Fourier epicycles thru nth: (d16) sum(((product((1 - (( - 1)^j * tan(((%pi * (2 * k + (1/3)))/(2^j)))/(sqrt(3)))),j,1,inf)) * (%e)^(%i * (6 * k + 1) * t)/((2 * k + (1/3))^2)),k, - n,n) See http://www.tweedledum.com/rwg/cog.htm Also http://gosper.org/fst.dvi Pix: http://gosper.org/fst1.png ,http://gosper.org/fst2.png , http://gosper.org/fst3.png ,http://gosper.org/fst4.png . (For parametric, take Re and Im.) Rich's base 4 recursion loops for rational arg. Find the fixed point of the pending stack ops to simulate infinite recursion, and get an exact val on the curve. Equate to this Fourier series for a *non-obvious* identity. --rwg
If you view the starting triangle as running from 0 to 3, and write the fraction base 4, you can use the fraction digits to select which part of the (sub)edge you are on. The orientation of the edge can be calculated from the 1s and 2s of the high order portion of the fraction, and that should give a somewhat kludgy sum for the (x,y) coordinates of the image point. RWG likely has some nice FFT clock sum expression.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of N. J. A. Sloane Sent: Thursday, February 25, 2010 12:43 PM To: math-fun@mailman.xmission.com Cc: njas@research.att.com Subject: Re: [math-fun] parametrizing the Koch snowflake curve
Jim Propp asked:
Does anyone (Bill Gosper, maybe?) know an explicit parametrization of the Koch snowflake curve?
The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that does not contain a 1. Is there a similar characterization of the Koch snowflake?
Neil Sloane