maple says b0 = a0^2-2*a1*a2)/(a0^3-6*a0*a1*a2+2*a1^3+4*a2^3, b1 = -(a0*a1-2*a2^2)/(a0^3-6*a0*a1*a2+2*a1^3+4*a2^3, b2 = -(a0*a2-a1^2)/(a0^3-6*a0*a1*a2+2*a1^3+4*a2^3. On Thu, Apr 5, 2018 at 11:09 AM, Henry Baker <hbaker1@pipeline.com> wrote:
BTW, "maxima" is free...
2 2 a1 a2 - a0 (%o2) [[b0 = - --------------------------------, 3 3 3 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 2 2 a2 - a0 a1 b1 = --------------------------------, 3 3 3 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 2 a0 a2 - a1 b2 = - --------------------------------]] 3 3 3 4 a2 - 6 a0 a1 a2 + 2 a1 + a0
At 08:32 AM 4/5/2018, Allan Wechsler wrote:
Suppose w is the real cube root of 2, w^3 = 2.
Then for a0, a1, a2 rational, numbers of the form a0 + a1*w + a1*w^2 form a field. What I want is the formula for the multiplicative inverse in this field. It's elementary algebra, but it is voluminous and I keep making errors when I try to do it by hand. Any computer algebra system could probably instantly solve for b0, b1, b2 subject to (a0 + a1w + a2ww)(b0 + b1w + b2ww) = 1. Can somebody do me a favor and give me formulas for the b's in terms of the a's?
Thank you!
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