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I'm trying to understand the concept of $x^{0.84}$ that Jeffrey Lagarias found for Collatz Conjecture. If I'm wrong, please correct me with an answer. I understand such that,

Suppose the interval $\left[1, 2^{1\,000\,000} \right]$ is given. In this interval we have at least $\lfloor{(2^{1\,000\,000}})^{0.84}\rfloor=2^{840\,000}$ natural numbers which is goes to $1$. In other words, we can choose at least such $2^{840\,000}$ natural numbers from this sequence $\left\{1,2,3,4,5,6,7,8,9\cdots 2^{1\,000\,000} \right\}$, which goes to $1$. Or, there are at least $2^{840\,000}$ natural numbers in this interval $\left[1, 2^{1\,000\,000} \right]$ (But, we don't know exactly what the numbers are ?), which gives result $1$. So, Collatz Conjecture is correct for at least $2^{840\,000}$ natural numbers. Is my understanding correct?

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    $\begingroup$ What do you mean with the "concept of $x^{0.84}$? $\endgroup$ – Wojowu Sep 14 '19 at 22:23
  • $\begingroup$ @Wojowu English is not my native language. I mean Lagarias has wrote it in your book. $\endgroup$ – Elementary Sep 14 '19 at 22:25
  • $\begingroup$ what do you mean by goes to 1 $\endgroup$ – Saketh Malyala Sep 14 '19 at 22:26
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    $\begingroup$ @SakethMalyala Do you know about Collatz Conjecture? $\endgroup$ – Elementary Sep 14 '19 at 22:27
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    $\begingroup$ @Wojowu I guess it's a reference to Bounds for the 3x+1 Problem using Difference Inequalities by Krasikov and Lagarias, which proves that at least x^0.84 of the integers (er, naturals) less than x eventually go to 1. $\endgroup$ – benrg Sep 14 '19 at 22:27
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(OP:)... So, Collatz Conjecture is correct for at least $2^{840\,000}$ natural numbers. Is my understanding correct?

As I've read Lagarias & Krasikov your take is correct. However in your sentence which I've cited here you should have included the clause:

... So, Collatz Conjecture is correct for at least $2^{840\,000}$ natural numbers below $2^{1\,000\,000}$. (...)

Otherwise you come in conflict with the property, that even infinitely many numbers are known to converge to $1$, for instance from the infinite sequence $\{1,2,4,8,\cdots,2^k, \cdots\}$ or from the infinite sequence $\{1,5,21,85, \cdots , {4^k-1 \over 3} , \cdots \}$ or from infinitely many infinite sequences generated in this style.

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  • $\begingroup$ Thank you answer and edit the question. Did you read Terence Tao`s partial proof? can I ask? $\endgroup$ – Elementary Sep 15 '19 at 9:20
  • $\begingroup$ @Elementary - yes, you can ask. On a nice sunny sunday morning as it is today here: I'm a really calm and happy human... I've read the T. Tao's proof but it is way over my head, I even did not grasp yet what he's talking about at all... :-( $\endgroup$ – Gottfried Helms Sep 15 '19 at 9:26
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    $\begingroup$ @Elementary - the arxiv-text was impossible for me to grasp. But the conversation on T.Tao's blog terrytao.wordpress.com/2019/09/10/… has a much enlightening factor. $\endgroup$ – Gottfried Helms Sep 15 '19 at 9:56
  • $\begingroup$ I agree...It is important to be happy on all Sunday mornings. A shaken/sad/heartbroken man can not do mathematics.. $\endgroup$ – Elementary Sep 15 '19 at 9:58
  • $\begingroup$ Yes, we can draw a conclusion from the discussions. $\endgroup$ – Elementary Sep 15 '19 at 10:01
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$x^{0.84} = x^\frac{21}{25} = (x^{21})^\frac{1}{25}$, which is the unique solution $y$ of $y^{25} = x^{21}$. For irrational $y$, $x^y = \sup_{z\in\mathbb{Q}\cap(-\infty,y)}x^z = \inf_{z\in\mathbb{Q}\cap(y,\infty)} x^z$.

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    $\begingroup$ Is this answer about the collatz conjecture? $\endgroup$ – Elementary Sep 14 '19 at 22:32

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