Since the solutions usually seem to have a lot of repetitions, how about introducing some intermediate symbols? The cubic solution is often expressed as a program, i.e. "Let Y be a solution of the quadratic Y^2+... = 0, then X = -B/3A + cbrt(...Y...) + ... ". This is also a good way to get consistency in choosing the Kth roots: The regular cubic formula involves two cube roots, with nominally 9 possible answers for X. Typically, the verbal condition is added that the product of the cube roots must be xxx, which eliminates 6 of the possible combinations. But you can also say Z = cbrt(...), and write the final answer as X = ... + Z/3A + (...)/3AZ Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] on behalf of Bill Gosper [billgosper@gmail.com] Sent: Friday, October 28, 2011 2:10 AM To: math-fun@mailman.xmission.com Subject: Re: [math-fun] the sextic for Cantrell's 5 circle packing Not so the sextic -4536767107584 + 2746424942208 x - 464573095920 x^2 + 14000936832 x^3 + 744041160 x^4 + 710712 x^5 + 169 x^6 for the 12 circle optimal packing http://www2.stetson.edu/~efriedma/cirRcir/12.gif, "6 dimes & 3 quarters around 3 nickels", which is what I Moebius transformed to make Twubblesome Twelve, naively complaining to this list at the time that the equation(s) seemed unduly complicated. So now, with the miracle of sextic solving, we have the solution in radicals, except I can't simplify it enough to fit on one page. --rwg On Thu, Oct 27, 2011 at 3:46 AM, Bill Gosper <billgosper@gmail.com> wrote:
maximizing sum(radii), http://www2.stetson.edu/~efriedma/cirRcir/5.gif, namely Root[-87131 + 32779 #1 + 18567 #1^2 + 10931 #1^3 - 14350 #1^4 + 2538 #1^5 + 162 #1^6 &, 2], is unsolvable in radicals, according to my (Ssexy) solver. --rwg
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