From: "Robin Chapman" Subject: Re: Rational approximation Date: Fri, 14 Dec 2001 13:32:33 +0000 (UTC) Newsgroups: sci.math Summary: How closely can real numbers be approximated by rationals? "Robert C. Helling" wrote in message news:slrna1jq0e.uj.helling@ariel.aei-potsdam.mpg.de... > I know that for irrational a there are infinitely many integer p and q such > that |a - p/q| < q^-2 > and that in general you cannot do better than the exponent -2. > > But can you do better in the following sense: Can I include an additional > integer N such that there is a solution (p,q) to > |a - p/q| < q^-2 / N ? > > What I am really interested in is: For given a, what is > > inf{ |pq - a q^2| for p and q non-zero integers} , > > are there irrational a for which it is larger than 0? The "worst" approximable real is the Golden ratio tau = (1 + sqrt(5))/2. The best you can do with that is p/q = 2 giving |pq - tau q^2| = 0.3819... The other good approximations are F_{n+1}/F_n for F_n Fibonacci numbers. They give |F_{n+1} F_n - tau F_n^2| -> 1/sqrt(5) as n -> infinity. If k < 1/sqrt(5) there are only finitely many p/q with |pq - tau q^2| < k. Similar is true for each quadratic irrational a. There is a K > 0 such that for each k < K there are only finitely many p/q with |pq - aq^2| < k. Robin Chapman -- Posted from olive.ulcc.wwwcache.ja.net [194.82.103.41] via Mailgate.ORG Server - http://www.Mailgate.ORG ============================================================================== From: stern@rowland.org (Alan Stern) Subject: Re: Rational approximation Date: 14 Dec 2001 09:31:36 -0800 Newsgroups: sci.math helling@ariel.physik.hu-berlin.de (Robert C. Helling) wrote in message news:... [quote of original message deleted --djr] Yes. Take, for example, a = sqrt(2). For positive integers p and q, we know that p^2 - 2 q^2 is non-zero. Suppose it is positive (the argument for when it is negative is similar), then it must be >= 1 since it is an integer. We get p^2 >= 2 q^2 + 1 and hence p >= sqrt(2) q + 1/(4 sqrt(2) q). You can check that by squaring the right-hand side; it yields 2 q^2 + (1/2) + 1/(32 q^2) < 2 q^2 + 1. Now multiply through by q to get pq - sqrt(2) q^2 >= 1/(4 sqrt(2)). No doubt this bound can be improved a little, but you only asked whether it was larger than 0. Alan Stern ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Rational approximation Date: 17 Dec 2001 23:04:32 GMT Newsgroups: sci.math In article , Robert C. Helling wrote: >I know that for irrational a there are infinitely many integer p and q such >that |a - p/q| < q^-2 >and that in general you cannot do better than the exponent -2. > >But can you do better in the following sense: Can I include an additional >integer N such that there is a solution (p,q) to >|a - p/q| < q^-2 / N ? For any irrational a there are infinitely many p and q such that |a - p/q| < q^(-2)/sqrt(5) and sqrt(5) is best possible. >What I am really interested in is: For given a, what is > >inf{ |pq - a q^2| for p and q non-zero integers} , > >are there irrational a for which it is larger than 0? This is greater than 0 if and only if the continued fraction representation of a has bounded elements. See A. Ya. Khinchin, "Continued Fractions", U. of Chicago Press 1964, Theorem 23. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Rational approximations of pi Date: 17 Dec 2001 02:13:48 GMT Newsgroups: sci.math In article , Axel Harvey wrote: >> Robert Israel wrote: >> >> > bound on how well pi can be approximated by rationals. By a result of >> > K. Mahler, |pi - p/q| > 1/q^42 for all integers p,q.... >Here's a supplementary question anyway... Obviously if q = 2 we don't >get anywhere close to a 2^(-42) discrepancy. The above result therefore >suggests that |pi - p/q| > 1/q^(some function of q) . >Or at any rate, something that cuts closer to reality for small q. Is >an inequality of this sort known? If pi is a "typical" real number, then for any epsilon > 0, |pi - p/q| > 1/q^(2+epsilon) with at most finitely many exceptions. There is no reason to believe that pi is not "typical" in this sense. On the other hand there are certainly infinitely many p/q with |pi - p/q| < 1/q^2. M. Hata showed |pi - p/q| > 1/q^8.016045... for sufficiently large q, but since I don't know how large is "sufficiently large" here I used Mahler's earlier result, which works for every q. It is extremely likely that Hata's result is true for all positive integers q. The least value of log_q(1/|pi - p/q|) that I could find is approximately 3.4293 for p/q = 22/7. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2