Re: [math-fun] Curious function
Hey thanks Fred! Your mind-reader machine managed to figure out how the word "asymptotic" applies to Dan's riddle. All I got was a recurrence relation, which of course is not closed-form. But I was stumped by "asymptotic". But since these are Lucas numbers, then course, there must be an *exact*, closed form formula too (just as with the Fibonacci numbers) viz.: ((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n This formula was contributed to A005248 by Creighton Dement back in Apr 19 2005. Remember "Floretions"? So we have a closed-form, exact not asymptotic, answer. On 1/31/13, Fred W. Helenius <fredh@ix.netcom.com> wrote:
[...]
2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, ...
This is the subsequence of the Lucas sequence consisting of the terms with even index. Since L(n) = phi^n + (-1/phi)^n, f(n) = L(2n) is asymptotic to phi^{2n}, where phi is the golden ratio.
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
Robert, and Fred H's machine, both successfully read my mind. The apparently unrelated problem involved counting the number of periodic points of period k for the self-homeomorphism H of the torus T^2 = R^2/Z^2 to itself H: T^2 -> T^2 defined by the matrix ((2 1),(1 1)) in SL(2,Z) (taken mod Z^2). Using the nifty Lefschetz fixed point theorem (< http://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem >), it's not hard to calculate that the number of periodic points of H having (not necessarily least) period = k is 2 - ((3-sqrt(5))^k + (3+sqrt(5))/2)^k for k >= 1. Calculating a few values of the second, subtracted term led to 2, 3, 7, 18, 47, 123 and the functional equation I asked about. So the asymptotic formula I had in mind was f(n) ~ ((3 + sqrt(5))/2)^n. Which is just phi^(2n), as Fred said. (Damn good machine you have!) But unfortunately I just e-blurted the first thing on my mind without carefully formulating the "Puzzle" so it had a unique answer. Sorry about that. --Dan On 2013-01-30, at 10:40 PM, Robert Munafo wrote:
Hey thanks Fred! Your mind-reader machine managed to figure out how the word "asymptotic" applies to Dan's riddle. All I got was a recurrence relation, which of course is not closed-form. But I was stumped by "asymptotic".
But since these are Lucas numbers, then course, there must be an *exact*, closed form formula too (just as with the Fibonacci numbers) viz.:
((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n
This formula was contributed to A005248 by Creighton Dement back in Apr 19 2005. Remember "Floretions"?
So we have a closed-form, exact not asymptotic, answer.
On 1/31/13, Fred W. Helenius <fredh@ix.netcom.com> wrote:
[...]
2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, ...
This is the subsequence of the Lucas sequence consisting of the terms with even index. Since L(n) = phi^n + (-1/phi)^n, f(n) = L(2n) is asymptotic to phi^{2n}, where phi is the golden ratio.
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
-
Dan Asimov -
Robert Munafo