[math-fun] number theory: continued fractions with constraints
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint. In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral. Suppose I used a continued fraction process to generate better & better approximations. Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ? I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral. So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions? Or perhaps this situation is exceedingly rare. If so, how rare is it?
You should look at semi convergents. Look at http://shreevatsa.wordpress.com/2011/01/10/not-all-best-rational-approximati... for a good account. Victor Sent from my iPhone On Aug 14, 2013, at 10:23, Henry Baker <hbaker1@pipeline.com> wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
Suppose I used a continued fraction process to generate better & better approximations.
Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ?
I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral.
So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions?
Or perhaps this situation is exceedingly rare.
If so, how rare is it?
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Thanks very much! This excellent draft paper by Cortzen is referenced in the discussion mentioned by Victor: From: http://lanco.host22.com/web_documents/continfrac.ver.4.pdf Size: 247 KB (252,129 bytes) At 10:40 AM 8/14/2013, Victor S. Miller wrote:
You should look at semi convergents. Look at http://shreevatsa.wordpress.com/2011/01/10/not-all-best-rational-approximati...
for a good account.
Victor
Sent from my iPhone
On Aug 14, 2013, at 10:23, Henry Baker <hbaker1@pipeline.com> wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
Suppose I used a continued fraction process to generate better & better approximations.
Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ?
I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral.
So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions?
Or perhaps this situation is exceedingly rare.
If so, how rare is it?
This is a very interesting question. We know that the rational points on the unit circle are all of the form (a/c, b/c) where (a,b,c) is a Pythagorean triple. And we know these are dense in the unit circle, so this approximation must be possible. (In fact these rational points form an interesting subgroup of the unit circle group SO(2).) --Dan On 2013-08-14, at 10:23 AM, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
Suppose I used a continued fraction process to generate better & better approximations.
Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ?
I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral.
So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions?
Or perhaps this situation is exceedingly rare.
If so, how rare is it?
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A (partial) answer is given here: http://www.cs.ubc.ca/~gyomalin/docs/Thesis_GuillaumeAlain_2006juillet29.pdf. In it he shows that an irrational real can be simultaneously approximated well by infinitely many algebraic number of degree n (for fixed n). On the other hand, on page 39 (section 5) he looks at the case of a fixed quadratic imaginary field, and shows that there can't be an infinite number of good approximations. Victor On Wed, Aug 14, 2013 at 11:04 AM, Dan Asimov <dasimov@earthlink.net> wrote:
This is a very interesting question. We know that the rational points on the unit circle are all of the form (a/c, b/c) where (a,b,c) is a Pythagorean triple. And we know these are dense in the unit circle, so this approximation must be possible. (In fact these rational points form an interesting subgroup of the unit circle group SO(2).)
--Dan
On 2013-08-14, at 10:23 AM, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
Suppose I used a continued fraction process to generate better & better approximations.
Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ?
I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral.
So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions?
Or perhaps this situation is exceedingly rare.
If so, how rare is it?
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hello, This message is properly read with FIXED fonts. I am working on arguments of complex numbers. For example, arg(1/2^(1/2+12*I) , if you use Maple or Mathematica or by hand you can expand this BUT, actually the expression is quite cumbersome and tricky. The proper answer is 2*Pi - 12*log(2). But that answer does not appear at fist hand, the <12> comes from an operation with { } and [ ]. the same with argument(2^I+3^I), it is a linear combination of log(2) and log(3). Again, the coefficients are coming from [ ] and { } functions. Strange, because if you try to simplify an expression like Arg of 1/2^(1/2+I)+1/3^(1/2+I) which is 1/2 1/2 2 sin(ln(2)) + 3 sin(ln(3)) arctan(---------------------------------) 1/2 1/2 2 cos(ln(2)) + 3 cos(ln(3)) Yes, but the arctan of sin of log should fall back to its feet but in this case it does not. This is tricky, in some very big expression with the Arctan [ Sin [ Log [ ]]] sometimes it does simplify to something more readable and some times not. in the first example the answer once cleaned is : sin(ln(2)) + sin(ln(3)) arctan(-----------------------) cos(ln(2)) + cos(ln(3)) That is the argument of 2^I+3^I which is a linear combination of log(2) and log(3) in disguise. So my question is : why is it that the argument does simplify and why ?? I looked at many trig identities and could not find the reason. Some of the cases where cracked using LLL algorithm and my big table of real numbers, so I cheated. If someone has a clue on that one ? NOTE: just try Arg[ ] with Mathematica, it won't work. In Maple, in some cases you will get an horrible answer and if you try the usual tricks of simplify(); expand(); evalc(); and such, the only thing it will do is inflate the thing to thousands of characters even with the simple one like 2^(1/2+I)+3^(1/2+I). Here is my question can someone find the closed expression of the ARGUMENT of that simple sum ??? Best regards, any answer would help. Simon Plouffe
P.S. As may have been clear, I was addressing only Henry's second sentence quoted below. But the existence of arbitrarily good approximations says nothing about how one might go about finding them. QUESTION: Given an irrational real number c, (say, given all the decimal digits of c), how would we explicitly find a sequence {p_n / q_n} of rationals approaching c in the limit as n -> oo, *such that* for each n there exists an integer r_n with (p_n)^2 + (q_n)^2 = (r_n)^2 ??? --Dan On 2013-08-14, at 11:04 AM, Dan Asimov wrote:
This is a very interesting question. We know that the rational points on the unit circle are all of the form (a/c, b/c) where (a,b,c) is a Pythagorean triple. And we know these are dense in the unit circle, so this approximation must be possible. (In fact these rational points form an interesting subgroup of the unit circle group SO(2).)
--Dan
On 2013-08-14, at 10:23 AM, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
P.P.S. This paper may shed some light on both the continued-fraction Pythagorean-triple question and the uncontinued-fraction Pythagorean-triple question: < http://www.fq.math.ca/Scanned/41-2/elsner.pdf >. --Dan On 2013-08-14, at 3:21 PM, Dan Asimov wrote:
P.S. As may have been clear, I was addressing only Henry's second sentence quoted below.
But the existence of arbitrarily good approximations says nothing about how one might go about finding them.
QUESTION: Given an irrational real number c, (say, given all the decimal digits of c), how would we explicitly find a sequence {p_n / q_n} of rationals approaching c in the limit as n -> oo, *such that* for each n there exists an integer r_n with
(p_n)^2 + (q_n)^2 = (r_n)^2
???
--Dan
On 2013-08-14, at 11:04 AM, Dan Asimov wrote:
This is a very interesting question. We know that the rational points on the unit circle are all of the form (a/c, b/c) where (a,b,c) is a Pythagorean triple. And we know these are dense in the unit circle, so this approximation must be possible. (In fact these rational points form an interesting subgroup of the unit circle group SO(2).)
--Dan
On 2013-08-14, at 10:23 AM, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
At first glance, this paper seems to be spot on! Carsten Eisner. ON RATIONAL APPROXIMATIONS BY PYTHAGOREAN NUMBERS. May, 2003. I'll go over it more carefully. Thanks for the link, Dan! At 03:36 PM 8/14/2013, Dan Asimov wrote:
P.P.S. This paper may shed some light on both the continued-fraction Pythagorean-triple question and the uncontinued-fraction Pythagorean-triple question:
< http://www.fq.math.ca/Scanned/41-2/elsner.pdf >.
--Dan
On 2013-08-14, at 3:21 PM, Dan Asimov wrote:
P.S. As may have been clear, I was addressing only Henry's second sentence quoted below.
But the existence of arbitrarily good approximations says nothing about how one might go about finding them.
QUESTION: Given an irrational real number c, (say, given all the decimal digits of c), how would we explicitly find a sequence {p_n / q_n} of rationals approaching c in the limit as n -> oo, *such that* for each n there exists an integer r_n with
(p_n)^2 + (q_n)^2 = (r_n)^2
???
--Dan
On 2013-08-14, at 11:04 AM, Dan Asimov wrote:
This is a very interesting question. We know that the rational points on the unit circle are all of the form (a/c, b/c) where (a,b,c) is a Pythagorean triple. And we know these are dense in the unit circle, so this approximation must be possible. (In fact these rational points form an interesting subgroup of the unit circle group SO(2).)
--Dan
On 2013-08-14, at 10:23 AM, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
On Wed, 14 Aug 2013, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
Suppose I used a continued fraction process to generate better & better approximations.
Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ?
I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral.
So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions?
Or perhaps this situation is exceedingly rare.
If so, how rare is it?
I don't know how rare it is, but it's certainly not impossible. For example, the continued fraction for .7501324981324 is [0 1 3 471 2 5 37 1 217 1 6 5 1 2 4 5 5] (or my program has a bug.) The first 3 terms are [0 1 3]=3/4 (and the next term is huge, so this is a pretty good approximation.) Of course sqrt(3^2+4^2)=5. I arrived at .7501324981324 by typing .750 and then banging on my keyboard. I think that every Pythagorean fraction has an interval containing it where the fraction itself is a convergent of the continued fractions of all the numbers in the interval. Without thinking about how to prove it, I bet that the density of those intervals is non-zero and that Pythagorean approximants aren't infinitely rare. But I've been wrong before... -- Tom Duff. I was typing and suddenly my laptop was in Iran.
participants (6)
-
Dan Asimov -
Henry Baker -
Simon Plouffe -
Tom Duff -
Victor Miller -
Victor S. Miller