Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$x \in \mathbb{R}$

$2^{500}<x<2^{501} $

How many significant figures are needed in base 2, to know in high approximation whether $2^x$ is integer?

share|improve this question
    
What does it mean to "know in high approximation whether $2^x$ is integer"? Whether $2^x$ is integer is either true or false, and supposing you don't consider "true" to be a high approximation of "false" or vice versa, you need to know exactly. Also, unless you know $x$ exactly (or you know for instance that $x$ is integer), you will never be able to tell for sure that $2^x$ is integer. –  Marc van Leeuwen Apr 18 '12 at 9:48
    
@MarcvanLeeuwen, I mean that $\exists n \in \mathbb{N}, 2^x - 2^{-500}<n<2^x + 2^{-500}$ –  Must Apr 18 '12 at 10:51
    
What is the motivation? –  lhf Apr 18 '12 at 13:30
    
@lhf, To know how many bit I will need to calculate of x. –  Must Apr 18 '12 at 14:57
1  
Question is very close to a cross-post on scicomp.SE. –  Geoff Oxberry Apr 18 '12 at 16:48

2 Answers 2

up vote 5 down vote accepted

Basically you want to estimate the $\delta$ such that $2^{x+\delta} - 2^x = 1$. This means $2^\delta = 1+\dfrac{1}{2^x}$, so that $\delta$ might be estimated as $\dfrac{1}{2^x \times \ln2}$, or, taking the upper bound for $x$, $\delta$ might be estimated as $\dfrac{1}{2^{501} \times \ln2}$.

The number of required significant digits (after the decimal point) is about $-\log_{10}\delta = \log_{10}(2^{501}) + \log_{10}\ln2$, which is about 151, plus-minus a digit. Or, if you're working in base 2, the number of required significant digits is about $-\log_{2}\delta = \log_{2}(2^{501}) + \log_{2}\ln2$, which is about 502.

Given such a number of digits after the decimal point, changing the remaining digits won't change $2^x$ by more than $1$, so that, if $2^x$ is an integer, you can say what integer it is.

However, it is impossible to tell for sure whether $2^x$ is an integer, given only its rounded value, independent of the accuracy, as adding a small value beyond the accuracy limits to $x$ will turn $2^x$ from integer to non-integer and vice versa.

share|improve this answer
    
$\drfac{x}{y}$ doesn't seem to work... –  penartur Apr 18 '12 at 9:01
    
Of course that \drfac would not work, but I was hoping that the obvious typo will be obvious to you as well (at least now when I re-read my comment and notice that typo). –  Asaf Karagila Apr 18 '12 at 9:16
    
You might want to consider using \dfrac instead of \frac because that would make the fraction somewhat more readable. –  Asaf Karagila Apr 18 '12 at 9:17
    
I'm just not a TeX expert :) –  penartur Apr 18 '12 at 11:56
    
$\dfrac{1}{2^x \times \ln2} \neq \dfrac{1}{2^{501} \times \ln2}$ –  Must Apr 20 '12 at 11:04

For $2^x$ to be an integer where x has a finite decimal representation, x must also be an integer. Am I missing something here?

share|improve this answer
3  
This is not correct... any positive integer has a logarithm to the base 2, which need not be an integer. –  Ted Apr 18 '12 at 8:53
    
Yes, but it will in general be an irrational number and you will not be able to distinguish it from a rational number in a finite decimal representation. –  Wonder Apr 18 '12 at 8:56
    
+1: @Wonder should edit his post to reflect the finite representation problem. –  Bill Barth Apr 18 '12 at 17:00
    
Ok I edited it accordingly. –  Wonder Apr 18 '12 at 17:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.