I haven't thought this through, but let me take a stab at this. We have a given target real number alpha, which we want to approximate with rational numbers. We have some sense of which rational numbers are "worse"; one possible such figure of (de)merit is just the denominator in lowest terms. Each alpha produces a monotonically decreasing function, the minimum error obtainable by an approximation no worse than x. We spend some time analyzing what kind of asymptotics to expect from such a function, and that allows us to classify different alphas by the asymptotics of these approximation functions. Surely, we could repeat this sort of reasoning with algebraic numbers. One possible figure of demerit for an algebraic number is the sum of the absolute values of the coefficients of the relevant polynomial. For each limit on this sum, there is a class of polynomials, and I am guessing that that class offers a well-defined best approximation to a given alpha. Again we would analyze the asymptotic behavior of such approximations as the sum of coefficients is allowed to increase; this seems to offer promise of some way to measure how badly a real is approximated by algebraics. On Sun, Jun 13, 2010 at 1:33 PM, Victor Miller <victorsmiller@gmail.com>wrote:
Whoops, I meant that mu(alpha) is the inf over all such numbers, not the sup
On Sun, Jun 13, 2010 at 1:20 PM, Victor Miller <victorsmiller@gmail.com
wrote:
If alpha is real, then the irrationality measure of alpha (written mu(alpha)) is the sup over mu > 0 such that
0 < | alpha - p/q | < 1/q^mu has an infinite number of solutions for relatively prime integers p and q. If alpha is rational then mu(alpha) = 1, otherwise mu(alpha) >= 2. Roth's famous theorem showed that mu(alpha) = 2 for alpha an algebraic number (which isn't rational). Liouville's result exhibited a number for which mu(alpha) = infinity. However, there has never been a case (that I know of) of a transcendental number occurring "in nature" (i.e. the value or root of some reasonable function, such as a hyper geometric function) which has been shown to have irrationality measure >
There have been a lot of results for numbers such as values of logarithms (which includes pi) or of the zeta function where explicit upper bounds for mu(alpha) have been exhibited, but none have been shown to be > 2. There is the whole study of the Markov Spectrum which involves the constant on top of 1/q^2. In that phi is the worst approximabile.
Victor
On Sun, Jun 13, 2010 at 11:46 AM, Mike Stay <metaweta@gmail.com> wrote:
Well, since all trancendentals are irrational, it can't be by the same metric: you can approximate any of them with rationals better than phi.
On Sun, Jun 13, 2010 at 12:27 AM, Kerry Mitchell <lkmitch@gmail.com> wrote:
I've read that phi (~ 1.618, (1+sqrt(5))/2) is the "most irrational" number because of how poorly it is approximated by rational numbers. I assume that this is because it's continued fraction expansion is all 1's. Is there a sense in which some number is the "most transcendental" number? If so, what would that number and that meaning be?
Kerry Mitchell -- lkmitch@gmail.com www.kerrymitchellart.com http://spacefilling.blogspot.com/ _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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