Ok, so if we are given c+is, c,s real, cc+ss=1, then we want to compute cos(acos(c)/3) for the real part of one of the cube roots of c+is. cos(acos(c)/3) = T_(1/3)(c) = d, so cos(3acos(d)) = T_3(d) = c, i.e., 4d^3-3d=c So we have reduced the problem to one of finding a real root of a real cubic 4d^3-3d-c=0. Once we have d, we can easily compute the imaginary part, since we know that the absolute value is 1. Question: Since we know that the real parts of all three of the complex cube roots of c+is are real (duh!), this would seem to imply that the equation 4d^3-3d-c=0 always has 3 real roots. Is this easy to see? Also, all three roots should lie within the range [-1,1]. Is this easy to see? At 05:27 PM 11/12/2009, Joerg Arndt wrote:
* Joerg Arndt <arndt@jjj.de> [Nov 13. 2009 12:21]:
* Henry Baker <hbaker1@pipeline.com> [Nov 13. 2009 08:29]:
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The real part of cbrt(a'+b'i) is cos(acos(a')/3), which looks like Chebyshev 1/3. Are there Chebyshev functions with non-integral n?
yes, take T_n(x) = cos( n*acos(x) ) or (hypergeometric 2F1) T_n(x) = F( -n, +n; 1/2; x )
correction: T_n(x) = F( -n, +n; 1/2; (1-x)/2 )
Note that T_n( T_{1/n}( x ) ) == x (because T_n( T_m( x ) ) == T_{n*m}(x) )
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