# Do I understand the concept of $x^{0.84}$ correctly?

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?

• What do you mean with the "concept of $x^{0.84}$? – Wojowu Sep 14 '19 at 22:23
• @Wojowu English is not my native language. I mean Lagarias has wrote it in your book. – Elementary Sep 14 '19 at 22:25
• what do you mean by goes to 1 – Saketh Malyala Sep 14 '19 at 22:26
• @SakethMalyala Do you know about Collatz Conjecture? – Elementary Sep 14 '19 at 22:27
• @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. – benrg Sep 14 '19 at 22:27

(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.

• Thank you answer and edit the question. Did you read Terence Tao`s partial proof? can I ask? – Elementary Sep 15 '19 at 9:20
• @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... :-( – Gottfried Helms Sep 15 '19 at 9:26
• @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. – Gottfried Helms Sep 15 '19 at 9:56
• I agree...It is important to be happy on all Sunday mornings. A shaken/sad/heartbroken man can not do mathematics.. – Elementary Sep 15 '19 at 9:58
• Yes, we can draw a conclusion from the discussions. – Elementary Sep 15 '19 at 10:01

$$x^{0.84} = x^\frac{21}{25} = (x^{21})^\frac{1}{25}$$, which is the unique solution $$y$$ of $$y^{25} = x^{21}$$. For irrational $$y$$, $$x^y = \sup_{z\in\mathbb{Q}\cap(-\infty,y)}x^z = \inf_{z\in\mathbb{Q}\cap(y,\infty)} x^z$$.

• Is this answer about the collatz conjecture? – Elementary Sep 14 '19 at 22:32