.

Here's how to use a similar "proof" to prove something false. Define f(n) to be n+5 if n is a multiple of 10, and n-1 otherwise. You could say that a random n has only a 10% chance of increasing by 5, and a 90% chance of decreasing by 1, and so on average it will decrease. And so this function shouldn't have any cycles at high numbers. But the truth is that EVERY starting number reaches a small cycle!

Why is the "proof" wrong? Because we are not handling the probabilities correctly.

First, you can't say that a random number is odd 50% of the time, unless you first describe how you plan to generate the random number. You have to say what the probability distribution is. And you can't just say "every number is equally likely". That would be a uniform probability distribution over an infinite set, which is mathematically impossible.

Second, even if you could define the probability distribution of n where you're starting, you'd also having to define the distribution after you've iterated the f() function many times. In my example, after iterating f() many times, you're guaranteed to be at a multiple of 10 or within 5 greater than it. Definitely not a number like 107. So the true probability distribution after many steps isn't as simple as it was on the first or second step.

Third, even if you could somehow prove that a random starting number n has a 100% probability of satisfying Collatz, that wouldn't necessarily prove that all numbers satisfy Collatz. This is something weird about probabilities over infinite sets. Just because something is true 100% of the time, that doesn't mean it's true all the time. Yes, that's strange. In math terms, there's a difference between "true with probability one" and "true". A mathematician would say that if you flip a fair coin repeatedly, forever, then you will eventually see heads, with probability 1 (which is a 100% probability), but that you are not guaranteed to ever flip heads. Because it is theoretically possible to flip only tails forever. Flipping tails forever has a probability of zero, but it is not impossible.

TL;DR: probabilities are tricky, which is why the "proof" was incorrect.
Guest
 

Re: I have the proof to Collatz conjecture.

Postby Guest » Sun Sep 30, 2018 10:04 am

Guest wrote:The "proof" isn't correct. People often say that the Collatz conjecture intuitively ought to be true, because half the time you multiple by 1.5 and half the time you divide by 2, so it seems like on average it should be going down. But that isn't a proof. It's just an intuitive argument. And that kind of argument sometimes gives the wrong result.

Here's how to use a similar "proof" to prove something false. Define f(n) to be n+5 if n is a multiple of 10, and n-1 otherwise. You could say that a random n has only a 10% chance of increasing by 5, and a 90% chance of decreasing by 1, and so on average it will decrease. And so this function shouldn't have any cycles at high numbers. But the truth is that EVERY starting number reaches a small cycle!

Why is the "proof" wrong? Because we are not handling the probabilities correctly.

First, you can't say that a random number is odd 50% of the time, unless you first describe how you plan to generate the random number. You have to say what the probability distribution is. And you can't just say "every number is equally likely". That would be a uniform probability distribution over an infinite set, which is mathematically impossible.

Second, even if you could define the probability distribution of n where you're starting, you'd also having to define the distribution after you've iterated the f() function many times. In my example, after iterating f() many times, you're guaranteed to be at a multiple of 10 or within 5 greater than it. Definitely not a number like 107. So the true probability distribution after many steps isn't as simple as it was on the first or second step.

Third, even if you could somehow prove that a random starting number n has a 100% probability of satisfying Collatz, that wouldn't necessarily prove that all numbers satisfy Collatz. This is something weird about probabilities over infinite sets. Just because something is true 100% of the time, that doesn't mean it's true all the time. Yes, that's strange. In math terms, there's a difference between "true with probability one" and "true". A mathematician would say that if you flip a fair coin repeatedly, forever, then you will eventually see heads, with probability 1 (which is a 100% probability), but that you are not guaranteed to ever flip heads. Because it is theoretically possible to flip only tails forever. Flipping tails forever has a probability of zero, but it is not impossible.

TL;DR: probabilities are tricky, which is why the "proof" was incorrect.


Your arguments are non-mathematical, illogical, and fictional. What we need is a sound mathematical argument (proof) or a conterexample which explains why the Collatz conjecture is true or false. You have not prove not disprove the conjecture... Your efforts are fruitless. Please stop.
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