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May 22nd, 2017, 11:18 PM   #191
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Alright then, I will admit I still don't understand your point if my reply doesn't sufficiently answer you.

Let's say we have the following polynomial

$$(a^pr)^n-(br) ^n=b^{n-1}+b^{n-2}a^{x_1}+b^{n-3}a^{x_2}+\cdots +ba^{x_{n-2}}+a^{x_{n-1}}$$
Then the polynomial must be regular. Otherwise the equation wouldn't hold.

we can factor

$$(a^pr)^n-(br)^n=(a^pr-br)(br^{n-1}+br^{n-2}a^p+br^{n-3}a^{2p}+\cdots +bra^{p(n-2)}+a^{p(n-1)}$$

to make the two polynomi(als equivalent, then multiply each term by $(a^pr-br)$
so that
$$(a^pr)^n-(br)^n=br^{n-1}(a^pr-br)+br^{n-2}a^p(a^pr-br)+br^{n-3}a^{2p}(a^pr-br)+\cdots +bra^{p(n-2)}(a^pr-br)+a^{p(n-1)}(a^pr-br)$$

since two polynomials are regular and have equal number of terms, then the first terms in both are equal, so are the rest corresponding terms.
we wont mind if $(a^pr-br)$ is $b^w$.
The fact is, $br^{n-1}=br^{n-1}(a^pr-br)$

Also, if $(a^pr-br)=b^w$ then the $(a^pr)^n-(br)^n$ would not be as it is but would be $(a^pr)^{n+w}-(br)^{n+w}$. That's the only way it would hold.
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May 22nd, 2017, 11:31 PM   #192
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the BIG different is that when you are saying "polynomial"

$$b^{n-1}+b^{n-2}a^{x_1}+b^{n-3}a^{x_2}+\cdots +ba^{x_{n-2}}+a^{x_{n-1}}$$

there is no other $$b^{n-1},b^{n-2},b^{n-3} \cdots $$

this is given to you as a fact!

after it is a given then you can use what you wanted to do

make sure that

$$a^{x_1},a^{x_2}, \cdots +a^{x_{n-2}},a^{x_{n-1}}$$

all are different from b first!

then you can use it
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May 23rd, 2017, 12:24 AM   #193
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Now I get your point. I would give you a thank you but I still don't know how they are given. I will just write it. Thank you.
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May 23rd, 2017, 05:01 AM   #194
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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), for example the first term in polynomial (1) is equal to either the first or last term in polynomial (2). The second term in polynomial (1) is equal to either the second or the second last term in polynomial (2) and so forth.
If the first term in polynomial (1) equals the last term in polynomial (2), then
\begin{align*}
3^{n-1}&=\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
\frac{3^{n-1}}{(2^{\frac{k}{n}})^{n-1}}&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\
(\frac{3}{2^{\frac{k}{n}}})^{n-1}&=x_0(2^{\frac{k}{n}}-3)\\
x_0&=(\frac{3}{2^{\frac{k}{n}}})^{n-1}\frac{1}{(2^{\frac{k}{n}}-3)}
\end{align*}
But $2^{\frac{k}{n}}>3$. Hence, $x_0<1$ in this case.

If the first term in polynomial (1) equals the first term in polynomial (2), then
\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 $x_0=1$.
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May 24th, 2017, 02:35 AM   #195
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Quote:
Originally Posted by Mariga View Post
Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2), for example the first term in polynomial (1) is equal to either the first or last term in polynomial (2). The second term in polynomial (1) is equal to either the second or the second last term in polynomial (2) and so forth.
If the first term in polynomial (1) equals the last term in polynomial (2)
It's the same mistake.

You need to check all of them, not 2.

You have infinitely many checks to do.

Last edited by skipjack; July 25th, 2017 at 04:32 AM.
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May 28th, 2017, 10:00 AM   #196
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Indeed. That step also didn't seem quite right to me too. It didn't make the proof completely rigorous. But I have a way around it. coming soon
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May 29th, 2017, 01:55 AM   #197
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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, the number of terms in the two polynomials is equal. The first term in the polynomial in equation (1) is equal to any term in the polynomial in equation (2). Each term in polynomial 2 can be defined by the following,
$$3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$
Where $r=1,2,3,...,n$
Hence,
\begin{align*}
3^{n-1}&=3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\
3^{n-1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\
x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}\\
x_0&=\{\frac{3}{2^{\frac{k}{n}}}\}^{r-1}\frac{1}{(2^{\frac{k}{n}}-3)}
\end{align*}
But $2^{\frac{k}{n}}>3$. Hence, $x_0<1$ for $r>1$.
When $r=1$,
\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 term in the polynomial in equation (1), is equal to any term in polynomial in equation (2), except $\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$. Therefore,
$$3^{n-2}2^{a_1}=\{3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$
Such that $r=2,3,4,...,n$
\begin{align*}
3^{n-2}2^{a_1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\
2^{a_1}&=\frac{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)}{3^{n-2}}
\end{align*}
We substitute (3)
\begin{align*}
2^{a_1}&=\frac{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}\frac{1}{(2^{\frac{k}{n}}-3)}(2^{\frac{k}{n}}-3)}{3^{n-2}}\\
2^{a_1}&=3^{2-r}\{2^{\frac{k}{n}}\}^{r-1}\\
\end{align*}
But $a_1$ is an integer. Therefore $r=2$ to eliminate $3^{2-r}$. Hence,
$$2^{a_1}=2^{\frac{k}{n}}$$
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 $x_0=1$.
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July 25th, 2017, 01:13 AM   #198
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I have found an error..

The statement
$2^{\frac{k}{n}}>3$. This implies $x_0<1$ for $r>1$.

is incorrect. $2^{\frac{k}{n}}$ can be greater than 3 but less than 4. Therefore $x_0$ can be greater than 1 for $r>1$.
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July 25th, 2017, 09:24 PM   #199
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Also, I will soon give the complete algebraic proof of collatz conjecture that doesn't rely on a probabilistic model.
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