Logarithmic identity

nsato

Logarithmic identity

**Posted:** Tue Mar 27, 2007 6:54 am

Rust

Let
$x=\frac{2}{2^{1/n}-1}=\frac{2}{e^{ln2/n}-1}=\frac{2n}{ln2}\frac{1}{1+(1/2!)(2y)+(1/3!)(2y)^{2}+(1/4!)(2y)^{3}+...}, y=\frac{l...$ .
Then
$1>z=\frac{1}{1+y+(2/3)y^{2}+(1/3)y^{3}+..}>\frac{1}{\frac{1}{1-y}}=1-y.$
Therefore $\frac{2n}{ln2}-1<x<\frac{2n}{ln2}.$

**Posted:** Tue Mar 27, 2007 7:23 am

Rust

#3

I prove only $[x+1]\ge [1/y]$ . But I don't chek, that can not be
(*) $[x]=[\frac{1}{y}]$ .
Exactly $\frac{1}{y}-1+\frac{1}{3}y-y^{3}<x<\frac{1}{y}-1+\frac{1}{3}y$ . Therefore if (*), then $n=P_{2k}$ .
Let $\frac{P_{k}}{Q_{k}}$ is good rational approach for $\frac{ln2}{2}$ .
We have $P_{1}=1,Q_{1}=2$ and $\frac{P_{2k}}{Q_{2k}}<\frac{ln2}{2}<\frac{P_{2k+1}}{Q_{2k+1}}$ . If $n=P_{2k}$ we get
$Q_{2k}-\frac{1}{P_{2k+1}}<\frac{1}{y}<Q_{2k}$ . Because x and y irrational, $Q_{2k}-\frac{1}{P_{2k+1}}<\frac{1}{y}<Q_{2k}-1$ and $x>Q_{2k}-1$ if $P_{2k+1}>3Q_{2k}$ .
When $P_{2k+1}>3Q_{2k}$ your statement is not true for $n=P_{2k}$ .
If $q_{2k+1}\ge 9$ , then $P_{2k+1}>3Q_{2k}$ and $[x]=[\frac{1}{y}]$ ,
were $\frac{P_{2k+1}}{Q_{2k+1}}=[0;q_{1},q_{2},...,q_{2k+1}]$ .

**Posted:** Tue Mar 27, 2007 9:52 am

maxal · Riemann Hypothesis Offline Joined: 09 Apr 2005 Posts: 369

#4

A counterexample is $n=777451915729368,$ for which
$\left \lceil \frac{2}{2^{1/n}-1}\right \rceil = \left \lfloor \frac{2n}{\log 2}\right \rfloor+1.$

**Posted:** Tue Mar 27, 2007 1:48 pm

maxal · Riemann Hypothesis Offline Joined: 09 Apr 2005 Posts: 369

#5

Further examples are listed in A129935 @ OEIS.

**Posted:** Fri Jun 08, 2007 9:47 pm