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2019-11-13, 22:48   #11
Dr Sardonicus

Feb 2017
Nowhere

22×5×179 Posts

Quote:
 Originally Posted by alpertron The answer shown by the application is wrong. I unchecked Pretty Print, copied the output to Notepad++, deleted all (xxx digits) and spaces, replaced i by I and finally copied that to gp. All five roots shown are non-real, so something is not good. Another error is that the digits in group is not working. The program always shows groups of 6 digits. I will work on that.
I think that $2\cos(\frac{2\pi}{11})$ and its conjugates are about as good as it gets for expressing the roots of this polynomial. There is no expression in terms of "real radicals."

It is somewhat similar to the minimum polynomial x^3 - 3*x - 1 for $2\cos(\frac{\pi}{3})$, which the calculator expresses in terms of cosines. Obviously, it "knows" the trigonometric solution for the casus irreducibilis of the cubic.