Thread: Polynomial
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Old 2005-09-16, 12:43   #10
R.D. Silverman
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Nov 2003

746010 Posts

Originally Posted by R.D. Silverman
Yes, I was looking for formula %9. I was hoping that it might be
symmetric. It is not, but I thought that if it isn't, it might have a
representation as a sextic not in (z+1/z), but rather [with Z = 2^53]
in (z + 2/z). It seems that it does, but the constant is much too large.
Perhaps we might try a sextic in (z + k/z) for some other value of k?
I am not familiar with the syntax of Maple.

The above calculation should solve for (c1, c2, c3, .....) where

(z + 2/z)^6 + c1 (z + 2/z)^5 + c2 (z + 2/z)^4 + c3 (z + 2/z)^3 + .... =

z^-6 *(z^12 + 2z^11 - 22z^10 - 44z^9 ........ etc [expression %9])
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