20170227, 14:39  #1 
Apr 2014
Marlow, UK
2^{3}×7 Posts 
Finding multiples of a real number that are close to a whole number
Given a real number r, and a small positive value e arbitrarily close to 0, does anyone know of a fast way to find integer multipliers m such that either:
{mr} < eor 1  {mr} < e(where {mr} is the fractional part of mr) Thanks in advance for any help, Mick. 
20170227, 15:08  #2 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{2}·3^{2}·179 Posts 
The search term you want is "continued fractions"

20170227, 15:20  #3 
Apr 2014
Marlow, UK
2^{3}×7 Posts 

20170227, 15:33  #4  
(loop (#_fork))
Feb 2006
Cambridge, England
14454_{8} Posts 
Quote:
Code:
default(realprecision,500) p=Pi() M=contfracpnqn(contfrac(p,21)) M[2,1]*p Last fiddled with by fivemack on 20170227 at 15:33 

20170227, 15:51  #5  
Apr 2014
Marlow, UK
38_{16} Posts 
Quote:
Regards, Mick. 

20170227, 16:46  #6 
Feb 2017
Nowhere
19·269 Posts 
Given a real number r, and a small positive value e arbitrarily close to 0,
It depends, of course, on how you are given the real number. A numerical approximation to some prescribed number of decimal or binary digits, would be amenable to conversion to a "simple continued fraction" (an alternative search term). Also pursuant to fivemack's mention of the infinite SCF whose partial quotients are all 1, I mention Roth's Theorem which shows that algebraic numbers are "almost always" hard to approximate very well by rational numbers. It is generally difficult to know both numerical and SCF representations of a given number. See, however, the following paper. Last fiddled with by Dr Sardonicus on 20170227 at 16:50 
20170227, 21:42  #7 
Apr 2014
Marlow, UK
38_{16} Posts 
Changing the question...
Having given some thought to this, I realise that what I really want is a way to find integer values for m such that
where s <= 1. I'm guessing this is a different kettle of fish... Last fiddled with by mickfrancis on 20170227 at 21:47 
20170227, 21:58  #8  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3·181 Posts 
Quote:
if r=0.5 and e=0.25 How can there be an integer m where m*0.5<0.25? 

20170227, 22:16  #9 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3·181 Posts 
If you are looking for an algebraic equivalence to ceil or floor function, then I can assure you that no such equivalence exists. I wasted years trying to find one, only to realize it is equivalent to a general and complete formula for primes. That means that if such an algebraic function could exist, then it could be used to reduce the series function consisting of floor()s to get a complete prime numbers formula (which would generate all prime numbers as function of n).
Last fiddled with by a1call on 20170227 at 22:27 
20170227, 22:27  #10  
Feb 2017
Nowhere
5111_{10} Posts 
Quote:
 N/m  r < c/m^(2  s) (c = positive constant), which is more easily achievable than having the exponent 2 in the denominator. There is a notion of "best rational approximations," i.e. fractional approximations which are closer than any fraction with a smaller denominator. These consist of the "intermediate convergents" for the SCF. I know it's somewhere in Chrystal's Textbook of Algebra, but I'm sure something can be found on line. 

20170227, 23:34  #11 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
well the equivalent inequalities depend on a few things if m and r are the same sign then all values are positive ( because a negative times a negative is a positive) if they are different sign then mr is negative and with s<=1 it could be an inequality where you ask if a real value is less than an imaginary or complex number. so if we don't want to ask that question it may be safer to stay with the m values that are of the same sign as r. also if mr were negative then in theory you get floor(abs(mr)) as an equivalent edit: to ceil
Last fiddled with by science_man_88 on 20170227 at 23:46 
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